W12-W15. Vector Calculus

Author

Mohammad Alkousa

Published

April 22, 2026

1. Theory

1.1 Vector Fields

In single-variable calculus, a function usually assigns a number to each input. In vector calculus, we often need a richer object: at each point of space, we want to attach a vector. This is how we describe gravitational force, electric force, velocity of a fluid, wind direction and speed, or the direction in which a quantity increases.

A vector field is a function whose output is a vector. In three-dimensional space, a typical vector field has the form

\[\mathbf{F}(x,y,z)=M(x,y,z)\hat{i}+N(x,y,z)\hat{j}+P(x,y,z)\hat{k},\]

where \(M\), \(N\), and \(P\) are scalar functions. The vector assigned to \((x,y,z)\) has components \(M(x,y,z)\), \(N(x,y,z)\), and \(P(x,y,z)\).

The most important habit is to interpret both the direction and the magnitude of each vector. For example, in a velocity field, the vector direction tells the direction of motion of a small particle of fluid, while the vector length tells its speed. In a gravitational field around a mass, vectors point toward the center of mass, and the vector length represents the strength of attraction.

Vector fields may be defined on all of \(\mathbb{R}^3\), on a smaller solid region, on a surface, or even only along a curve. For example, if \(\mathbf{r}(t)\) is a space curve, then its tangent vectors \(\mathbf{T}\) and normal vectors \(\mathbf{N}\) form vector fields along that curve. If \(f(x,y,z)\) is a scalar field, then attaching \(\nabla f\) to every point creates a vector field called the gradient field.

As with ordinary functions, a vector field has a domain. If no domain is explicitly stated, the natural domain is the largest set of points where all component functions are defined. This matters because operations such as curl and divergence are only valid at points where the required partial derivatives exist.

1.2 Gradient Fields and Conservative Fields

1.2.1 Gradient Fields

If \(f(x,y,z)\) is a differentiable scalar function, its gradient is the vector field

\[\nabla f=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}.\]

For a function of \(n\) variables, the same idea becomes

\[\nabla f=\left(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\dots,\frac{\partial f}{\partial x_n}\right).\]

The gradient is not just a list of partial derivatives. It has a geometric meaning: at each point, \(\nabla f\) points in the direction where \(f\) increases fastest, and its magnitude gives the maximum rate of increase. This connects vector calculus with the earlier study of directional derivatives.

For instance, if \(f\) represents temperature, then \(\nabla f\) points in the direction of fastest temperature increase. If a particle moves perpendicular to \(\nabla f\), then, to first order, the temperature does not change. That is why gradients are normal to level curves and level surfaces.

1.2.2 Conservative Vector Fields

A vector field \(\mathbf{F}\) is called a conservative vector field if it is the gradient of some scalar function:

\[\mathbf{F}=\nabla f.\]

The scalar function \(f\) is called a potential function for \(\mathbf{F}\). In coordinates, if

\[\mathbf{F}=M\hat{i}+N\hat{j}+P\hat{k},\]

then \(\mathbf{F}\) is conservative when there is a function \(f\) such that

\[f_x=M,\qquad f_y=N,\qquad f_z=P.\]

The word “potential” is important in applications. In physics, conservative force fields are forces whose work can be recovered from a potential energy function. The practical consequence, studied later with line integrals, is that the work done by a conservative field depends only on the starting and ending points, not on the path.

To find a potential function, integrate one component and then use the remaining components to determine the missing functions. For example, if \(M=f_x\), first integrate \(M\) with respect to \(x\):

\[f(x,y,z)=\int M(x,y,z)\,dx+\phi(y,z).\]

The term \(\phi(y,z)\) appears because integration with respect to \(x\) treats \(y\) and \(z\) as constants. Then differentiate this expression with respect to \(y\) and \(z\), compare with \(N\) and \(P\), and solve for the remaining unknown parts.

1.3 Curl

1.3.1 Meaning and Formula

The curl measures the local rotational tendency of a vector field. If a tiny paddle wheel were placed in a fluid flow, curl describes the axis and strength of its rotation. A field with zero curl has no local spinning behavior, although this alone does not always guarantee that the field is conservative.

For a vector field

\[\mathbf{F}=M\hat{i}+N\hat{j}+P\hat{k},\]

the curl is

\[\operatorname{curl}(\mathbf{F})=\nabla\times\mathbf{F} =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ M & N & P \end{vmatrix}.\]

Expanding the determinant gives

\[\operatorname{curl}(\mathbf{F}) =\left(P_y-N_z\right)\hat{i}-\left(P_x-M_z\right)\hat{j}+\left(N_x-M_y\right)\hat{k}.\]

Equivalently,

\[\operatorname{curl}(\mathbf{F}) =\left(P_y-N_z\right)\hat{i}+\left(M_z-P_x\right)\hat{j}+\left(N_x-M_y\right)\hat{k}.\]

The two displayed versions are the same; the second one simply distributes the minus sign in the \(\hat{j}\) component.

1.3.2 Curl of a Gradient

If \(f\) has continuous second-order partial derivatives, then

\[\operatorname{curl}(\nabla f)=\mathbf{0}.\]

This follows from equality of mixed partial derivatives:

\[f_{yz}=f_{zy},\qquad f_{xz}=f_{zx},\qquad f_{xy}=f_{yx}.\]

Therefore, every continuously differentiable conservative vector field must have zero curl. In practical terms:

\[\mathbf{F}\text{ conservative}\quad\Longrightarrow\quad \operatorname{curl}(\mathbf{F})=\mathbf{0}.\]

This gives a fast way to disprove conservativeness. If the curl is not zero, the field cannot be conservative.

In component form, \(\nabla\times\mathbf{F}=\mathbf{0}\) means three separate compatibility equations:

\[P_y=N_z,\qquad M_z=P_x,\qquad N_x=M_y.\]

These equations say that the mixed partial derivatives that would come from a potential function are consistent. They are the practical test used in many exercises before trying to find \(f\).

1.3.3 When Zero Curl Is Enough

The converse statement needs a condition on the domain. If \(\mathbf{F}\) is defined on an open simply connected domain, and its component functions have continuous first partial derivatives, then

\[\mathbf{F}\text{ is conservative}\quad\Longleftrightarrow\quad \operatorname{curl}(\mathbf{F})=\mathbf{0}.\]

A simply connected domain is, informally, a domain with no holes. The condition matters. A vector field can have zero curl everywhere in its natural domain and still fail to be conservative if the domain has a hole. The standard example is a rotational field around the origin in the punctured plane; its curl is zero away from the origin, but its circulation around the origin is nonzero.

The typical workflow is:

Check the domain. If it is all of \(\mathbb{R}^3\), or another open simply connected region, the curl test is reliable.
Compute \(\nabla\times\mathbf{F}\).
If the curl is nonzero, the field is not conservative.
If the curl is zero and the domain condition is satisfied, find a potential function \(f\) by integration.

1.4 Divergence

1.4.1 Meaning and Formula

The divergence measures the local source or sink behavior of a vector field. If a vector field describes fluid velocity, positive divergence means fluid is locally spreading outward from a point, while negative divergence means fluid is locally flowing inward toward a point.

For

\[\mathbf{F}=M\hat{i}+N\hat{j}+P\hat{k},\]

the divergence is the scalar field

\[\operatorname{div}(\mathbf{F})=\nabla\cdot\mathbf{F}=M_x+N_y+P_z.\]

Unlike curl, which produces a vector field in three dimensions, divergence produces a scalar function. It adds the rate of change of each component in its own coordinate direction.

For example, if

\[\mathbf{F}(x,y,z)=y^2z^3\hat{i}+2xyz^3\hat{j}+3xy^2z^2\hat{k},\]

then

\[\operatorname{div}(\mathbf{F})=0+2xz^3+6xy^2z.\]

The first term is zero because \(y^2z^3\) does not depend on \(x\).

1.4.2 Divergence of Curl

If \(\mathbf{F}\) has continuous second-order partial derivatives, then

\[\operatorname{div}(\operatorname{curl}(\mathbf{F}))=0.\]

This identity is another consequence of equality of mixed partial derivatives. It is useful as a consistency check: if some vector field \(\mathbf{G}\) is actually the curl of another field, then \(\operatorname{div}(\mathbf{G})\) must be zero.

1.5 The Laplace Operator

When divergence is applied to a gradient field, we obtain a very important second-order operator:

\[\operatorname{div}(\nabla f)=\nabla\cdot\nabla f=f_{xx}+f_{yy}+f_{zz}.\]

This expression is called the Laplacian of \(f\) and is denoted by

\[\nabla^2 f=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2}.\]

The equation

\[\nabla^2 f=0\]

is called Laplace’s equation. Solutions of Laplace’s equation are called harmonic functions. They appear in heat conduction, electrostatics, gravitational potential theory, and fluid flow.

The conceptual chain is important:

\(\nabla f\) turns a scalar field into a vector field showing fastest increase.
\(\operatorname{curl}(\mathbf{F})\) measures rotation of a vector field.
\(\operatorname{div}(\mathbf{F})\) measures source or sink strength of a vector field.
\(\nabla^2 f=\operatorname{div}(\nabla f)\) measures how a scalar field bends or spreads locally.

1.6 Standard Identities and Common Pitfalls

Vector calculus notation is compact, but it is easy to misuse. The operator \(\nabla\) behaves like a vector of partial derivatives, but products must still be interpreted carefully.

For scalar fields \(f\) and vector fields \(\mathbf{F},\mathbf{G}\) with sufficiently smooth components, the standard linearity rules are

\[\operatorname{div}(\mathbf{F}+\mathbf{G})=\operatorname{div}(\mathbf{F})+\operatorname{div}(\mathbf{G}),\]

and

\[\operatorname{curl}(\mathbf{F}+\mathbf{G})=\operatorname{curl}(\mathbf{F})+\operatorname{curl}(\mathbf{G}).\]

The product rules are

\[\operatorname{div}(f\mathbf{F})=f\operatorname{div}(\mathbf{F})+\mathbf{F}\cdot\nabla f,\]

and

\[\operatorname{curl}(f\mathbf{F})=\nabla f\times\mathbf{F}+f\operatorname{curl}(\mathbf{F}).\]

Two common errors should be avoided. First, do not write \(\nabla f\times\operatorname{curl}(\mathbf{F})\) in the product rule for curl; the cross product is with \(\mathbf{F}\) itself. Second, do not assume that zero curl always means conservative without checking the domain. Holes in the domain can break path independence.

When computing curl and divergence, always check which variable is being differentiated. For example, in \(M_x\), only \(x\) varies; all other variables are treated as constants. Most algebraic mistakes in this topic come from differentiating the wrong component with respect to the wrong variable.

2. Definitions

Vector field: A function that assigns a vector to each point in its domain.
Scalar field: A function that assigns a real number to each point in its domain.
Component functions: The scalar functions \(M\), \(N\), and \(P\) in \(\mathbf{F}=M\hat{i}+N\hat{j}+P\hat{k}\).
Natural domain: The largest set of points where all component functions of a field are defined.
Gradient field: A vector field of the form \(\nabla f\) for some differentiable scalar function \(f\).
Gradient: The vector of first partial derivatives of a scalar function, pointing in the direction of greatest increase.
Conservative vector field: A vector field that can be written as the gradient of a scalar potential function.
Potential function: A scalar function \(f\) such that \(\mathbf{F}=\nabla f\).
Curl: The vector field \(\nabla\times\mathbf{F}\) that measures local rotational tendency of \(\mathbf{F}\).
Rotation: Another name for the curl of a vector field.
Divergence: The scalar field \(\nabla\cdot\mathbf{F}\) that measures local source or sink behavior of \(\mathbf{F}\).
Simply connected domain: A domain with no holes; every closed curve inside it can be continuously shrunk to a point without leaving the domain.
Open domain: A domain in which every point has a small surrounding ball contained completely in the domain.
Laplacian: The operator \(\nabla^2 f=f_{xx}+f_{yy}+f_{zz}\), equal to \(\operatorname{div}(\nabla f)\).
Laplace’s equation: The equation \(\nabla^2 f=0\).
Harmonic function: A function satisfying Laplace’s equation on its domain.

3. Formulas

Vector Field in Space: \(\mathbf{F}(x,y,z)=M(x,y,z)\hat{i}+N(x,y,z)\hat{j}+P(x,y,z)\hat{k}\).
Gradient: \(\nabla f=f_x\hat{i}+f_y\hat{j}+f_z\hat{k}\).
Conservative Field Condition: \(\mathbf{F}=\nabla f\iff M=f_x,\ N=f_y,\ P=f_z\).
Curl Determinant: \(\nabla\times\mathbf{F}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ M&N&P\end{vmatrix}\).
Curl Expanded: \(\nabla\times\mathbf{F}=(P_y-N_z)\hat{i}+(M_z-P_x)\hat{j}+(N_x-M_y)\hat{k}\).
Zero Curl Component Test: \(\nabla\times\mathbf{F}=\mathbf{0}\iff P_y=N_z,\ M_z=P_x,\ N_x=M_y\).
Divergence: \(\nabla\cdot\mathbf{F}=M_x+N_y+P_z\).
Curl of a Gradient: \(\nabla\times(\nabla f)=\mathbf{0}\), assuming continuous second-order partial derivatives.
Divergence of a Curl: \(\nabla\cdot(\nabla\times\mathbf{F})=0\), assuming continuous second-order partial derivatives.
Laplacian: \(\nabla^2 f=\nabla\cdot\nabla f=f_{xx}+f_{yy}+f_{zz}\).
Curl Test on Simply Connected Domains: \(\mathbf{F}\text{ is conservative}\iff\nabla\times\mathbf{F}=\mathbf{0}\), when \(\mathbf{F}\) is defined on an open simply connected domain and has continuous first partial derivatives.
Divergence Linearity: \(\operatorname{div}(\mathbf{F}+\mathbf{G})=\operatorname{div}(\mathbf{F})+\operatorname{div}(\mathbf{G})\).
Curl Linearity: \(\operatorname{curl}(\mathbf{F}+\mathbf{G})=\operatorname{curl}(\mathbf{F})+\operatorname{curl}(\mathbf{G})\).
Divergence Product Rule: \(\operatorname{div}(f\mathbf{F})=f\operatorname{div}(\mathbf{F})+\mathbf{F}\cdot\nabla f\).
Curl Product Rule: \(\operatorname{curl}(f\mathbf{F})=\nabla f\times\mathbf{F}+f\operatorname{curl}(\mathbf{F})\).

4. Practice

4.1. Calculate Triple Integrals (Lab 12, Task 1)

Calculate:

a) \(\displaystyle \int_{0}^{7} \int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} \frac{y}{1 + z} \,dx \,dy \,dz\).

b) \(\displaystyle \int_{0}^{4} \int_{0}^{2} \int_{2y}^{4} \frac{4 \cos(x^2)}{2\sqrt{z}} \,dx \,dy \,dz\).

Click to see the solution

Key Concept: In a triple integral, integrate from the inside out unless changing the order makes a non-elementary integral disappear.

a) The integrand does not depend on \(x\), so the inner integral only multiplies by the length of the \(x\)-interval:

\[\int_0^{\sqrt{4-y^2}}\frac{y}{1+z}\,dx=\frac{y\sqrt{4-y^2}}{1+z}.\]

Thus

\[I_a=\int_0^7\int_0^2 \frac{y\sqrt{4-y^2}}{1+z}\,dy\,dz.\]

For the \(y\)-integral, set \(u=4-y^2\), so \(du=-2y\,dy\):

\[\int_0^2 y\sqrt{4-y^2}\,dy =-\frac12\int_4^0 u^{1/2}\,du =\frac12\int_0^4 u^{1/2}\,du =\frac12\cdot \frac{2}{3}u^{3/2}\Big|_0^4 =\frac{8}{3}.\]

Therefore

\[I_a=\frac{8}{3}\int_0^7\frac{dz}{1+z} =\frac{8}{3}\ln(1+z)\Big|_0^7 =\frac{8}{3}\ln 8.\]

b) First simplify the integrand:

\[\frac{4\cos(x^2)}{2\sqrt z}=\frac{2\cos(x^2)}{\sqrt z}.\]

The region in the \(xy\)-plane is

\[0\le y\le 2,\qquad 2y\le x\le 4.\]

Equivalently,

\[0\le x\le 4,\qquad 0\le y\le \frac{x}{2}.\]

This change is useful because \(\int \cos(x^2)\,dx\) has no elementary antiderivative, while integrating in \(y\) first is easy. Then

\[I_b=\int_0^4\int_0^4\int_0^{x/2}\frac{2\cos(x^2)}{\sqrt z}\,dy\,dx\,dz.\]

Integrate with respect to \(y\):

\[\int_0^{x/2}\frac{2\cos(x^2)}{\sqrt z}\,dy =\frac{2\cos(x^2)}{\sqrt z}\cdot \frac{x}{2} =\frac{x\cos(x^2)}{\sqrt z}.\]

Thus

\[I_b=\int_0^4\frac{dz}{\sqrt z}\int_0^4 x\cos(x^2)\,dx.\]

Now

\[\int_0^4 z^{-1/2}\,dz=2\sqrt z\Big|_0^4=4,\]

and, with \(u=x^2\), \(du=2x\,dx\),

\[\int_0^4 x\cos(x^2)\,dx =\frac12\int_0^{16}\cos u\,du =\frac12\sin 16.\]

Therefore

\[I_b=4\cdot \frac12\sin 16=2\sin 16.\]

Answer: a) \(\displaystyle \frac{8}{3}\ln 8\); b) \(\displaystyle 2\sin 16\).

4.2. Volume Bounded by Five Planes (Lab 12, Task 2)

Find the volume of the solid region bounded by the planes \(z = x\), \(x + z = 8\), \(z = y\), \(y = 8\), and \(z = 0\).

Click to see the solution

Key Concept: Translate the bounding planes into inequalities and integrate \(1\) over the resulting solid.

The plane \(z=0\) gives the bottom. For a fixed height \(z\), the plane \(z=x\) gives \(x=z\), and the plane \(x+z=8\) gives \(x=8-z\). Hence

\[z\le x\le 8-z.\]

This interval is nonempty when \(z\le 8-z\), so \(0\le z\le 4\).

Similarly, the plane \(z=y\) gives \(y=z\), and the plane \(y=8\) gives

\[z\le y\le 8.\]

Therefore the volume is

\[V=\int_0^4\int_z^8\int_z^{8-z}1\,dx\,dy\,dz.\]

Compute the inner lengths:

\[\int_z^{8-z}1\,dx=8-2z,\]

\[V=\int_0^4(8-z)(8-2z)\,dz =\int_0^4(64-24z+2z^2)\,dz.\]

Thus

\[V=\left[64z-12z^2+\frac{2}{3}z^3\right]_0^4 =256-192+\frac{128}{3} =\frac{320}{3}.\]

Answer: \(\displaystyle V=\frac{320}{3}\).

4.3. Integral Over a Tetrahedron (Lab 12, Task 3)

Set up and evaluate \(\displaystyle \iiint_G (x + y + z) \,dV\), where \(G\) is the tetrahedron in the first octant bounded by the plane \(x + y + z = 1\).

Click to see the solution

Key Concept: The first-octant tetrahedron is described by \(x,y,z\ge0\) and \(x+y+z\le1\).

Use the order \(dz\,dy\,dx\):

\[0\le x\le1,\qquad 0\le y\le1-x,\qquad 0\le z\le1-x-y.\]

Thus

\[I=\int_0^1\int_0^{1-x}\int_0^{1-x-y}(x+y+z)\,dz\,dy\,dx.\]

A quick symmetric computation is possible. The tetrahedron has volume

\[\operatorname{Vol}(G)=\frac{1}{6}.\]

Its centroid is

\[\left(\frac14,\frac14,\frac14\right),\]

so the average value of \(x+y+z\) over the tetrahedron is

\[\frac14+\frac14+\frac14=\frac34.\]

Therefore

\[I=\operatorname{Vol}(G)\cdot \frac34=\frac16\cdot\frac34=\frac18.\]

Answer: \(\displaystyle \iiint_G(x+y+z)\,dV=\frac18\).

4.4. Integral Over a Cylindrical Wedge (Lab 12, Task 4)

Let \(G\) be the wedge in the first octant that is cut from the cylindrical solid \(y^2 + z^2 \leq 1\) by the planes \(y = x\) and \(x = 0\). Evaluate \(\displaystyle \iiint_G z \,dV\).

Click to see the solution

Key Concept: The cylinder is in the \(yz\)-variables, while the planes bound \(x\).

Because the solid lies in the first octant,

\[y\ge0,\qquad z\ge0,\qquad y^2+z^2\le1.\]

The planes \(x=0\) and \(y=x\) give

\[0\le x\le y.\]

Thus

\[I=\iint_{D}\int_0^y z\,dx\,dA_{yz},\]

where \(D\) is the quarter disk \(y^2+z^2\le1\) in the first quadrant of the \(yz\)-plane.

The inner integral is

\[\int_0^y z\,dx=zy.\]

Use polar coordinates in the \(yz\)-plane:

\[y=r\cos\theta,\qquad z=r\sin\theta,\qquad 0\le r\le1,\quad 0\le\theta\le\frac{\pi}{2}.\]

Then \(dA_{yz}=r\,dr\,d\theta\), and

\[I=\int_0^{\pi/2}\int_0^1(r\cos\theta)(r\sin\theta)r\,dr\,d\theta.\]

\[I=\left(\int_0^1r^3\,dr\right)\left(\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta\right) =\frac14\cdot\frac12=\frac18.\]

Answer: \(\displaystyle \frac18\).

4.5. Integral Between a Plane and a Paraboloid (Lab 12, Task 5)

Let \(G\) be the solid bounded above by the plane \(z = 4\) and below by the paraboloid \(z = x^2 + y^2\), with projection \(R\) being the disk \(x^2 + y^2 \leq 4\) in the xy-plane. Evaluate \(\displaystyle \iiint_G (x^2 + y^2) \,dV\).

Click to see the solution

Key Concept: Because the projection is a disk and the integrand is \(x^2+y^2\), cylindrical coordinates are natural.

Use

\[x^2+y^2=r^2,\qquad dV=r\,dz\,dr\,d\theta.\]

The bounds are

\[0\le r\le2,\qquad 0\le\theta\le2\pi,\qquad r^2\le z\le4.\]

Therefore

\[I=\int_0^{2\pi}\int_0^2\int_{r^2}^4 r^2\cdot r\,dz\,dr\,d\theta.\]

Integrate with respect to \(z\):

\[I=\int_0^{2\pi}\int_0^2 r^3(4-r^2)\,dr\,d\theta.\]

Now

\[\int_0^2 r^3(4-r^2)\,dr =\int_0^2(4r^3-r^5)\,dr =4\cdot\frac{2^4}{4}-\frac{2^6}{6} =16-\frac{64}{6} =\frac{16}{3}.\]

Thus

\[I=2\pi\cdot\frac{16}{3}=\frac{32\pi}{3}.\]

Answer: \(\displaystyle \frac{32\pi}{3}\).

4.6. Double Integral by the Transformation \(x=u/v\), \(y=uv\) (Lab 12, Task 6)

Use the transformation \(x = u/v\), \(y = uv\) to evaluate \(\displaystyle \iint_R xy \,dA\) where \(R\) is bounded by \(xy = 1\), \(xy = 2\), \(y = x\), \(y = 2x\) in the first quadrant.

Click to see the solution

Key Concept: Choose the transformation because it turns the four boundary curves into constant-\(u\) and constant-\(v\) lines.

From

\[x=\frac{u}{v},\qquad y=uv,\]

we get

\[xy=u^2,\qquad \frac{y}{x}=v^2.\]

Since the region is in the first quadrant, \(u>0\) and \(v>0\). The boundaries become

\[1\le u^2\le2\quad\Rightarrow\quad 1\le u\le\sqrt2,\]

and

\[1\le v^2\le2\quad\Rightarrow\quad 1\le v\le\sqrt2.\]

Compute the Jacobian:

\[\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} 1/v & -u/v^2\\ v & u \end{vmatrix} =\frac{u}{v}+\frac{u}{v} =\frac{2u}{v}.\]

Also, \(xy=u^2\). Hence

\[\iint_R xy\,dA =\int_1^{\sqrt2}\int_1^{\sqrt2}u^2\cdot\frac{2u}{v}\,dv\,du.\]

Separate the variables:

\[I=\left(\int_1^{\sqrt2}2u^3\,du\right)\left(\int_1^{\sqrt2}\frac{dv}{v}\right).\]

Now

\[\int_1^{\sqrt2}2u^3\,du=\frac{u^4}{2}\Big|_1^{\sqrt2}=\frac{4-1}{2}=\frac32,\]

and

\[\int_1^{\sqrt2}\frac{dv}{v}=\ln(\sqrt2)=\frac12\ln2.\]

Therefore

\[I=\frac32\cdot\frac12\ln2=\frac34\ln2.\]

Answer: \(\displaystyle \frac34\ln2\).

4.7. Cylindrical Coordinates for a Triple Integral (Lab 12, Task 7)

Calculate the integral by changing to cylindrical coordinates:

\[\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{9-x^2-y^2} \sqrt{x^2 + y^2} \,dz \,dy \,dx.\]

Click to see the solution

Key Concept: The \(xy\)-projection is the upper half disk \(x^2+y^2\le9\), \(y\ge0\).

Use cylindrical coordinates:

\[x=r\cos\theta,\qquad y=r\sin\theta,\qquad \sqrt{x^2+y^2}=r,\qquad dV=r\,dz\,dr\,d\theta.\]

The bounds become

\[0\le r\le3,\qquad 0\le\theta\le\pi,\qquad 0\le z\le9-r^2.\]

Thus

\[I=\int_0^\pi\int_0^3\int_0^{9-r^2}r\cdot r\,dz\,dr\,d\theta.\]

Integrate with respect to \(z\):

\[I=\int_0^\pi\int_0^3 r^2(9-r^2)\,dr\,d\theta.\]

Now

\[\int_0^3 r^2(9-r^2)\,dr =\int_0^3(9r^2-r^4)\,dr =9\cdot\frac{27}{3}-\frac{243}{5} =81-\frac{243}{5} =\frac{162}{5}.\]

Therefore

\[I=\pi\cdot\frac{162}{5}=\frac{162\pi}{5}.\]

Answer: \(\displaystyle \frac{162\pi}{5}\).

4.8. Spherical Coordinates for a Cone-Sphere Solid (Lab 12, Task 8)

Use spherical coordinates to find the volume of the solid that lies above the cone \(z = \sqrt{x^2 + y^2}\) and below the sphere \(x^2 + y^2 + z^2 = z\).

Click to see the solution

Key Concept: In spherical coordinates, cones have constant \(\phi\) and spheres through the origin often have simple \(\rho\) bounds.

The cone \(z=\sqrt{x^2+y^2}\) becomes

\[\rho\cos\phi=\rho\sin\phi,\]

\[\phi=\frac{\pi}{4}.\]

“Above the cone” means

\[0\le\phi\le\frac{\pi}{4}.\]

The sphere is

\[x^2+y^2+z^2=z.\]

In spherical coordinates this is

\[\rho^2=\rho\cos\phi,\]

so, for \(\rho\ge0\),

\[0\le\rho\le\cos\phi.\]

Also \(0\le\theta\le2\pi\). Therefore

\[V=\int_0^{2\pi}\int_0^{\pi/4}\int_0^{\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.\]

Integrate in \(\rho\):

\[V=2\pi\int_0^{\pi/4}\frac{\cos^3\phi}{3}\sin\phi\,d\phi.\]

Use \(u=\cos\phi\), \(du=-\sin\phi\,d\phi\):

\[\int_0^{\pi/4}\cos^3\phi\sin\phi\,d\phi =\frac{1-\left(\frac{\sqrt2}{2}\right)^4}{4} =\frac{1-\frac14}{4} =\frac{3}{16}.\]

Thus

\[V=\frac{2\pi}{3}\cdot\frac{3}{16}=\frac{\pi}{8}.\]

Answer: \(\displaystyle \frac{\pi}{8}\).

4.9. Volume of a Paraboloid-Capped Cylinder (Lab 12, Task 9)

Describe the solid and compute its volume:

\[G = \{(x, y, z) : x^2 + y^2 \leq 4,\ 0 \leq z \leq 5 - x^2 - y^2\}.\]

Which coordinate system is most natural?

Click to see the solution

Key Concept: The inequalities use \(x^2+y^2\), so cylindrical coordinates are the natural choice.

The solid lies above the plane \(z=0\), below the downward-opening paraboloid

\[z=5-x^2-y^2,\]

and above the disk

\[x^2+y^2\le4.\]

Use cylindrical coordinates:

\[0\le r\le2,\qquad 0\le\theta\le2\pi,\qquad 0\le z\le5-r^2.\]

Then

\[V=\int_0^{2\pi}\int_0^2\int_0^{5-r^2}r\,dz\,dr\,d\theta.\]

Integrate:

\[V=2\pi\int_0^2(5-r^2)r\,dr =2\pi\left[\frac{5r^2}{2}-\frac{r^4}{4}\right]_0^2.\]

\[V=2\pi(10-4)=12\pi.\]

Answer: The solid is a cylinder of radius \(2\) capped by the paraboloid \(z=5-r^2\) and cut below by \(z=0\); the most natural coordinates are cylindrical; \(\displaystyle V=12\pi\).

4.10. Volume of an Ellipsoid by Scaling (Lab 12, Task 10)

Use a scaling change of variables to find the volume of the ellipsoid

\[\frac{x^2}{4} + y^2 + \frac{z^2}{9} \leq 1.\]

Click to see the solution

Key Concept: Scale the ellipsoid to the unit ball.

Let

\[x=2u,\qquad y=v,\qquad z=3w.\]

Then

\[\frac{x^2}{4}+y^2+\frac{z^2}{9}=u^2+v^2+w^2,\]

so the ellipsoid maps from the unit ball

\[u^2+v^2+w^2\le1.\]

The Jacobian is the determinant of the diagonal scaling matrix:

\[\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|=2\cdot1\cdot3=6.\]

Therefore

\[V_{\text{ellipsoid}}=6\cdot V_{\text{unit ball}} =6\cdot\frac{4\pi}{3} =8\pi.\]

Answer: \(\displaystyle 8\pi\).

4.11. Triple Integrals Over Rectangular Boxes (Lab 12, Homework 1)

Calculate:

a) \(\displaystyle \iiint_G (x + y^2 + z^3) \,dV\) over \(G = [0, 1] \times [0, 2] \times [0, 3]\).

b) \(\displaystyle \iiint_G xyz \,dV\) where \(G = [1, 2] \times [0, 1] \times [0, 4]\).

c) \(\displaystyle \iiint_G e^{x+y+z} \,dV\) over \(G = [0, 1] \times [0, 1] \times [0, 1]\).

d) \(\displaystyle \iiint_G 6x^2yz^2 \,dV\) over \(G = [-1, 1] \times [0, 2] \times [1, 3]\).

Click to see the solution

Key Concept: On rectangular boxes, sums and products often separate into one-variable integrals.

a) Split the integral term by term:

\[\iiint_G x\,dV=\left(\int_0^1x\,dx\right)(2)(3)=\frac12\cdot6=3,\]

\[\iiint_G y^2\,dV=(1)\left(\int_0^2y^2\,dy\right)(3)=\frac{8}{3}\cdot3=8,\]

and

\[\iiint_G z^3\,dV=(1)(2)\left(\int_0^3z^3\,dz\right)=2\cdot\frac{81}{4}=\frac{81}{2}.\]

Therefore

\[I_a=3+8+\frac{81}{2}=\frac{103}{2}.\]

b) The integrand separates:

\[I_b=\left(\int_1^2x\,dx\right)\left(\int_0^1y\,dy\right)\left(\int_0^4z\,dz\right) =\frac32\cdot\frac12\cdot8=6.\]

c) Since \(e^{x+y+z}=e^xe^ye^z\),

\[I_c=\left(\int_0^1e^x\,dx\right)^3=(e-1)^3.\]

d) Separate all factors:

\[I_d=6\left(\int_{-1}^1x^2\,dx\right)\left(\int_0^2y\,dy\right)\left(\int_1^3z^2\,dz\right).\]

Compute each:

\[\int_{-1}^1x^2\,dx=\frac23,\qquad \int_0^2y\,dy=2,\qquad \int_1^3z^2\,dz=\frac{26}{3}.\]

Thus

\[I_d=6\cdot\frac23\cdot2\cdot\frac{26}{3}=\frac{208}{3}.\]

Answer: a) \(\displaystyle \frac{103}{2}\); b) \(6\); c) \((e-1)^3\); d) \(\displaystyle \frac{208}{3}\).

4.12. Integral Over a Solid Under a Plane (Lab 12, Homework 2)

The solid \(G\) is bounded by the planes \(z = 0\), \(z = 1 - y\), and the parabolic cylinder \(x = y^2\), and by the planes \(x = 0\) and \(y = 1\) (where \(x \geq 0\), \(y \geq 0\)). The projection \(R\) onto the xy-plane is the region \(0 \leq y \leq 1\), \(0 \leq x \leq y^2\). Compute \(\displaystyle \iiint_G 2z \,dV\).

Click to see the solution

Key Concept: Use the given projection and place \(z\) between the lower and upper planes.

The bounds are

\[0\le y\le1,\qquad 0\le x\le y^2,\qquad 0\le z\le1-y.\]

Therefore

\[I=\int_0^1\int_0^{y^2}\int_0^{1-y}2z\,dz\,dx\,dy.\]

The \(z\)-integral is

\[\int_0^{1-y}2z\,dz=z^2\Big|_0^{1-y}=(1-y)^2.\]

\[I=\int_0^1\int_0^{y^2}(1-y)^2\,dx\,dy =\int_0^1 y^2(1-y)^2\,dy.\]

Expand:

\[y^2(1-y)^2=y^2-2y^3+y^4.\]

Then

\[I=\left[\frac{y^3}{3}-\frac{y^4}{2}+\frac{y^5}{5}\right]_0^1 =\frac13-\frac12+\frac15 =\frac{1}{30}.\]

Answer: \(\displaystyle \frac{1}{30}\).

4.13. Polar Integral Over a Sector of the Unit Disk (Lab 12, Homework 3)

Evaluate \(\displaystyle \iint_R \frac{1}{\sqrt{x^2 + y^2}} \,dA\) over the region inside \(x^2 + y^2 = 1\) and above \(y = |x|\) using polar coordinates.

Click to see the solution

Key Concept: The factor \(\sqrt{x^2+y^2}\) becomes \(r\), and \(dA=r\,dr\,d\theta\).

The region above \(y=|x|\) inside the unit disk is the sector

\[0\le r\le1,\qquad \frac{\pi}{4}\le\theta\le\frac{3\pi}{4}.\]

The integrand becomes

\[\frac{1}{\sqrt{x^2+y^2}}=\frac1r.\]

Thus

\[I=\int_{\pi/4}^{3\pi/4}\int_0^1\frac1r\cdot r\,dr\,d\theta =\int_{\pi/4}^{3\pi/4}\int_0^1 1\,dr\,d\theta.\]

Therefore

\[I=\left(\frac{3\pi}{4}-\frac{\pi}{4}\right)(1)=\frac{\pi}{2}.\]

Answer: \(\displaystyle \frac{\pi}{2}\).

4.14. Exponential Integral Over the Upper Half Disk (Lab 12, Homework 4)

Evaluate \(\displaystyle \iint_R e^{x^2+y^2} \,dA\) where \(R\) is the upper half of the unit disk using polar coordinates.

Click to see the solution

Key Concept: The expression \(x^2+y^2\) becomes \(r^2\).

For the upper half of the unit disk,

\[0\le r\le1,\qquad 0\le\theta\le\pi.\]

Therefore

\[I=\int_0^\pi\int_0^1 e^{r^2}r\,dr\,d\theta.\]

Use \(u=r^2\), \(du=2r\,dr\):

\[\int_0^1 e^{r^2}r\,dr=\frac12\int_0^1e^u\,du=\frac{e-1}{2}.\]

Thus

\[I=\pi\cdot\frac{e-1}{2}=\frac{\pi(e-1)}{2}.\]

Answer: \(\displaystyle \frac{\pi(e-1)}{2}\).

4.15. Change of Variables with a Malformed Bound (Lab 12, Homework 5)

Evaluate \(\displaystyle \int_{1}^{2} \int_{y/x}^{\sqrt{y}} e^{\sqrt{xy}} \,dx \,dy\) using \(u = \sqrt{xy}\), \(v = \sqrt{y/x}\).

Click to see the solution

Key Concept: The printed inner lower limit \(y/x\) contains the same variable \(x\) that is being integrated with respect to. Therefore the integral, as written, is not a standard iterated integral.

If we read the lower condition literally as the curve

\[x=\frac{y}{x},\]

then it implies

\[x^2=y.\]

Since the upper limit is also

\[x=\sqrt y,\]

the lower and upper boundaries coincide. Under this literal interpretation, every inner integral has zero width:

\[\int_{\sqrt y}^{\sqrt y} e^{\sqrt{xy}}\,dx=0.\]

Therefore

\[\int_1^2 0\,dy=0.\]

The intended problem likely has a typo in the lower limit. With the text exactly as given, the only consistent literal value is zero.

Answer: \(\displaystyle 0\) under the literal interpretation of the printed bounds.

4.16. Transformation \(u=x+y\), \(v=x-y\) Over a Square (Lab 12, Homework 6)

Use the transformation \(u = x + y\), \(v = x - y\) to evaluate \(\displaystyle \iint_R e^{x+y} \sin(x - y) \,dA\) where \(R\) is the square with vertices \((0, 0)\), \((\pi, 0)\), \((\pi, \pi)\), \((0, \pi)\).

Click to see the solution

Key Concept: The integrand becomes \(e^u\sin v\), and the square becomes a symmetric diamond in the \(uv\)-plane.

The inverse transformation is

\[x=\frac{u+v}{2},\qquad y=\frac{u-v}{2}.\]

The Jacobian is

\[\left|\frac{\partial(x,y)}{\partial(u,v)}\right| =\left| \begin{vmatrix} 1/2 & 1/2\\ 1/2 & -1/2 \end{vmatrix} \right| =\frac12.\]

The square \(0\le x\le\pi\), \(0\le y\le\pi\) becomes

\[0\le u+v\le2\pi,\qquad 0\le u-v\le2\pi.\]

For each fixed \(u\), the allowed \(v\)-interval is symmetric about \(0\). Since \(\sin v\) is odd, the inner integral in \(v\) is zero:

\[\int_{-a}^{a}e^u\sin v\cdot\frac12\,dv=0.\]

Therefore the whole double integral is zero.

Answer: \(\displaystyle 0\).

4.17. Transforming the Region Bounded by \(xy\) and \(y/x\) (Lab 12, Homework 7)

Evaluate \(\displaystyle \iint_R (x^2 + y^2) \,dA\) over the region in the first quadrant bounded by \(xy = 1\), \(xy = 2\), \(y = x\), and \(y = 2x\) using an appropriate transformation.

Click to see the solution

Key Concept: Use \(u=xy\) and \(v=y/x\), because the boundary curves become \(u=\text{constant}\) and \(v=\text{constant}\).

In the first quadrant,

\[1\le u\le2,\qquad 1\le v\le2.\]

Solving for \(x\) and \(y\) gives

\[x=\sqrt{\frac{u}{v}},\qquad y=\sqrt{uv}.\]

The Jacobian is

\[\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{1}{2v}.\]

Also,

\[x^2+y^2=\frac{u}{v}+uv=u\left(\frac1v+v\right).\]

Therefore

\[I=\int_1^2\int_1^2 u\left(\frac1v+v\right)\frac{1}{2v}\,dv\,du.\]

Simplify:

\[I=\int_1^2u\,du\cdot\frac12\int_1^2\left(\frac{1}{v^2}+1\right)\,dv.\]

Now

\[\int_1^2u\,du=\frac32,\]

and

\[\frac12\int_1^2\left(1+\frac1{v^2}\right)\,dv =\frac12\left[v-\frac1v\right]_1^2 =\frac12\left(\frac32\right) =\frac34.\]

Thus

\[I=\frac32\cdot\frac34=\frac98.\]

Answer: \(\displaystyle \frac98\).

4.18. Integral Inside a Cylinder Between Two Planes (Lab 12, Homework 8)

Evaluate \(\displaystyle \iiint_G z \,dV\), where \(G\) is the solid inside the cylinder \(x^2 + y^2 \leq 1\) and between the planes \(z = 0\) and \(z = 2 + y\).

Click to see the solution

Key Concept: Integrating \(z\) first produces a function over the disk.

The bounds are

\[x^2+y^2\le1,\qquad 0\le z\le2+y.\]

Thus

\[I=\iint_{x^2+y^2\le1}\int_0^{2+y}z\,dz\,dA =\frac12\iint_D(2+y)^2\,dA.\]

Expand:

\[\frac12(2+y)^2=\frac12(4+4y+y^2).\]

Over the disk, \(\iint_D y\,dA=0\) by symmetry, \(\iint_D1\,dA=\pi\), and

\[\iint_D y^2\,dA=\frac{\pi}{4}.\]

Therefore

\[I=\frac12\left(4\pi+\frac{\pi}{4}\right)=\frac{17\pi}{8}.\]

Answer: \(\displaystyle \frac{17\pi}{8}\).

4.19. Volume Inside a Sphere Above a Cone (Lab 12, Homework 9)

Let \(G\) be the solid enclosed by the sphere \(x^2 + y^2 + z^2 = 9\) and above the cone \(z = \sqrt{x^2 + y^2}\). Compute the volume of \(G\).

Click to see the solution

Key Concept: Use spherical coordinates because the boundary is a sphere and a cone.

The sphere is \(\rho=3\). The cone is

\[z=\sqrt{x^2+y^2}\quad\Longrightarrow\quad \phi=\frac{\pi}{4}.\]

Above the cone means

\[0\le\phi\le\frac{\pi}{4}.\]

Hence

\[V=\int_0^{2\pi}\int_0^{\pi/4}\int_0^3\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.\]

Compute:

\[V=2\pi\cdot\frac{3^3}{3}\cdot\left[-\cos\phi\right]_0^{\pi/4} =18\pi\left(1-\frac{\sqrt2}{2}\right).\]

Equivalently,

\[V=9\pi(2-\sqrt2).\]

Answer: \(\displaystyle 9\pi(2-\sqrt2)\).

4.20. Integral Over the Upper Half Ball (Lab 12, Homework 10)

Evaluate \(\displaystyle \iiint_G (x^2 + y^2) \,dV\), where \(G\) is the upper half of the ball \(x^2 + y^2 + z^2 \leq 4\) (that is, \(z \geq 0\)).

Click to see the solution

Key Concept: In spherical coordinates, \(x^2+y^2=\rho^2\sin^2\phi\).

The upper half ball of radius \(2\) has bounds

\[0\le\rho\le2,\qquad 0\le\phi\le\frac{\pi}{2},\qquad 0\le\theta\le2\pi.\]

Then

\[I=\int_0^{2\pi}\int_0^{\pi/2}\int_0^2 \rho^2\sin^2\phi\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.\]

\[I=\left(\int_0^{2\pi}d\theta\right) \left(\int_0^{\pi/2}\sin^3\phi\,d\phi\right) \left(\int_0^2\rho^4\,d\rho\right).\]

Use

\[\int_0^{\pi/2}\sin^3\phi\,d\phi=\frac23,\qquad \int_0^2\rho^4\,d\rho=\frac{32}{5}.\]

Therefore

\[I=2\pi\cdot\frac23\cdot\frac{32}{5}=\frac{128\pi}{15}.\]

Answer: \(\displaystyle \frac{128\pi}{15}\).

4.21. Accumulated Signal in a Cylindrical Sensor (Lab 12, Homework 11)

A cylindrical sensor covers the region \(G = \{(r, \theta, z) : 0 \leq r \leq 2, 0 \leq \theta \leq 2\pi, 0 \leq z \leq 4 - r\}\). Assume the signal intensity is \(I(x, y, z) = z\). Find the accumulated signal

\[S = \iiint_G I(x, y, z) \,dV.\]

Click to see the solution

Key Concept: In cylindrical coordinates, \(dV=r\,dz\,dr\,d\theta\).

Since \(I=z\),

\[S=\int_0^{2\pi}\int_0^2\int_0^{4-r}z\,r\,dz\,dr\,d\theta.\]

Integrate in \(z\):

\[\int_0^{4-r}z\,r\,dz=\frac{r(4-r)^2}{2}.\]

Thus

\[S=2\pi\int_0^2\frac{r(4-r)^2}{2}\,dr =\pi\int_0^2r(16-8r+r^2)\,dr.\]

Compute:

\[S=\pi\left[8r^2-\frac{8}{3}r^3+\frac{r^4}{4}\right]_0^2 =\pi\left(32-\frac{64}{3}+4\right) =\frac{44\pi}{3}.\]

Answer: \(\displaystyle \frac{44\pi}{3}\).

4.22. Triple Change of Variables for a Solid Defined by Inequalities (Lab 12, Homework 12)

Let \(G\) be the solid in 3-space defined by the inequalities \(1 - e^x \leq y \leq 3 - e^x\), \(1 - y \leq 2z \leq 2 - y\), \(y \leq e^x \leq y + 4\).

a) Using the coordinate transformation \(u = e^x + y\), \(v = y + 2z\), \(w = e^x - y\), calculate the Jacobian \(\frac{\partial(x,y,z)}{\partial(u,v,w)}\). Express your answer in terms of \(u, v, w\).

b) Using a triple integral and the change of variables given in part (a), find the volume of \(G\).

Click to see the solution

Key Concept: The transformation was chosen so that each inequality becomes a simple interval.

From

\[u=e^x+y,\qquad w=e^x-y,\]

we get

\[e^x=\frac{u+w}{2},\qquad y=\frac{u-w}{2},\qquad x=\ln\left(\frac{u+w}{2}\right).\]

Also,

\[v=y+2z\quad\Rightarrow\quad z=\frac{v-y}{2}=\frac{2v-u+w}{4}.\]

\[x=\ln\left(\frac{u+w}{2}\right),\qquad y=\frac{u-w}{2},\qquad z=\frac{2v-u+w}{4}.\]

The Jacobian matrix is

\[ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix} \frac{1}{u+w} & 0 & \frac{1}{u+w}\\ \frac12 & 0 & -\frac12\\ -\frac14 & \frac12 & \frac14 \end{vmatrix}. \]

Expanding along the middle column,

\[J=-\frac12 \begin{vmatrix} \frac{1}{u+w} & \frac{1}{u+w}\\ \frac12 & -\frac12 \end{vmatrix} =-\frac12\left(-\frac{1}{u+w}\right) =\frac{1}{2(u+w)}.\]

The inequalities become

\[1\le u\le3,\qquad 1\le v\le2,\qquad 0\le w\le4.\]

Therefore the volume is

\[V=\int_1^3\int_1^2\int_0^4\frac{1}{2(u+w)}\,dw\,dv\,du.\]

The \(v\)-interval has length \(1\), so

\[V=\frac12\int_1^3\left[\ln(u+w)\right]_{w=0}^{4}\,du =\frac12\int_1^3\ln\left(\frac{u+4}{u}\right)\,du.\]

An antiderivative is

\[\int\ln\left(\frac{u+4}{u}\right)\,du =(u+4)\ln(u+4)-u\ln u-4.\]

Thus

\[V=\frac12\left[7\ln7-3\ln3-5\ln5\right].\]

Answer: a) \(\displaystyle J=\frac{1}{2(u+w)}\); b) \(\displaystyle V=\frac12(7\ln7-3\ln3-5\ln5)\).

4.23. Linear Change of Variables in a Box (Lab 12, Homework 13)

Use the change of variables \(x = u + v\), \(y = u - v\), \(z = w\) to evaluate \(\displaystyle \iiint_G (x - y) \,dV\), where \(G\) is the image of the box \(0 \leq u \leq 1\), \(0 \leq v \leq 2\), \(0 \leq w \leq 3\).

Click to see the solution

Key Concept: Since \(G\) is already described as the image of a box, integrate in the parameter variables.

First,

\[x-y=(u+v)-(u-v)=2v.\]

The Jacobian is

\[\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| = \left| \begin{vmatrix} 1 & 1 & 0\\ 1 & -1 & 0\\ 0 & 0 & 1 \end{vmatrix} \right|=2.\]

Therefore

\[I=\int_0^1\int_0^2\int_0^3 (2v)(2)\,dw\,dv\,du.\]

Thus

\[I=\int_0^1du\int_0^2 4v\,dv\int_0^3dw =1\cdot 8\cdot3=24.\]

Answer: \(\displaystyle 24\).

4.24. Integral Over a Transformed Solid (Lab 12, Homework 14)

Let \(D\) be the solid region in xyz-space defined by the inequalities \(1 \leq x \leq 2\), \(0 \leq xy \leq 2\), \(0 \leq z \leq 1\). Calculate the integral \(\displaystyle \iiint_D (x^2y + 3xyz) \,dx \,dy \,dz\) by applying the transformation \(u = x\), \(v = xy\), \(w = 3z\).

Click to see the solution

Key Concept: The variables \(u=x\), \(v=xy\), and \(w=3z\) turn the inequalities into a rectangular box.

The inverse transformation is

\[x=u,\qquad y=\frac{v}{u},\qquad z=\frac{w}{3}.\]

The bounds become

\[1\le u\le2,\qquad 0\le v\le2,\qquad 0\le w\le3.\]

The Jacobian is

\[\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|=\frac{1}{3u}.\]

The integrand transforms as

\[x^2y+3xyz=u^2\cdot\frac{v}{u}+3u\cdot\frac{v}{u}\cdot\frac{w}{3}=uv+vw=v(u+w).\]

Therefore

\[I=\int_1^2\int_0^2\int_0^3 v(u+w)\frac{1}{3u}\,dw\,dv\,du.\]

First integrate with respect to \(v\):

\[\int_0^2 v\,dv=2.\]

\[I=\frac{2}{3}\int_1^2\int_0^3\frac{u+w}{u}\,dw\,du =\frac{2}{3}\int_1^2\left(3+\frac{9}{2u}\right)\,du.\]

Thus

\[I=\frac{2}{3}\left(3+\frac92\ln2\right)=2+3\ln2.\]

Answer: \(\displaystyle 2+3\ln2\).

4.25. Average Value of \(z\) on an Upper Half Ball (Lab 12, Homework 15)

Find the average value of \(f(x, y, z) = z\) on the solid \(G = \{(x, y, z) : x^2 + y^2 + z^2 \leq 4, z \geq 0\}\). Recall that \(\displaystyle f_{\text{avg}} = \frac{1}{\text{Vol}(G)} \iiint_G f \,dV\).

Click to see the solution

Key Concept: Average value equals total integral divided by volume.

Use spherical coordinates. The upper half ball of radius \(2\) has

\[0\le\rho\le2,\qquad 0\le\phi\le\frac{\pi}{2},\qquad 0\le\theta\le2\pi.\]

Since \(z=\rho\cos\phi\) and \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\),

\[\iiint_G z\,dV =\int_0^{2\pi}\int_0^{\pi/2}\int_0^2\rho^3\cos\phi\sin\phi\,d\rho\,d\phi\,d\theta.\]

Compute:

\[\int_0^2\rho^3\,d\rho=4,\qquad \int_0^{\pi/2}\cos\phi\sin\phi\,d\phi=\frac12,\qquad \int_0^{2\pi}d\theta=2\pi.\]

Thus

\[\iiint_G z\,dV=4\pi.\]

The volume of a half ball of radius \(2\) is

\[\operatorname{Vol}(G)=\frac12\cdot\frac43\pi(2^3)=\frac{16\pi}{3}.\]

Therefore

\[f_{\text{avg}}=\frac{4\pi}{16\pi/3}=\frac34.\]

Answer: \(\displaystyle \frac34\).

4.26. Sketch the Rotational Field \(\mathbf{F}(x,y)=-y\mathbf{i}+x\mathbf{j}\) (Lab 13, Example 1)

Describe the vector field \(\mathbf{F}(x, y) = -y\mathbf{i} + x\mathbf{j}\) by sketching some vectors.

Click to see the solution

Key Concept: Evaluate the field at representative points and look for a geometric pattern.

At \((1,0)\),

\[\mathbf{F}(1,0)=0\mathbf{i}+1\mathbf{j}=\mathbf{j},\]

so the vector points upward.

At \((0,1)\),

\[\mathbf{F}(0,1)=-1\mathbf{i}+0\mathbf{j}=-\mathbf{i},\]

so the vector points left.

At \((-1,0)\),

\[\mathbf{F}(-1,0)=-\mathbf{j},\]

so the vector points downward.

At \((0,-1)\),

\[\mathbf{F}(0,-1)=\mathbf{i},\]

so the vector points right.

Each vector is perpendicular to the radius vector \(\langle x,y\rangle\), because

\[\langle x,y\rangle\cdot\langle -y,x\rangle=-xy+xy=0.\]

Also,

\[\|\mathbf{F}(x,y)\|=\sqrt{(-y)^2+x^2}=\sqrt{x^2+y^2},\]

so vectors farther from the origin are longer. The field rotates counterclockwise around the origin.

Answer: The sketch is a counterclockwise rotational field with vector length proportional to distance from the origin.

4.27. Sketch the Space Field \(\mathbf{F}(x,y,z)=z\mathbf{k}\) (Lab 13, Example 2)

Sketch the vector field on \(\mathbb{R}^3\) given by \(\mathbf{F}(x, y, z) = z\mathbf{k}\).

Click to see the solution

Key Concept: The vector has only a \(z\)-component.

At every point,

\[\mathbf{F}(x,y,z)=\langle0,0,z\rangle.\]

If \(z>0\), the vector points upward. If \(z<0\), it points downward. If \(z=0\), the vector is zero. The magnitude is

\[\|\mathbf{F}(x,y,z)\|=|z|.\]

Thus the vectors are vertical everywhere, and their lengths increase with distance from the \(xy\)-plane.

Answer: Vertical arrows point upward above the \(xy\)-plane, downward below it, and vanish on the \(xy\)-plane.

4.28. Sketch Several Vector Fields (Lab 13, Task 1)

Sketch the following vector fields:

a) \(\mathbf{F}(x, y) = 0.3\mathbf{i} - 0.4\mathbf{j}\).

b) \(\mathbf{F}(x, y) = -\frac{1}{2}\mathbf{i} + (y - x)\mathbf{j}\).

c) \(\mathbf{F}(x, y) = \dfrac{y\mathbf{i} + x\mathbf{j}}{\sqrt{x^2 + y^2}}\).

d) \(\mathbf{F}(x, y, z) = \mathbf{i}\) in \(\mathbb{R}^3\).

Click to see the solution

Key Concept: A vector field sketch is built from representative arrows and symmetry.

a) The vector is constant:

\[\mathbf{F}(x,y)=\langle0.3,-0.4\rangle.\]

Every arrow points right and downward with the same length

\[\sqrt{0.3^2+(-0.4)^2}=0.5.\]

b) The horizontal component is always \(-1/2\), so every vector points left. The vertical component is \(y-x\):

on the line \(y=x\), vectors are purely horizontal and point left;
above the line \(y=x\), vectors point left and upward;
below the line \(y=x\), vectors point left and downward.

c) The field is undefined at \((0,0)\). Away from the origin,

\[\|\mathbf{F}(x,y)\|=\frac{\sqrt{y^2+x^2}}{\sqrt{x^2+y^2}}=1.\]

So all arrows have unit length. Representative values:

\[\mathbf{F}(1,0)=\mathbf{j},\qquad \mathbf{F}(0,1)=\mathbf{i},\qquad \mathbf{F}(1,1)=\frac{1}{\sqrt2}(\mathbf{i}+\mathbf{j}).\]

This field swaps the roles of \(x\) and \(y\) in the direction vector and normalizes the result.

d) The field is constant in space:

\[\mathbf{F}(x,y,z)=\langle1,0,0\rangle.\]

Every arrow points in the positive \(x\)-direction with length \(1\).

Answer: a) constant southeast field; b) leftward field tilted up or down depending on \(y-x\); c) unit field undefined at the origin with swapped-coordinate directions; d) constant positive \(x\)-direction field.

4.29. Gradient of \(f(x,y)=x^2y-y^3\) (Lab 13, Example 3)

Find the gradient vector field of \(f(x, y) = x^2y - y^3\).

Click to see the solution

Key Concept: The gradient is the vector of first partial derivatives.

For a two-variable scalar field,

\[\nabla f=f_x\mathbf{i}+f_y\mathbf{j}.\]

Differentiate with respect to \(x\), treating \(y\) as constant:

\[f_x=\frac{\partial}{\partial x}(x^2y-y^3)=2xy.\]

Differentiate with respect to \(y\), treating \(x\) as constant:

\[f_y=\frac{\partial}{\partial y}(x^2y-y^3)=x^2-3y^2.\]

Therefore

\[\nabla f(x,y)=2xy\mathbf{i}+(x^2-3y^2)\mathbf{j}.\]

Answer: \(\displaystyle \nabla f=2xy\mathbf{i}+(x^2-3y^2)\mathbf{j}\).

4.30. Curl and Divergence of \(xz\mathbf{i}+xyz\mathbf{j}-y^2\mathbf{k}\) (Lab 13, Example 4)

Let \(\mathbf{F}(x, y, z) = xz\mathbf{i} + xyz\mathbf{j} - y^2\mathbf{k}\). Find \(\operatorname{curl}\mathbf{F}\) and \(\operatorname{div}\mathbf{F}\).

Click to see the solution

Key Concept: For \(\mathbf{F}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k}\),

\[\nabla\times\mathbf{F}=(R_y-Q_z)\mathbf{i}+(P_z-R_x)\mathbf{j}+(Q_x-P_y)\mathbf{k},\]

and

\[\nabla\cdot\mathbf{F}=P_x+Q_y+R_z.\]

Here

\[P=xz,\qquad Q=xyz,\qquad R=-y^2.\]

Compute the curl:

\[R_y=-2y,\qquad Q_z=xy,\]

so the \(\mathbf{i}\)-component is

\[R_y-Q_z=-2y-xy=-y(2+x).\]

Next,

\[P_z=x,\qquad R_x=0,\]

so the \(\mathbf{j}\)-component is

\[P_z-R_x=x.\]

Finally,

\[Q_x=yz,\qquad P_y=0,\]

so the \(\mathbf{k}\)-component is

\[Q_x-P_y=yz.\]

Therefore

\[\nabla\times\mathbf{F}=-y(2+x)\mathbf{i}+x\mathbf{j}+yz\mathbf{k}.\]

For divergence,

\[P_x=z,\qquad Q_y=xz,\qquad R_z=0.\]

Thus

\[\nabla\cdot\mathbf{F}=z+xz=z(1+x).\]

Answer: \(\displaystyle \operatorname{curl}\mathbf{F}=-y(2+x)\mathbf{i}+x\mathbf{j}+yz\mathbf{k}\) and \(\displaystyle \operatorname{div}\mathbf{F}=z(1+x)\).

4.31. Gradient Vector Fields (Lab 13, Task 2)

Find the gradient vector field of the following functions:

a) \(f(x, y) = y \sin(xy)\).

b) \(f(x, y, z) = \sqrt{x^2 + y^2 + z^2}\).

Click to see the solution

Key Concept: Differentiate with respect to one variable at a time and treat all other variables as constants.

a) For \(f(x,y)=y\sin(xy)\),

\[f_x=y\cos(xy)\cdot y=y^2\cos(xy).\]

For \(f_y\), use the product rule:

\[f_y=\sin(xy)+y\cdot x\cos(xy)=\sin(xy)+xy\cos(xy).\]

Therefore

\[\nabla f=y^2\cos(xy)\mathbf{i}+\left(\sin(xy)+xy\cos(xy)\right)\mathbf{j}.\]

b) Let

\[f(x,y,z)=(x^2+y^2+z^2)^{1/2}.\]

Then

\[f_x=\frac{x}{\sqrt{x^2+y^2+z^2}},\qquad f_y=\frac{y}{\sqrt{x^2+y^2+z^2}},\qquad f_z=\frac{z}{\sqrt{x^2+y^2+z^2}}.\]

Therefore, away from the origin,

\[\nabla f=\frac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{x^2+y^2+z^2}}.\]

Answer: a) \(\displaystyle y^2\cos(xy)\mathbf{i}+(\sin(xy)+xy\cos(xy))\mathbf{j}\); b) \(\displaystyle \frac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{x^2+y^2+z^2}}\) for \((x,y,z)\ne(0,0,0)\).

4.32. Curl of \(yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}\) (Lab 13, Task 3)

Compute the curl of \(\mathbf{F}(x, y, z) = yz\mathbf{i} + xz\mathbf{j} + xy\mathbf{k}\).

Click to see the solution

Key Concept: Use \(\nabla\times\mathbf{F}=(R_y-Q_z)\mathbf{i}+(P_z-R_x)\mathbf{j}+(Q_x-P_y)\mathbf{k}\).

Here

\[P=yz,\qquad Q=xz,\qquad R=xy.\]

Compute:

\[R_y=x,\qquad Q_z=x,\]

\[R_y-Q_z=0.\]

Next,

\[P_z=y,\qquad R_x=y,\]

\[P_z-R_x=0.\]

Finally,

\[Q_x=z,\qquad P_y=z,\]

\[Q_x-P_y=0.\]

Therefore

\[\nabla\times\mathbf{F}=\mathbf{0}.\]

Answer: \(\displaystyle \operatorname{curl}\mathbf{F}=\mathbf{0}\).

4.33. Divergence and the Identity \(\operatorname{div}(\operatorname{curl}\mathbf{F})=0\) (Lab 13, Task 4)

Compute the divergence of \(\mathbf{F}(x, y, z) = yz\mathbf{i} + xz\mathbf{j} + xy\mathbf{k}\) and verify that for any \(\mathbf{F}\) with continuous second partials, \(\operatorname{div}(\operatorname{curl}\mathbf{F}) = 0\).

Click to see the solution

Key Concept: Divergence differentiates each component with respect to its matching coordinate.

For

\[\mathbf{F}=yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k},\]

we have

\[\nabla\cdot\mathbf{F}=\frac{\partial}{\partial x}(yz)+\frac{\partial}{\partial y}(xz)+\frac{\partial}{\partial z}(xy)=0+0+0=0.\]

Now let a general field be

\[\mathbf{F}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k}.\]

Then

\[\nabla\times\mathbf{F}=(R_y-Q_z)\mathbf{i}+(P_z-R_x)\mathbf{j}+(Q_x-P_y)\mathbf{k}.\]

Taking divergence,

\[\nabla\cdot(\nabla\times\mathbf{F}) =\frac{\partial}{\partial x}(R_y-Q_z) \frac{\partial}{\partial y}(P_z-R_x) \frac{\partial}{\partial z}(Q_x-P_y).\]

\[\nabla\cdot(\nabla\times\mathbf{F}) =R_{yx}-Q_{zx}+P_{zy}-R_{xy}+Q_{xz}-P_{yz}.\]

If second partial derivatives are continuous, mixed partials are equal:

\[R_{yx}=R_{xy},\qquad Q_{zx}=Q_{xz},\qquad P_{zy}=P_{yz}.\]

All terms cancel, hence

\[\nabla\cdot(\nabla\times\mathbf{F})=0.\]

Answer: For the given field, \(\operatorname{div}\mathbf{F}=0\); in general, \(\operatorname{div}(\operatorname{curl}\mathbf{F})=0\) when the required second partials are continuous.

4.34. Sketch \(\mathbf{F}(x,y)=\frac12x\mathbf{i}+y\mathbf{j}\) (Lab 13, Homework 1)

Sketch the vector field \(\mathbf{F}(x, y) = \frac{1}{2}x\mathbf{i} + y\mathbf{j}\).

Click to see the solution

Key Concept: The vector at \((x,y)\) is \(\langle x/2,y\rangle\).

On the positive \(x\)-axis, vectors point right. On the negative \(x\)-axis, they point left. On the positive \(y\)-axis, they point upward. On the negative \(y\)-axis, they point downward.

The origin is the only zero vector:

\[\mathbf{F}(0,0)=\mathbf{0}.\]

The magnitude is

\[\|\mathbf{F}(x,y)\|=\sqrt{\frac{x^2}{4}+y^2},\]

so arrows grow farther from the origin. The \(y\)-component grows twice as strongly as the \(x\)-component for the same coordinate size.

Answer: The field points outward from the origin, with vertical stretching stronger than horizontal stretching.

4.35. Describe a CAS Plot of \(\langle \ln(1+y^2),\ln(1+x^2)\rangle\) (Lab 13, Homework 2)

Use a Computer Algebra System (CAS) to plot the vector field \(\mathbf{F}(x, y) = \langle \ln(1 + y^2), \ln(1 + x^2) \rangle\) and describe its appearance.

Click to see the solution

Key Concept: The components are always nonnegative because \(\ln(1+t^2)\ge0\).

The horizontal component is

\[P(x,y)=\ln(1+y^2),\]

so it depends only on the distance from the \(x\)-axis. It is zero when \(y=0\) and grows as \(|y|\) grows.

The vertical component is

\[Q(x,y)=\ln(1+x^2),\]

so it depends only on the distance from the \(y\)-axis. It is zero when \(x=0\) and grows as \(|x|\) grows.

Thus every vector points into the first-quadrant direction, except:

on the \(x\)-axis, vectors are vertical upward;
on the \(y\)-axis, vectors are horizontal rightward;
at the origin, the vector is zero.

The field is symmetric with respect to both coordinate axes because both components contain squares.

Answer: The plot shows first-quadrant-pointing arrows, symmetric across both axes, with arrow lengths increasing away from the coordinate axes.

4.36. Gradient of \(f(s,t)=\sqrt{2s+3t}\) (Lab 13, Homework 3)

Find the gradient vector field of \(f(s, t) = \sqrt{2s + 3t}\).

Click to see the solution

Key Concept: Apply the chain rule to \((2s+3t)^{1/2}\).

Write

\[f(s,t)=(2s+3t)^{1/2}.\]

Then

\[f_s=\frac12(2s+3t)^{-1/2}\cdot2=\frac{1}{\sqrt{2s+3t}},\]

and

\[f_t=\frac12(2s+3t)^{-1/2}\cdot3=\frac{3}{2\sqrt{2s+3t}}.\]

Therefore

\[\nabla f=\frac{1}{\sqrt{2s+3t}}\mathbf{i}+\frac{3}{2\sqrt{2s+3t}}\mathbf{j},\]

where \(2s+3t>0\).

Answer: \(\displaystyle \nabla f=\left\langle \frac{1}{\sqrt{2s+3t}},\frac{3}{2\sqrt{2s+3t}}\right\rangle\).

4.37. Curl and Divergence of \(y\mathbf{i}+z\mathbf{j}+x\mathbf{k}\) (Lab 13, Homework 4)

Compute both the curl and divergence of \(\mathbf{F}(x, y, z) = y\mathbf{i} + z\mathbf{j} + x\mathbf{k}\).

Click to see the solution

Key Concept: Use the component formulas directly.

Here

\[P=y,\qquad Q=z,\qquad R=x.\]

The curl is

\[\nabla\times\mathbf{F}=(R_y-Q_z)\mathbf{i}+(P_z-R_x)\mathbf{j}+(Q_x-P_y)\mathbf{k}.\]

Compute:

\[R_y=0,\quad Q_z=1,\quad P_z=0,\quad R_x=1,\quad Q_x=0,\quad P_y=1.\]

Thus

\[\nabla\times\mathbf{F}=-\mathbf{i}-\mathbf{j}-\mathbf{k}.\]

The divergence is

\[\nabla\cdot\mathbf{F}=P_x+Q_y+R_z=0+0+0=0.\]

Answer: \(\displaystyle \operatorname{curl}\mathbf{F}=-\mathbf{i}-\mathbf{j}-\mathbf{k}\) and \(\displaystyle \operatorname{div}\mathbf{F}=0\).

4.38. Verify \(\operatorname{div}(\nabla f)=\nabla^2 f\) (Lab 13, Homework 5)

Verify that \(\operatorname{div}(\nabla f) = \nabla^2f\) for \(f(x, y, z) = x^2y e^{y/z}\).

Click to see the solution

Key Concept: By definition, \(\operatorname{div}(\nabla f)=f_{xx}+f_{yy}+f_{zz}\), which is the Laplacian.

Let

\[f=x^2y e^{y/z}.\]

First,

\[f_x=2xy e^{y/z},\qquad f_{xx}=2y e^{y/z}.\]

Next,

\[f_y=x^2e^{y/z}+x^2y\frac{1}{z}e^{y/z} =x^2e^{y/z}\left(1+\frac{y}{z}\right).\]

Differentiate again with respect to \(y\):

\[f_{yy}=x^2e^{y/z}\left(\frac{2}{z}+\frac{y}{z^2}\right).\]

Finally,

\[f_z=x^2y e^{y/z}\left(-\frac{y}{z^2}\right) =-\frac{x^2y^2}{z^2}e^{y/z}.\]

Differentiate again with respect to \(z\):

\[f_{zz}=x^2y^2e^{y/z}\left(\frac{2}{z^3}+\frac{y}{z^4}\right).\]

Therefore

\[\operatorname{div}(\nabla f)=f_{xx}+f_{yy}+f_{zz}\]

equals

\[2y e^{y/z} x^2e^{y/z}\left(\frac{2}{z}+\frac{y}{z^2}\right) x^2y^2e^{y/z}\left(\frac{2}{z^3}+\frac{y}{z^4}\right).\]

This is exactly \(\nabla^2f\).

Answer: \(\displaystyle \operatorname{div}(\nabla f)=\nabla^2f=f_{xx}+f_{yy}+f_{zz}\), with the explicit value above.

4.39. Show That \(e^{y+2z}(\mathbf{i}+x\mathbf{j}+2x\mathbf{k})\) Is Conservative (Lab 13, Homework 6)

Show that \(\mathbf{F}(x, y, z) = e^{y+2z}(\mathbf{i} + x\mathbf{j} + 2x\mathbf{k})\) is conservative by computing its curl.

Click to see the solution

Key Concept: On a simply connected domain such as \(\mathbb{R}^3\), a continuously differentiable field with zero curl is conservative.

Write

\[P=e^{y+2z},\qquad Q=xe^{y+2z},\qquad R=2xe^{y+2z}.\]

Compute the curl components:

\[R_y=2xe^{y+2z},\qquad Q_z=2xe^{y+2z},\]

\[R_y-Q_z=0.\]

Next,

\[P_z=2e^{y+2z},\qquad R_x=2e^{y+2z},\]

\[P_z-R_x=0.\]

Finally,

\[Q_x=e^{y+2z},\qquad P_y=e^{y+2z},\]

\[Q_x-P_y=0.\]

Therefore

\[\nabla\times\mathbf{F}=\mathbf{0}.\]

Since the field is defined and continuously differentiable on all of \(\mathbb{R}^3\), it is conservative.

A potential function is visible by integrating \(P\) with respect to \(x\):

\[f(x,y,z)=xe^{y+2z}+C.\]

Indeed,

\[\nabla f=e^{y+2z}\mathbf{i}+xe^{y+2z}\mathbf{j}+2xe^{y+2z}\mathbf{k}.\]

Answer: \(\operatorname{curl}\mathbf{F}=\mathbf{0}\), so \(\mathbf{F}\) is conservative; one potential is \(\displaystyle f=xe^{y+2z}+C\).

4.40. Explain Why the Gravitational Field Is Conservative (Lab 13, Homework 7)

Explain why the gravitational field \(\mathbf{F}(\mathbf{x}) = -\dfrac{mMG}{|\mathbf{x}|^3}\mathbf{x}\) is conservative.

Click to see the solution

Key Concept: A field is conservative if it is the gradient of a scalar potential function.

Let

\[r=|\mathbf{x}|=\sqrt{x^2+y^2+z^2}.\]

The scalar function

\[f(\mathbf{x})=\frac{mMG}{r}\]

is defined on the natural domain \(\mathbb{R}^3\setminus\{\mathbf{0}\}\). Since

\[\nabla\left(\frac1r\right)=-\frac{\mathbf{x}}{r^3},\]

we have

\[\nabla f=mMG\nabla\left(\frac1r\right) =-\frac{mMG}{r^3}\mathbf{x} =\mathbf{F}(\mathbf{x}).\]

Thus \(\mathbf{F}\) is a gradient field on its natural domain, so it is conservative there.

Answer: \(\mathbf{F}=\nabla\left(\dfrac{mMG}{|\mathbf{x}|}\right)\) on \(\mathbb{R}^3\setminus\{\mathbf{0}\}\), hence it is conservative.

4.41. Curl and Divergence of Logarithmic and Arctangent Fields (Lab 14, Task 1)

Find the curl and the divergence of the following vector fields:

a) \(\mathbf{F}=\ln(2y+3z)\mathbf{i}+\ln(x+3z)\mathbf{j}+\ln(x+2y)\mathbf{k}\).

b) \(\mathbf{F}=\arctan(xy)\mathbf{i}+\arctan(yz)\mathbf{j}+\arctan(xz)\mathbf{k}\).

Click to see the solution

Key Concept: For \(\mathbf{F}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k}\),

\[\nabla\times\mathbf{F}=(R_y-Q_z)\mathbf{i}+(P_z-R_x)\mathbf{j}+(Q_x-P_y)\mathbf{k},\]

and

\[\nabla\cdot\mathbf{F}=P_x+Q_y+R_z.\]

a) Here

\[P=\ln(2y+3z),\qquad Q=\ln(x+3z),\qquad R=\ln(x+2y).\]

Compute the derivatives needed for the curl:

\[R_y=\frac{2}{x+2y},\qquad Q_z=\frac{3}{x+3z},\]

\[P_z=\frac{3}{2y+3z},\qquad R_x=\frac{1}{x+2y},\]

\[Q_x=\frac{1}{x+3z},\qquad P_y=\frac{2}{2y+3z}.\]

Therefore

\[\nabla\times\mathbf{F} =\left(\frac{2}{x+2y}-\frac{3}{x+3z}\right)\mathbf{i} +\left(\frac{3}{2y+3z}-\frac{1}{x+2y}\right)\mathbf{j} +\left(\frac{1}{x+3z}-\frac{2}{2y+3z}\right)\mathbf{k}.\]

For the divergence, each component is independent of its matching variable:

\[P_x=0,\qquad Q_y=0,\qquad R_z=0.\]

Hence

\[\nabla\cdot\mathbf{F}=0.\]

b) Now

\[P=\arctan(xy),\qquad Q=\arctan(yz),\qquad R=\arctan(xz).\]

The curl derivatives are

\[R_y=0,\qquad Q_z=\frac{y}{1+y^2z^2},\]

\[P_z=0,\qquad R_x=\frac{z}{1+x^2z^2},\]

\[Q_x=0,\qquad P_y=\frac{x}{1+x^2y^2}.\]

Thus

\[\nabla\times\mathbf{F} =-\frac{y}{1+y^2z^2}\mathbf{i} -\frac{z}{1+x^2z^2}\mathbf{j} -\frac{x}{1+x^2y^2}\mathbf{k}.\]

For the divergence,

\[P_x=\frac{y}{1+x^2y^2},\qquad Q_y=\frac{z}{1+y^2z^2},\qquad R_z=\frac{x}{1+x^2z^2}.\]

Therefore

\[\nabla\cdot\mathbf{F} =\frac{y}{1+x^2y^2}+\frac{z}{1+y^2z^2}+\frac{x}{1+x^2z^2}.\]

4.42. Determine Whether Vector Fields Are Conservative (Lab 14, Task 2)

Determine whether each vector field is conservative. If it is conservative, find a potential function \(f\) such that \(\mathbf{F}=\nabla f\).

a) \(\mathbf{F}(x,y)=x\mathbf{i}-y\mathbf{j}\).

b) \(\mathbf{F}(x,y)=(2x^3y^4+x)\mathbf{i}+(2x^4y^3+y)\mathbf{j}\).

c) \(\mathbf{F}(x,y,z)=xyz^4\mathbf{i}+x^2z^4\mathbf{j}+4x^2yz^3\mathbf{k}\).

d) \(\mathbf{F}(x,y,z)=z\cos y\,\mathbf{i}+xz\sin y\,\mathbf{j}+x\cos y\,\mathbf{k}\).

Click to see the solution

Key Concept: In the plane, a continuously differentiable field \(P\mathbf{i}+Q\mathbf{j}\) on a simply connected domain is conservative exactly when \(P_y=Q_x\). In space, use the curl test: a nonzero curl means the field is not conservative.

a) Here \(P=x\) and \(Q=-y\). Then

\[P_y=0,\qquad Q_x=0.\]

The domain is all of \(\mathbb{R}^2\), so the field is conservative. Find \(f\) from

\[f_x=x,\qquad f_y=-y.\]

Integrating \(f_x=x\) with respect to \(x\) gives

\[f=\frac{x^2}{2}+g(y).\]

Then

\[f_y=g'(y)=-y,\]

\[g(y)=-\frac{y^2}{2}+C.\]

Thus

\[f(x,y)=\frac{x^2}{2}-\frac{y^2}{2}+C.\]

b) Let

\[P=2x^3y^4+x,\qquad Q=2x^4y^3+y.\]

Then

\[P_y=8x^3y^3,\qquad Q_x=8x^3y^3.\]

So the field is conservative on \(\mathbb{R}^2\). Integrate \(P\) with respect to \(x\):

\[f=\int(2x^3y^4+x)\,dx=\frac{x^4y^4}{2}+\frac{x^2}{2}+g(y).\]

Differentiate with respect to \(y\):

\[f_y=2x^4y^3+g'(y).\]

Compare with \(Q=2x^4y^3+y\):

\[g'(y)=y,\qquad g(y)=\frac{y^2}{2}+C.\]

Therefore

\[f(x,y)=\frac{x^4y^4}{2}+\frac{x^2}{2}+\frac{y^2}{2}+C.\]

c) Let

\[P=xyz^4,\qquad Q=x^2z^4,\qquad R=4x^2yz^3.\]

Already one compatibility condition fails:

\[P_y=xz^4,\qquad Q_x=2xz^4.\]

Since \(P_y\ne Q_x\) in general, \(\nabla\times\mathbf{F}\ne\mathbf{0}\), and the field is not conservative.

d) Let

\[P=z\cos y,\qquad Q=xz\sin y,\qquad R=x\cos y.\]

Again, check one curl component:

\[P_y=-z\sin y,\qquad Q_x=z\sin y.\]

These are not equal in general. Hence the curl is not zero, so the field is not conservative.

Answer: a) conservative, \(f=\frac{x^2-y^2}{2}+C\); b) conservative, \(f=\frac{x^4y^4+x^2+y^2}{2}+C\); c) not conservative; d) not conservative.

4.43. Scalar Line Integral Over a Line Segment (Lab 14, Task 3)

Calculate

\[\int_C (x-y+z-2)\,ds,\]

where \(C\) is the straight-line segment from \((0,1,1)\) to \((1,0,1)\).

Click to see the solution

Key Concept: For a scalar line integral, parametrize the curve and use \(ds=|\mathbf{r}'(t)|\,dt\).

A convenient parametrization of the line segment is

\[\mathbf{r}(t)=\langle t,1-t,1\rangle,\qquad 0\le t\le1.\]

Then

\[\mathbf{r}'(t)=\langle1,-1,0\rangle,\qquad |\mathbf{r}'(t)|=\sqrt2.\]

Along the curve,

\[x-y+z-2=t-(1-t)+1-2=2t-2.\]

Therefore

\[\int_C(x-y+z-2)\,ds =\int_0^1(2t-2)\sqrt2\,dt.\]

Compute:

\[\sqrt2\left[t^2-2t\right]_0^1=-\sqrt2.\]

Answer: \(\displaystyle -\sqrt2\).

4.44. Scalar Line Integral Along a Helix (Lab 14, Task 4)

Calculate

\[\int_C\sqrt{x^2+y^2}\,ds\]

along the curve

\[\mathbf{r}(t)=4\cos t\,\mathbf{i}+4\sin t\,\mathbf{j}+3t\,\mathbf{k},\qquad -2\pi\le t\le2\pi.\]

Click to see the solution

Key Concept: Along a parametrized curve, substitute the parametrization into the integrand and multiply by the speed.

Here

\[x=4\cos t,\qquad y=4\sin t,\]

\[\sqrt{x^2+y^2}=\sqrt{16\cos^2t+16\sin^2t}=4.\]

Also

\[\mathbf{r}'(t)=\langle-4\sin t,4\cos t,3\rangle,\]

and therefore

\[|\mathbf{r}'(t)|=\sqrt{16\sin^2t+16\cos^2t+9}=5.\]

Thus

\[\int_C\sqrt{x^2+y^2}\,ds =\int_{-2\pi}^{2\pi}4\cdot5\,dt =20(4\pi)=80\pi.\]

Answer: \(\displaystyle 80\pi\).

4.45. Vector Line Integral Along Three Paths (Lab 14, Task 5)

Find the line integral of

\[\mathbf{F}=(3x^2-3x)\mathbf{i}+3z\mathbf{j}+\mathbf{k}\]

from \((0,0,0)\) to \((1,1,1)\) over each path:

a) \(C_1:\mathbf{r}(t)=t\mathbf{i}+t\mathbf{j}+t\mathbf{k}\), \(0\le t\le1\).

b) \(C_2:\mathbf{r}(t)=t\mathbf{i}+t^2\mathbf{j}+t^4\mathbf{k}\), \(0\le t\le1\).

c) \(C_3\cup C_4\), where \(C_3\) is the segment from \((0,0,0)\) to \((1,1,0)\) and \(C_4\) is the segment from \((1,1,0)\) to \((1,1,1)\).

Click to see the solution

Key Concept: For a vector line integral,

\[\int_C\mathbf{F}\cdot d\mathbf{r}=\int_a^b\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt.\]

a) For \(C_1\), \(\mathbf{r}(t)=\langle t,t,t\rangle\) and \(\mathbf{r}'(t)=\langle1,1,1\rangle\). Then

\[\mathbf{F}(\mathbf{r}(t))=\langle3t^2-3t,3t,1\rangle.\]

\[\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=3t^2-3t+3t+1=3t^2+1.\]

Hence

\[\int_{C_1}\mathbf{F}\cdot d\mathbf{r} =\int_0^1(3t^2+1)\,dt=2.\]

b) For \(C_2\), \(\mathbf{r}(t)=\langle t,t^2,t^4\rangle\) and \(\mathbf{r}'(t)=\langle1,2t,4t^3\rangle\). Then

\[\mathbf{F}(\mathbf{r}(t))=\langle3t^2-3t,3t^4,1\rangle.\]

The dot product is

\[3t^2-3t+6t^5+4t^3.\]

Therefore

\[\int_{C_2}\mathbf{F}\cdot d\mathbf{r} =\int_0^1(3t^2-3t+4t^3+6t^5)\,dt.\]

Compute term by term:

\[1-\frac32+1+1=\frac32.\]

c) First parametrize \(C_3\) by \(\mathbf{r}_3(t)=\langle t,t,0\rangle\), \(0\le t\le1\). Then \(\mathbf{r}_3'(t)=\langle1,1,0\rangle\), and

\[\mathbf{F}(\mathbf{r}_3(t))=\langle3t^2-3t,0,1\rangle.\]

Thus

\[\int_{C_3}\mathbf{F}\cdot d\mathbf{r} =\int_0^1(3t^2-3t)\,dt =-\frac12.\]

Next parametrize \(C_4\) by \(\mathbf{r}_4(t)=\langle1,1,t\rangle\), \(0\le t\le1\). Then \(\mathbf{r}_4'(t)=\langle0,0,1\rangle\), and

\[\mathbf{F}(\mathbf{r}_4(t))=\langle0,3t,1\rangle.\]

\[\int_{C_4}\mathbf{F}\cdot d\mathbf{r}=\int_0^1 1\,dt=1.\]

Therefore

\[\int_{C_3\cup C_4}\mathbf{F}\cdot d\mathbf{r}=-\frac12+1=\frac12.\]

Answer: a) \(2\); b) \(\frac32\); c) \(\frac12\).

4.46. Circulation and Flux of \(\mathbf{F}=x\mathbf{i}+y\mathbf{j}\) (Lab 14, Task 6)

Find the circulation and flux of \(\mathbf{F}=x\mathbf{i}+y\mathbf{j}\) around and across each curve:

a) The circle \(\mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}\), \(0\le t\le2\pi\).

b) The ellipse \(\mathbf{r}(t)=\cos t\,\mathbf{i}+4\sin t\,\mathbf{j}\), \(0\le t\le2\pi\).

Click to see the solution

Key Concept: Green’s theorem gives circulation from \(Q_x-P_y\) and outward flux from \(P_x+Q_y\).

Here \(P=x\) and \(Q=y\). Therefore

\[Q_x-P_y=0-0=0,\]

so the circulation around any positively oriented simple closed curve is \(0\).

For flux,

\[P_x+Q_y=1+1=2.\]

Thus the outward flux equals

\[\iint_D2\,dA=2\operatorname{Area}(D).\]

a) The unit circle has area \(\pi\), so

\[\text{flux}=2\pi.\]

b) The ellipse has semiaxes \(1\) and \(4\), so its area is \(4\pi\). Therefore

\[\text{flux}=8\pi.\]

Answer: a) circulation \(0\), flux \(2\pi\); b) circulation \(0\), flux \(8\pi\).

4.47. Potential Function and Line Integral by the Fundamental Theorem (Lab 14, Task 7)

Find a function \(f\) such that \(\mathbf{F}=\nabla f\), and use it to evaluate \(\int_C\mathbf{F}\cdot d\mathbf{r}\) for

\[\mathbf{F}(x,y,z)=(y^2z+2xz^2)\mathbf{i}+2xyz\mathbf{j}+(xy^2+2x^2z)\mathbf{k},\]

where

\[C:\quad x=\sqrt t,\qquad y=t+1,\qquad z=t^2,\qquad 0\le t\le1.\]

Click to see the solution

Key Concept: If \(\mathbf{F}=\nabla f\), then \(\int_C\mathbf{F}\cdot d\mathbf{r}=f(B)-f(A)\).

Let

\[P=y^2z+2xz^2,\qquad Q=2xyz,\qquad R=xy^2+2x^2z.\]

Start with \(f_x=P\):

\[f=\int(y^2z+2xz^2)\,dx=xy^2z+x^2z^2+g(y,z).\]

Differentiate this expression with respect to \(y\):

\[f_y=2xyz+g_y(y,z).\]

Since \(f_y\) must equal \(Q=2xyz\), we get \(g_y=0\). Differentiate with respect to \(z\):

\[f_z=xy^2+2x^2z+g_z(y,z).\]

Since \(f_z\) must equal \(R=xy^2+2x^2z\), we get \(g_z=0\). Thus one potential is

\[f(x,y,z)=xy^2z+x^2z^2.\]

The curve starts at

\[A=(0,1,0)\]

and ends at

\[B=(1,2,1).\]

Therefore

\[\int_C\mathbf{F}\cdot d\mathbf{r}=f(1,2,1)-f(0,1,0).\]

Compute:

\[f(1,2,1)=1\cdot4\cdot1+1^2\cdot1^2=5,\qquad f(0,1,0)=0.\]

\[\int_C\mathbf{F}\cdot d\mathbf{r}=5.\]

Answer: \(\displaystyle f=xy^2z+x^2z^2+C\) and \(\displaystyle \int_C\mathbf{F}\cdot d\mathbf{r}=5\).

4.48. Green’s Theorem on a Rectangle (Lab 15, Task 1)

Evaluate

\[\int_C ye^x\,dx+2e^x\,dy,\]

where \(C\) is the positively oriented rectangle with vertices \((0,0)\), \((3,0)\), \((3,4)\), and \((0,4)\).

Click to see the solution

Key Concept: Green’s theorem converts \(\oint_C P\,dx+Q\,dy\) into a double integral of \(Q_x-P_y\) over the enclosed region.

Here

\[P=ye^x,\qquad Q=2e^x.\]

Then

\[Q_x=2e^x,\qquad P_y=e^x.\]

\[Q_x-P_y=e^x.\]

The rectangle is \(0\le x\le3\), \(0\le y\le4\). Hence

\[\oint_C ye^x\,dx+2e^x\,dy =\int_0^3\int_0^4 e^x\,dy\,dx.\]

Integrating first in \(y\) gives

\[\int_0^3 4e^x\,dx=4(e^3-1).\]

Answer: \(\displaystyle 4(e^3-1)\).

4.49. Green’s Theorem on an Ellipse With Odd Symmetry (Lab 15, Task 2)

Evaluate

\[\int_C y^4\,dx+2xy^3\,dy,\]

where \(C\) is the positively oriented ellipse \(x^2+2y^2=2\).

Click to see the solution

Key Concept: Green’s theorem often exposes symmetry that is hidden in the boundary integral.

Let

\[P=y^4,\qquad Q=2xy^3.\]

Then

\[Q_x=2y^3,\qquad P_y=4y^3.\]

Therefore

\[Q_x-P_y=-2y^3.\]

The ellipse is symmetric about the \(x\)-axis. The function \(-2y^3\) is odd in \(y\), so its integral over this symmetric region is zero:

\[\iint_D -2y^3\,dA=0.\]

Answer: \(\displaystyle 0\).

4.50. Work Around a Triangular Path by Green’s Theorem (Lab 15, Task 3)

Use Green’s theorem to find the work done by

\[\mathbf{F}=x(x+y)\hat{i}+xy^2\hat{j}\]

moving a particle counterclockwise along the closed triangle

\[ (0,0)\to(1,0)\to(0,1)\to(0,0).\]

Click to see the solution

Key Concept: Work around a closed plane curve is the circulation \(\oint_C P\,dx+Q\,dy\).

Here

\[P=x(x+y)=x^2+xy,\qquad Q=xy^2.\]

By Green’s theorem,

\[\oint_C P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA.\]

Compute the derivatives:

\[Q_x=y^2,\qquad P_y=x.\]

So the integrand is

\[y^2-x.\]

The triangle is

\[0\le y\le1,\qquad 0\le x\le1-y.\]

Thus

\[W=\int_0^1\int_0^{1-y}(y^2-x)\,dx\,dy.\]

Separate the two pieces:

\[\iint_D y^2\,dA=\int_0^1y^2(1-y)\,dy=\frac13-\frac14=\frac1{12},\]

and

\[\iint_D x\,dA=\int_0^1\frac{(1-y)^2}{2}\,dy=\frac16.\]

Therefore

\[W=\frac1{12}-\frac16=-\frac1{12}.\]

Answer: \(\displaystyle -\frac{1}{12}\).

4.51. Surface Integral Over a Parabolic Cylinder (Lab 15, Task 4)

Integrate \(G(x,y,z)=x\) over the parabolic cylinder

\[y=x^2,\qquad 0\le x\le2,\qquad 0\le z\le3.\]

Click to see the solution

Key Concept: Parametrize the surface and use \(d\sigma=|\mathbf{r}_u\times\mathbf{r}_v|\,du\,dv\).

Use the parametrization

\[\mathbf{r}(x,z)=\langle x,x^2,z\rangle,\qquad 0\le x\le2,\quad 0\le z\le3.\]

Then

\[\mathbf{r}_x=\langle1,2x,0\rangle,\qquad \mathbf{r}_z=\langle0,0,1\rangle.\]

Their cross product has magnitude

\[|\mathbf{r}_x\times\mathbf{r}_z|=\sqrt{1+4x^2}.\]

Since \(G=x\) on the surface,

\[\iint_S G\,d\sigma=\int_0^3\int_0^2x\sqrt{1+4x^2}\,dx\,dz.\]

Use \(u=1+4x^2\), so \(du=8x\,dx\):

\[\int_0^2x\sqrt{1+4x^2}\,dx =\frac18\int_1^{17}u^{1/2}\,du =\frac{1}{12}\left(17^{3/2}-1\right).\]

Multiplying by the \(z\)-length \(3\) gives

\[\iint_SG\,d\sigma=\frac14(17^{3/2}-1)=\frac{17\sqrt{17}-1}{4}.\]

Answer: \(\displaystyle \frac{17\sqrt{17}-1}{4}\).

4.52. Surface Integral of \(x^2\) Over the Unit Sphere (Lab 15, Task 5)

Integrate \(G(x,y,z)=x^2\) over the unit sphere

\[x^2+y^2+z^2=1.\]

Click to see the solution

Key Concept: Use symmetry on the sphere: the integrals of \(x^2\), \(y^2\), and \(z^2\) over the full sphere are equal.

On the unit sphere,

\[x^2+y^2+z^2=1.\]

By symmetry,

\[\iint_Sx^2\,d\sigma=\iint_Sy^2\,d\sigma=\iint_Sz^2\,d\sigma.\]

Adding these three equal integrals gives

\[3\iint_Sx^2\,d\sigma=\iint_S(x^2+y^2+z^2)\,d\sigma=\iint_S1\,d\sigma.\]

The surface area of the unit sphere is \(4\pi\), so

\[3\iint_Sx^2\,d\sigma=4\pi.\]

Therefore

\[\iint_Sx^2\,d\sigma=\frac{4\pi}{3}.\]

Answer: \(\displaystyle \frac{4\pi}{3}\).

4.53. Surface Integral Over a First-Octant Plane (Lab 15, Task 6)

Integrate \(G(x,y,z)=x+y+z\) over the portion of the plane

\[2x+2y+z=2\]

that lies in the first octant.

Click to see the solution

Key Concept: Write the plane as a graph \(z=f(x,y)\) over its triangular projection.

Solve for \(z\):

\[z=2-2x-2y.\]

In the first octant, \(z\ge0\), so

\[x\ge0,\qquad y\ge0,\qquad x+y\le1.\]

The surface element for \(z=f(x,y)\) is

\[d\sigma=\sqrt{1+f_x^2+f_y^2}\,dA.\]

Here \(f_x=-2\) and \(f_y=-2\), so

\[d\sigma=\sqrt{1+4+4}\,dA=3\,dA.\]

On the plane,

\[x+y+z=x+y+2-2x-2y=2-x-y.\]

Therefore

\[\iint_SG\,d\sigma=3\iint_R(2-x-y)\,dA,\]

where \(R\) is the triangle \(0\le y\le1\), \(0\le x\le1-y\). Now

\[\iint_R2\,dA=1,\qquad \iint_Rx\,dA=\iint_Ry\,dA=\frac16.\]

Thus

\[\iint_SG\,d\sigma=3\left(1-\frac16-\frac16\right)=2.\]

Answer: \(\displaystyle 2\).

4.54. Flux Through a Parabolic Cylinder (Lab 15, Task 7)

Find the flux of

\[\mathbf{F}=z^2\hat{i}+x\hat{j}-3z\hat{k}\]

through the surface cut from the parabolic cylinder \(z=4-y^2\) by the planes \(x=0\), \(x=1\), and \(z=0\), directed away from the \(x\)-axis.

Click to see the solution

Key Concept: For an oriented parametrized surface, flux is \(\iint \mathbf{F}(\mathbf{r})\cdot(\mathbf{r}_u\times\mathbf{r}_v)\,du\,dv\) with the cross product chosen in the required direction.

The surface can be parametrized by

\[\mathbf{r}(x,y)=\langle x,y,4-y^2\rangle.\]

Since \(z\ge0\), we have

\[-2\le y\le2,\qquad 0\le x\le1.\]

Compute

\[\mathbf{r}_x=\langle1,0,0\rangle,\qquad \mathbf{r}_y=\langle0,1,-2y\rangle.\]

Choose

\[\mathbf{r}_x\times\mathbf{r}_y=\langle0,2y,1\rangle.\]

This normal points away from the \(x\)-axis because its projection in the \(yz\)-plane points outward from the axis.

On the surface,

\[\mathbf{F}(\mathbf{r}(x,y))=\langle(4-y^2)^2,x,-3(4-y^2)\rangle.\]

Therefore

\[\mathbf{F}\cdot(\mathbf{r}_x\times\mathbf{r}_y) =2xy-3(4-y^2)=2xy-12+3y^2.\]

The flux is

\[\int_{-2}^{2}\int_0^1(2xy-12+3y^2)\,dx\,dy.\]

Integrate in \(x\):

\[\int_0^1(2xy-12+3y^2)\,dx=y-12+3y^2.\]

The odd term \(y\) integrates to zero over \([-2,2]\), so

\[\text{Flux}=\int_{-2}^{2}(-12+3y^2)\,dy=-48+16=-32.\]

Answer: \(\displaystyle -32\).

4.55. Flux of \(z\hat{k}\) Through a First-Octant Sphere (Lab 15, Task 8)

Find the flux of \(\mathbf{F}=z\hat{k}\) across the portion of the sphere

\[x^2+y^2+z^2=a^2\]

in the first octant, directed away from the origin.

Click to see the solution

Key Concept: On a sphere of radius \(a\), the outward unit normal is \(\mathbf{n}=\langle x,y,z\rangle/a\).

Since \(\mathbf{F}=\langle0,0,z\rangle\),

\[\mathbf{F}\cdot\mathbf{n}=\frac{z^2}{a}.\]

Use spherical coordinates on the sphere:

\[z=a\cos\phi,\qquad d\sigma=a^2\sin\phi\,d\phi\,d\theta.\]

The first octant corresponds to

\[0\le\theta\le\frac{\pi}{2},\qquad 0\le\phi\le\frac{\pi}{2}.\]

Therefore

\[\text{Flux} =\int_0^{\pi/2}\int_0^{\pi/2}\frac{a^2\cos^2\phi}{a}\cdot a^2\sin\phi\,d\phi\,d\theta.\]

This simplifies to

\[a^3\int_0^{\pi/2}d\theta\int_0^{\pi/2}\cos^2\phi\sin\phi\,d\phi.\]

The angular integrals are

\[\int_0^{\pi/2}d\theta=\frac{\pi}{2},\qquad \int_0^{\pi/2}\cos^2\phi\sin\phi\,d\phi=\frac13.\]

Hence

\[\text{Flux}=\frac{\pi a^3}{6}.\]

Answer: \(\displaystyle \frac{\pi a^3}{6}\).

4.56. Outward Flux Through a Closed Paraboloid Surface (Lab 15, Task 9)

Find the flux of

\[\mathbf{F}=x\hat{i}+y\hat{j}+z\hat{k}\]

outward through the closed surface consisting of the paraboloid

\[z=x^2+y^2,\qquad 0\le z\le1,\]

and the disk

\[x^2+y^2\le1,\qquad z=1.\]

Click to see the solution

Key Concept: Because the surface is closed, the divergence theorem is the fastest method.

The divergence is

\[\nabla\cdot\mathbf{F}=1+1+1=3.\]

The enclosed solid is

\[x^2+y^2\le z\le1.\]

In cylindrical coordinates,

\[0\le r\le1,\qquad 0\le\theta\le2\pi,\qquad r^2\le z\le1.\]

Its volume is

\[V=\int_0^{2\pi}\int_0^1(1-r^2)r\,dr\,d\theta.\]

Compute:

\[V=2\pi\left[\frac{r^2}{2}-\frac{r^4}{4}\right]_0^1=\frac{\pi}{2}.\]

By the divergence theorem,

\[\text{Flux}=\iiint_G3\,dV=3V=\frac{3\pi}{2}.\]

Answer: \(\displaystyle \frac{3\pi}{2}\).

4.57. Circulation and Flux Around a Square (Lab 15, Homework 1)

Find the counterclockwise circulation and the outward flux of

\[\mathbf{F}=(-\sin y)\hat{i}+(x\cos y)\hat{j}\]

around and across the boundary of the square

\[0\le x\le\frac{\pi}{2},\qquad 0\le y\le\frac{\pi}{2}.\]

Click to see the solution

Key Concept: For \(\mathbf{F}=P\mathbf{i}+Q\mathbf{j}\), Green’s theorem gives

\[\text{circulation}=\iint_D(Q_x-P_y)\,dA,\]

and the flux form gives

\[\text{outward flux}=\iint_D(P_x+Q_y)\,dA.\]

Here

\[P=-\sin y,\qquad Q=x\cos y.\]

For circulation,

\[Q_x=\cos y,\qquad P_y=-\cos y,\]

\[Q_x-P_y=2\cos y.\]

Thus

\[\text{circulation} =\int_0^{\pi/2}\int_0^{\pi/2}2\cos y\,dx\,dy =\frac{\pi}{2}\cdot2\int_0^{\pi/2}\cos y\,dy =\pi.\]

For flux,

\[P_x=0,\qquad Q_y=-x\sin y.\]

Therefore

\[\text{flux} =\int_0^{\pi/2}\int_0^{\pi/2}-x\sin y\,dx\,dy.\]

Separate the variables:

\[\text{flux} =-\left(\int_0^{\pi/2}x\,dx\right)\left(\int_0^{\pi/2}\sin y\,dy\right) =-\frac{\pi^2}{8}\cdot1 =-\frac{\pi^2}{8}.\]

Answer: circulation \(\displaystyle \pi\); outward flux \(\displaystyle -\frac{\pi^2}{8}\).

4.58. Surface Integral Over a Spherical Cap Above a Cone (Lab 15, Homework 2)

Integrate \(G(x,y,z)=yz\) over the part of the sphere

\[x^2+y^2+z^2=4\]

that lies above the cone

\[z=\sqrt{x^2+y^2}.\]

Click to see the solution

Key Concept: Symmetry can make a surface integral zero even before doing the full calculation.

The sphere has radius \(2\). The cone \(z=\sqrt{x^2+y^2}\) corresponds to \(\phi=\pi/4\) in spherical coordinates. The part above the cone has

\[0\le\phi\le\frac{\pi}{4},\qquad 0\le\theta\le2\pi.\]

On the sphere,

\[y=2\sin\phi\sin\theta,\qquad z=2\cos\phi.\]

Thus

\[yz=4\sin\phi\cos\phi\sin\theta.\]

The surface element on the sphere is

\[d\sigma=4\sin\phi\,d\phi\,d\theta.\]

So the integrand contains a factor \(\sin\theta\):

\[yz\,d\sigma=16\sin^2\phi\cos\phi\sin\theta\,d\phi\,d\theta.\]

Since

\[\int_0^{2\pi}\sin\theta\,d\theta=0,\]

the whole surface integral is zero.

Answer: \(\displaystyle 0\).

4.59. Upward Flux Through a Plane Above a Square (Lab 15, Homework 3)

Find the flux of

\[\mathbf{F}=2xy\hat{i}+2yz\hat{j}+2xz\hat{k}\]

upward across the portion of the plane

\[x+y+z=2a\]

that lies above the square

\[0\le x\le a,\qquad 0\le y\le a.\]

Click to see the solution

Key Concept: For a graph \(z=f(x,y)\) with upward orientation, use the vector surface element \(\langle-f_x,-f_y,1\rangle\,dA\).

Solve the plane equation for \(z\):

\[z=2a-x-y.\]

Then

\[f_x=-1,\qquad f_y=-1,\]

so the upward vector surface element is

\[\langle-f_x,-f_y,1\rangle\,dA=\langle1,1,1\rangle\,dA.\]

On the plane,

\[\mathbf{F}=\langle2xy,2y(2a-x-y),2x(2a-x-y)\rangle.\]

Therefore

\[\mathbf{F}\cdot\langle1,1,1\rangle =2xy+2y(2a-x-y)+2x(2a-x-y).\]

Simplify:

\[2xy+4ay-2xy-2y^2+4ax-2x^2-2xy =4a(x+y)-2x^2-2y^2-2xy.\]

The flux is

\[\int_0^a\int_0^a\left(4a(x+y)-2x^2-2y^2-2xy\right)\,dx\,dy.\]

Compute the terms:

\[\iint 4a(x+y)\,dA=4a\left(\frac{a^3}{2}+\frac{a^3}{2}\right)=4a^4,\]

\[\iint 2x^2\,dA=\frac{2a^4}{3},\qquad \iint 2y^2\,dA=\frac{2a^4}{3},\]

and

\[\iint 2xy\,dA=2\left(\frac{a^2}{2}\right)^2=\frac{a^4}{2}.\]

Therefore

\[\text{Flux}=4a^4-\frac{2a^4}{3}-\frac{2a^4}{3}-\frac{a^4}{2} =\frac{13a^4}{6}.\]

Answer: \(\displaystyle \frac{13a^4}{6}\).

4.60. Interpret Basic Vector Fields (Chapter 4, Illustrative Examples)

The source gives three basic examples of vector fields:

The tangent vectors \(\mathbf{T}\) and normal vectors \(\mathbf{N}\) along a space curve \(\mathbf{r}(t)=f(t)\hat{i}+g(t)\hat{j}+h(t)\hat{k}\).
The gradient field \(\nabla f\) attached to points of a level surface of a scalar function \(f(x,y,z)\).
The velocity field of a flowing fluid.

Click to see the solution

Key Concept: A vector field assigns a vector to each point of its domain. The domain may be a region in space, a surface, or only a curve.

1. Tangent and normal fields along a curve. If a particle moves along a curve \(\mathbf{r}(t)\), then its tangent direction is determined by \(\mathbf{r}'(t)\). Where \(\mathbf{r}'(t)\ne\mathbf{0}\), the unit tangent vector is

\[\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}.\]

As \(t\) changes, each point of the curve receives a tangent vector. That assignment is a vector field along the curve. When the curve bends smoothly and \(\mathbf{T}'(t)\ne\mathbf{0}\), the unit normal vector

\[\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|}\]

also assigns a direction to each point of the curve. This is not usually a field on all of space; it is a field whose domain is the curve itself.

2. Gradient field on level surfaces. If \(f(x,y,z)\) is differentiable, then

\[\nabla f=f_x\hat{i}+f_y\hat{j}+f_z\hat{k}.\]

At a point on a level surface \(f(x,y,z)=c\), the gradient is perpendicular to that level surface, provided \(\nabla f\ne\mathbf{0}\). Therefore attaching \(\nabla f\) to points of the surface gives a normal vector field on the surface.

3. Velocity field of a fluid. In a fluid, the vector attached to a point is the instantaneous velocity of the small particle of fluid passing through that point. Its direction gives the local direction of motion, and its magnitude gives speed. This is why vector fields are the natural language for flow, circulation, and flux.

Answer: These are all vector fields because each construction attaches a vector to each point of a chosen domain: a curve, a level surface, or a spatial flow region.

4.61. Check a Conservative Field and Its Potential (Chapter 4, Example 1)

The field

\[\mathbf{F} = (e^x \cos y + yz)\hat{i} + (xz - e^x \sin y)\hat{j} + (xy + z)\hat{k}\]

is conservative over its natural domain, and

\[f(x, y, z) = e^x \cos y + xyz + \frac{z^2}{2}\]

is a potential function for it. Check why.

Click to see the solution

Key Concept: To check that \(f\) is a potential for \(\mathbf{F}\), compute \(\nabla f\) and compare it with \(\mathbf{F}\).

Differentiate \(f\) with respect to \(x\):

\[f_x=\frac{\partial}{\partial x}\left(e^x\cos y+xyz+\frac{z^2}{2}\right)=e^x\cos y+yz.\]

Differentiate with respect to \(y\):

\[f_y=\frac{\partial}{\partial y}\left(e^x\cos y+xyz+\frac{z^2}{2}\right)=-e^x\sin y+xz.\]

Differentiate with respect to \(z\):

\[f_z=\frac{\partial}{\partial z}\left(e^x\cos y+xyz+\frac{z^2}{2}\right)=xy+z.\]

Thus

\[\nabla f=(e^x\cos y+yz)\hat{i}+(xz-e^x\sin y)\hat{j}+(xy+z)\hat{k}=\mathbf{F}.\]

Since \(\mathbf{F}=\nabla f\), the field is conservative.

Answer: \(\mathbf{F}\) is conservative, with potential \(\displaystyle f=e^x\cos y+xyz+\frac{z^2}{2}+C\).

4.62. Conservative Polynomial Vector Field (Chapter 4, Example 2)

Show that \(\mathbf{F}(x, y, z) = y^2 z^3 \hat{i} + 2xyz^3 \hat{j} + 3xy^2 z^2 \hat{k}\) is conservative and find a potential function.

Click to see the solution

Key Concept: Either check the curl or directly recognize the components as partial derivatives of one scalar function.

We want \(f_x=y^2z^3\). Integrating with respect to \(x\) gives

\[f=xy^2z^3+\phi(y,z),\]

where \(\phi\) is independent of \(x\).

Differentiate this expression with respect to \(y\):

\[f_y=2xyz^3+\phi_y(y,z).\]

The field requires \(f_y=2xyz^3\), so

\[\phi_y(y,z)=0.\]

Therefore \(\phi\) depends only on \(z\): \(\phi(y,z)=\psi(z)\).

Now differentiate with respect to \(z\):

\[f_z=3xy^2z^2+\psi'(z).\]

The field requires \(f_z=3xy^2z^2\), so

\[\psi'(z)=0.\]

Thus \(\psi\) is constant.

Therefore a potential function is

\[f(x,y,z)=xy^2z^3+C.\]

Answer: \(\mathbf{F}\) is conservative, and \(\displaystyle f=xy^2z^3+C\) is a potential.

4.63. Divergence of a Polynomial Vector Field (Chapter 4, Example 3)

For the field \(\mathbf{F}(x, y, z) = y^2 z^3 \hat{i} + 2xyz^3 \hat{j} + 3xy^2 z^2 \hat{k}\), compute \(\operatorname{div}\mathbf{F}\).

Click to see the solution

Key Concept: Divergence differentiates each component with respect to its own coordinate.

Here

\[M=y^2z^3,\qquad N=2xyz^3,\qquad P=3xy^2z^2.\]

Therefore

\[\operatorname{div}\mathbf{F}=M_x+N_y+P_z.\]

Compute:

\[M_x=0,\]

because \(M\) does not contain \(x\);

\[N_y=2xz^3,\]

and

\[P_z=6xy^2z.\]

Thus

\[\operatorname{div}\mathbf{F}=2xz^3+6xy^2z.\]

Answer: \(\displaystyle \operatorname{div}\mathbf{F}=2xz^3+6xy^2z\).

4.64. Curl and Divergence of Four Vector Fields (Chapter 4, Task 1)

Find the curl and the divergence of each vector field:

\(\mathbf{F} = xy^2 z^2 \hat{i} + yx^2 z^2 \hat{j} + zx^2 y^2 \hat{k}\).
\(\mathbf{F} = \dfrac{\sqrt{x}}{1 + z}\hat{i} + \dfrac{\sqrt{y}}{1 + x}\hat{j} + \dfrac{\sqrt{z}}{1 + y}\hat{k}\).
\(\mathbf{F} = \ln(2y + 3z)\hat{i} + \ln(x + 3z)\hat{j} + \ln(x + 2y)\hat{k}\).
\(\mathbf{F} = \arctan(xy)\hat{i} + \arctan(yz)\hat{j} + \arctan(xz)\hat{k}\).

Click to see the solution

Key Concept: For \(\mathbf{F}=M\hat{i}+N\hat{j}+P\hat{k}\),

\[\operatorname{curl}\mathbf{F}=(P_y-N_z)\hat{i}+(M_z-P_x)\hat{j}+(N_x-M_y)\hat{k},\]

and

\[\operatorname{div}\mathbf{F}=M_x+N_y+P_z.\]

1. Let

\[M=xy^2z^2,\qquad N=yx^2z^2,\qquad P=zx^2y^2.\]

Then

\[P_y=2x^2yz,\quad N_z=2x^2yz,\]

\[M_z=2xy^2z,\quad P_x=2xy^2z,\]

and

\[N_x=2xyz^2,\quad M_y=2xyz^2.\]

\[\operatorname{curl}\mathbf{F}=\mathbf{0}.\]

The divergence is

\[\operatorname{div}\mathbf{F}=y^2z^2+x^2z^2+x^2y^2.\]

2. Let

\[M=\frac{\sqrt{x}}{1+z},\qquad N=\frac{\sqrt y}{1+x},\qquad P=\frac{\sqrt z}{1+y}.\]

Then

\[P_y=-\frac{\sqrt z}{(1+y)^2},\qquad N_z=0,\]

\[M_z=-\frac{\sqrt x}{(1+z)^2},\qquad P_x=0,\]

and

\[N_x=-\frac{\sqrt y}{(1+x)^2},\qquad M_y=0.\]

Hence

\[\operatorname{curl}\mathbf{F} =-\frac{\sqrt z}{(1+y)^2}\hat{i} -\frac{\sqrt x}{(1+z)^2}\hat{j} -\frac{\sqrt y}{(1+x)^2}\hat{k}.\]

The divergence is

\[\operatorname{div}\mathbf{F} =\frac{1}{2\sqrt{x}(1+z)} +\frac{1}{2\sqrt{y}(1+x)} +\frac{1}{2\sqrt{z}(1+y)}.\]

3. Let

\[M=\ln(2y+3z),\quad N=\ln(x+3z),\quad P=\ln(x+2y).\]

Then

\[P_y=\frac{2}{x+2y},\qquad N_z=\frac{3}{x+3z},\]

\[M_z=\frac{3}{2y+3z},\qquad P_x=\frac{1}{x+2y},\]

and

\[N_x=\frac{1}{x+3z},\qquad M_y=\frac{2}{2y+3z}.\]

\[\operatorname{curl}\mathbf{F} =\left(\frac{2}{x+2y}-\frac{3}{x+3z}\right)\hat{i} +\left(\frac{3}{2y+3z}-\frac{1}{x+2y}\right)\hat{j} +\left(\frac{1}{x+3z}-\frac{2}{2y+3z}\right)\hat{k}.\]

Since each component is independent of its own matching variable,

\[\operatorname{div}\mathbf{F}=0+0+0=0.\]

4. Let

\[M=\arctan(xy),\quad N=\arctan(yz),\quad P=\arctan(xz).\]

Then

\[P_y=0,\qquad N_z=\frac{y}{1+y^2z^2},\]

\[M_z=0,\qquad P_x=\frac{z}{1+x^2z^2},\]

and

\[N_x=0,\qquad M_y=\frac{x}{1+x^2y^2}.\]

Therefore

\[\operatorname{curl}\mathbf{F} =-\frac{y}{1+y^2z^2}\hat{i} -\frac{z}{1+x^2z^2}\hat{j} -\frac{x}{1+x^2y^2}\hat{k}.\]

The divergence is

\[\operatorname{div}\mathbf{F} =\frac{y}{1+x^2y^2} +\frac{z}{1+y^2z^2} +\frac{x}{1+x^2z^2}.\]

4.65. Determine Whether Vector Fields Are Conservative (Chapter 4, Task 2)

Determine whether each vector field is conservative. If it is conservative, find \(f\) such that \(\mathbf{F}=\nabla f\).

\(\mathbf{F}(x, y, z) = xyz^4 \hat{i} + x^2 z^4 \hat{j} + 4x^2 yz^3 \hat{k}\).
\(\mathbf{F}(x, y, z) = z \cos y \hat{i} + xz \sin y \hat{j} + x \cos y \hat{k}\).
\(\mathbf{F}(x, y, z) = e^{yz}\hat{i} + xze^{yz}\hat{j} + xye^{yz}\hat{k}\).
\(\mathbf{F}(x, y, z) = e^x \sin yz \hat{i} + ze^x \cos yz \hat{j} + ye^x \cos yz \hat{k}\).

Click to see the solution

Key Concept: If the domain is simply connected, a continuously differentiable vector field is conservative exactly when its curl is zero. To find a potential, integrate one component and match the others.

1. Here

\[M=xyz^4,\qquad N=x^2z^4,\qquad P=4x^2yz^3.\]

Check one compatibility condition:

\[M_y=xz^4,\qquad N_x=2xz^4.\]

These are not equal in general. Therefore the field is not conservative.

2. Here

\[M=z\cos y,\qquad N=xz\sin y,\qquad P=x\cos y.\]

Check:

\[M_y=-z\sin y,\qquad N_x=z\sin y.\]

These are not equal in general. Therefore the field is not conservative.

3. Here

\[M=e^{yz},\qquad N=xze^{yz},\qquad P=xye^{yz}.\]

Integrate \(M\) with respect to \(x\):

\[f=xe^{yz}+\phi(y,z).\]

Then

\[f_y=xze^{yz}+\phi_y.\]

Since this must equal \(N=xze^{yz}\), we get \(\phi_y=0\). Also,

\[f_z=xye^{yz}+\phi_z.\]

Since this must equal \(P=xye^{yz}\), we get \(\phi_z=0\). Hence \(\phi\) is constant, and

\[f=xe^{yz}+C.\]

So the field is conservative.

4. Here

\[M=e^x\sin(yz),\qquad N=ze^x\cos(yz),\qquad P=ye^x\cos(yz).\]

Integrate \(M\) with respect to \(x\):

\[f=e^x\sin(yz)+\phi(y,z).\]

Then

\[f_y=ze^x\cos(yz)+\phi_y.\]

Matching \(N\) gives \(\phi_y=0\). Also,

\[f_z=ye^x\cos(yz)+\phi_z.\]

Matching \(P\) gives \(\phi_z=0\). Hence

\[f=e^x\sin(yz)+C.\]

So the field is conservative.

Answer: 1) not conservative; 2) not conservative; 3) conservative with \(\displaystyle f=xe^{yz}+C\); 4) conservative with \(\displaystyle f=e^x\sin(yz)+C\).

4.66. Prove Basic Vector Calculus Identities (Chapter 4, Task 3)

Let \(f\) be a scalar field, and let \(\mathbf{F}\) and \(\mathbf{G}\) be vector fields. Prove the standard linearity and product identities for divergence and curl.

Click to see the solution

Key Concept: The identities follow from component formulas and ordinary product rules for partial derivatives.

The source slide contains repeated \(\mathbf{G}\) terms in the first two formulas and writes the curl product rule incorrectly. The corrected standard identities are:

\[\operatorname{div}(\mathbf{F}+\mathbf{G})=\operatorname{div}\mathbf{F}+\operatorname{div}\mathbf{G},\]

\[\operatorname{curl}(\mathbf{F}+\mathbf{G})=\operatorname{curl}\mathbf{F}+\operatorname{curl}\mathbf{G},\]

\[\operatorname{div}(f\mathbf{F})=f\operatorname{div}\mathbf{F}+\mathbf{F}\cdot\nabla f,\]

and

\[\operatorname{curl}(f\mathbf{F})=\nabla f\times\mathbf{F}+f\operatorname{curl}\mathbf{F}.\]

Let

\[\mathbf{F}=\langle P,Q,R\rangle,\qquad \mathbf{G}=\langle A,B,C\rangle.\]

For divergence linearity,

\[\operatorname{div}(\mathbf{F}+\mathbf{G}) =(P+A)_x+(Q+B)_y+(R+C)_z.\]

Using linearity of partial derivatives,

\[\operatorname{div}(\mathbf{F}+\mathbf{G}) =P_x+A_x+Q_y+B_y+R_z+C_z =\operatorname{div}\mathbf{F}+\operatorname{div}\mathbf{G}.\]

Curl linearity is proved the same way by applying linearity to each component of

\[\nabla\times(\mathbf{F}+\mathbf{G}).\]

For the divergence product rule,

\[\operatorname{div}(f\mathbf{F})=(fP)_x+(fQ)_y+(fR)_z.\]

Apply the product rule:

\[\operatorname{div}(f\mathbf{F}) =f_xP+fP_x+f_yQ+fQ_y+f_zR+fR_z.\]

Group terms:

\[\operatorname{div}(f\mathbf{F}) =f(P_x+Q_y+R_z)+(P,Q,R)\cdot(f_x,f_y,f_z).\]

Thus

\[\operatorname{div}(f\mathbf{F})=f\operatorname{div}\mathbf{F}+\mathbf{F}\cdot\nabla f.\]

For the curl product rule, compute

\[\nabla\times(f\mathbf{F}) =((fR)_y-(fQ)_z,\ (fP)_z-(fR)_x,\ (fQ)_x-(fP)_y).\]

Expanding each component by the product rule gives one group equal to \(f\nabla\times\mathbf{F}\) and the remaining group equal to \(\nabla f\times\mathbf{F}\). Hence

\[\operatorname{curl}(f\mathbf{F})=\nabla f\times\mathbf{F}+f\operatorname{curl}\mathbf{F}.\]

Answer: The four corrected identities are proved componentwise using linearity and the product rule.

4.67. Scalar Line Integral Over an Upper Semicircle (Chapter 4, Example 4)

Calculate \(\displaystyle \int_C (2 + x^2 y)\,ds\), where \(C\) is the upper half of the unit circle \(x^2 + y^2 = 1\).

Click to see the solution

Key Concept: For a scalar line integral, parametrize the curve and use \(ds=\|\mathbf{r}'(t)\|\,dt\).

The upper half of the unit circle is

\[x=\cos t,\qquad y=\sin t,\qquad 0\le t\le\pi.\]

Then

\[\mathbf{r}'(t)=\langle-\sin t,\cos t\rangle,\]

\[ds=\sqrt{\sin^2t+\cos^2t}\,dt=dt.\]

Substitute into the integral:

\[\int_C(2+x^2y)\,ds =\int_0^\pi(2+\cos^2t\sin t)\,dt.\]

Compute the two terms separately:

\[\int_0^\pi2\,dt=2\pi.\]

For the second term, let \(u=\cos t\), so \(du=-\sin t\,dt\):

\[\int_0^\pi\cos^2t\sin t\,dt =-\int_1^{-1}u^2\,du =\int_{-1}^1u^2\,du =\frac{2}{3}.\]

Therefore

\[\int_C(2+x^2y)\,ds=2\pi+\frac23.\]

Answer: \(\displaystyle 2\pi+\frac23\).

4.68. Scalar Line Integral Along Two Different Paths (Chapter 4, Example 5)

Calculate \(\displaystyle \int_C (x - 3y^2 + z)\,ds\), where:

\(C\) is the line segment joining the origin to \((1,1,1)\).
\(C=C_1\cup C_2\), where \(C_1\) joins the origin to \((1,1,0)\) and \(C_2\) joins \((1,1,0)\) to \((1,1,1)\).

Click to see the solution

Key Concept: Scalar line integrals depend on the path because both the function values and the arc length element can change.

1. Direct segment. Parametrize the line by

\[\mathbf{r}(t)=\langle t,t,t\rangle,\qquad 0\le t\le1.\]

Then

\[ds=\|\mathbf{r}'(t)\|\,dt=\sqrt{1^2+1^2+1^2}\,dt=\sqrt3\,dt.\]

The integrand becomes

\[x-3y^2+z=t-3t^2+t=2t-3t^2.\]

Hence

\[\int_C(x-3y^2+z)\,ds =\sqrt3\int_0^1(2t-3t^2)\,dt =\sqrt3[t^2-t^3]_0^1=0.\]

2. Piecewise path. For \(C_1\) use

\[\mathbf{r}_1(t)=\langle t,t,0\rangle,\qquad 0\le t\le1,\]

\[ds=\sqrt2\,dt,\qquad x-3y^2+z=t-3t^2.\]

Thus

\[\int_{C_1}f\,ds=\sqrt2\int_0^1(t-3t^2)\,dt =\sqrt2\left(\frac12-1\right) =-\frac{\sqrt2}{2}.\]

For \(C_2\) use

\[\mathbf{r}_2(t)=\langle1,1,t\rangle,\qquad 0\le t\le1,\]

\[ds=dt,\qquad x-3y^2+z=1-3+t=t-2.\]

Thus

\[\int_{C_2}f\,ds=\int_0^1(t-2)\,dt=\frac12-2=-\frac32.\]

Therefore

\[\int_{C_1\cup C_2}f\,ds=-\frac{\sqrt2}{2}-\frac32=-\frac{\sqrt2+3}{2}.\]

Answer: Direct segment: \(0\); piecewise path: \(\displaystyle -\frac{\sqrt2+3}{2}\).

4.69. Line Integral with Respect to \(dx\), \(dy\), and \(dz\) (Chapter 4, Example 6)

Calculate \(I = \int_C y\,dx + z\,dy + x\,dz\), where \(C\) consists of the line segment \(C_1\) from \((2, 0, 0)\) to \((3, 4, 5)\), followed by the line segment \(C_2\) from \((3, 4, 5)\) to \((3, 4, 0)\).

Click to see the solution

Key Concept: For integrals with \(dx\), \(dy\), and \(dz\), substitute the parametrization and use \(dx=x'(t)dt\), \(dy=y'(t)dt\), \(dz=z'(t)dt\).

For \(C_1\), use

\[x=2+t,\qquad y=4t,\qquad z=5t,\qquad 0\le t\le1.\]

Then

\[dx=dt,\qquad dy=4dt,\qquad dz=5dt.\]

\[y\,dx+z\,dy+x\,dz =4t\,dt+(5t)(4dt)+(2+t)(5dt).\]

This simplifies to

\[4t+20t+10+5t=10+29t.\]

Therefore

\[\int_{C_1}y\,dx+z\,dy+x\,dz =\int_0^1(10+29t)\,dt =10+\frac{29}{2} =\frac{49}{2}.\]

For \(C_2\), use

\[x=3,\qquad y=4,\qquad z=5-5t,\qquad 0\le t\le1.\]

Then

\[dx=0,\qquad dy=0,\qquad dz=-5dt.\]

Thus

\[y\,dx+z\,dy+x\,dz=3(-5dt)=-15dt.\]

\[\int_{C_2}y\,dx+z\,dy+x\,dz=-15.\]

Hence

\[I=\frac{49}{2}-15=\frac{19}{2}.\]

Answer: \(\displaystyle \frac{19}{2}\).

4.70. Scalar Line Integral Exercises (Chapter 4, Task 4)

Compute the following scalar line integrals:

\(\displaystyle \int_C (x-y+z-2)\,ds\), where \(C\) is the straight-line segment from \((0,1,1)\) to \((1,0,1)\).
\(\displaystyle \int_C \sqrt{x^2+y^2}\,ds\) along \(C:\mathbf{r}(t)=4\cos t\,\hat{i}+4\sin t\,\hat{j}+3t\,\hat{k}\), \(-2\pi\le t\le2\pi\).
Integrate \(f(x,y,z)=x+\sqrt y-z^2\) over the path \(C_1\) followed by \(C_2\) followed by \(C_3\) from \((0,0,0)\) to \((1,1,1)\):

\[C_1:\mathbf{r}(t)=t\hat{k},\quad C_2:\mathbf{r}(t)=t\hat{j}+\hat{k},\quad C_3:\mathbf{r}(t)=t\hat{i}+\hat{j}+\hat{k},\quad 0\le t\le1.\]

Calculate \(\displaystyle \int_C (x+\sqrt y)\,ds\), where \(C\) goes along the parabola \(y=x^2\) from \((0,0)\) to \((1,1)\), then returns to the origin along \(y=x\).
Calculate \(\displaystyle \int_C \frac{ds}{1+x^2+y^2}\), where \(C\) is the unit square traversed counterclockwise.

Click to see the solution

Key Concept: Parametrize each curve piece and replace \(ds\) by speed times \(dt\).

1. The segment is

\[\mathbf{r}(t)=\langle t,1-t,1\rangle,\qquad 0\le t\le1.\]

Then \(ds=\sqrt2\,dt\), and

\[x-y+z-2=t-(1-t)+1-2=2t-2.\]

Hence

\[\int_C(x-y+z-2)\,ds =\sqrt2\int_0^1(2t-2)\,dt=-\sqrt2.\]

2. Here

\[\sqrt{x^2+y^2}=\sqrt{16\cos^2t+16\sin^2t}=4.\]

Also,

\[\mathbf{r}'(t)=\langle-4\sin t,4\cos t,3\rangle,\]

\[ds=\sqrt{16\sin^2t+16\cos^2t+9}\,dt=5\,dt.\]

Therefore

\[\int_C\sqrt{x^2+y^2}\,ds =\int_{-2\pi}^{2\pi}4\cdot5\,dt =20(4\pi)=80\pi.\]

3. On \(C_1\), \((x,y,z)=(0,0,t)\) and \(ds=dt\), so

\[\int_{C_1}f\,ds=\int_0^1(-t^2)\,dt=-\frac13.\]

On \(C_2\), \((x,y,z)=(0,t,1)\) and \(ds=dt\), so

\[\int_{C_2}f\,ds=\int_0^1(\sqrt t-1)\,dt=\frac23-1=-\frac13.\]

On \(C_3\), \((x,y,z)=(t,1,1)\) and \(ds=dt\), so

\[\int_{C_3}f\,ds=\int_0^1(t+1-1)\,dt=\frac12.\]

Adding,

\[\int_C f\,ds=-\frac13-\frac13+\frac12=-\frac16.\]

4. For the parabola \(y=x^2\), use \(x=t\), \(y=t^2\), \(0\le t\le1\). Then

\[x+\sqrt y=t+t=2t,\qquad ds=\sqrt{1+4t^2}\,dt.\]

\[\int_{\text{parabola}}(x+\sqrt y)\,ds =\int_0^1 2t\sqrt{1+4t^2}\,dt =\frac{5\sqrt5-1}{6}.\]

For the line \(y=x\), scalar line integrals do not depend on orientation, so use \(x=t\), \(y=t\), \(0\le t\le1\). Then

\[x+\sqrt y=t+\sqrt t,\qquad ds=\sqrt2\,dt.\]

Thus

\[\int_{\text{line}}(x+\sqrt y)\,ds =\sqrt2\int_0^1(t+\sqrt t)\,dt =\sqrt2\left(\frac12+\frac23\right) =\frac{7\sqrt2}{6}.\]

Therefore

\[\int_C(x+\sqrt y)\,ds=\frac{5\sqrt5-1+7\sqrt2}{6}.\]

5. Break the square into four sides. On the bottom side \(y=0\), \(0\le x\le1\):

\[I_1=\int_0^1\frac{dx}{1+x^2}=\frac{\pi}{4}.\]

On the right side \(x=1\), \(0\le y\le1\):

\[I_2=\int_0^1\frac{dy}{2+y^2}=\frac{1}{\sqrt2}\arctan\frac{1}{\sqrt2}.\]

On the top side \(y=1\), the same scalar integral gives

\[I_3=\int_0^1\frac{dx}{2+x^2}=\frac{1}{\sqrt2}\arctan\frac{1}{\sqrt2}.\]

On the left side \(x=0\):

\[I_4=\int_0^1\frac{dy}{1+y^2}=\frac{\pi}{4}.\]

\[I=\frac{\pi}{2}+\sqrt2\arctan\frac{1}{\sqrt2}.\]

Answer: 1) \(-\sqrt2\); 2) \(80\pi\); 3) \(-\frac16\); 4) \(\displaystyle \frac{5\sqrt5-1+7\sqrt2}{6}\); 5) \(\displaystyle \frac{\pi}{2}+\sqrt2\arctan\frac{1}{\sqrt2}\).

4.71. Work Along a Space Curve (Chapter 4, Example 7)

Find the work done by \(\mathbf{F}=(y-x^2)\hat{i}+(z-y^2)\hat{j}+(x-z^2)\hat{k}\) along \(C:\mathbf{r}(t)=t\hat{i}+t^2\hat{j}+t^3\hat{k}\), \(0\le t\le1\).

Click to see the solution

Key Concept: Work is \(\displaystyle \int_C\mathbf{F}\cdot d\mathbf{r}=\int_a^b\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt\).

We have

\[\mathbf{r}'(t)=\langle1,2t,3t^2\rangle.\]

Along the curve,

\[x=t,\qquad y=t^2,\qquad z=t^3.\]

Thus

\[\mathbf{F}(\mathbf{r}(t)) =\langle t^2-t^2,\ t^3-t^4,\ t-t^6\rangle =\langle0,\ t^3-t^4,\ t-t^6\rangle.\]

Dot with \(\mathbf{r}'(t)\):

\[\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =2t(t^3-t^4)+3t^2(t-t^6) =2t^4-2t^5+3t^3-3t^8.\]

Therefore

\[W=\int_0^1(3t^3+2t^4-2t^5-3t^8)\,dt.\]

Compute:

\[W=\frac34+\frac25-\frac13-\frac13=\frac{29}{60}.\]

Answer: \(\displaystyle \frac{29}{60}\).

4.72. Vector Line Integral Along a Parametric Curve (Chapter 4, Example 8)

Calculate \(\displaystyle \int_C \mathbf{F}\cdot d\mathbf{r}\), where

\[\mathbf{F}(x,y,z)=\sin x\,\hat{i}+\cos y\,\hat{j}+xz\,\hat{k},\]

and

\[\mathbf{r}(t)=t^3\hat{i}-t^2\hat{j}+t\hat{k},\qquad 0\le t\le1.\]

Click to see the solution

Key Concept: Substitute the parametrization into the field and dot with the velocity vector.

We have

\[\mathbf{r}'(t)=\langle3t^2,-2t,1\rangle.\]

Along the curve,

\[x=t^3,\qquad y=-t^2,\qquad z=t.\]

\[\mathbf{F}(\mathbf{r}(t))=\langle\sin(t^3),\cos(-t^2),t^4\rangle =\langle\sin(t^3),\cos(t^2),t^4\rangle.\]

Then

\[\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =3t^2\sin(t^3)-2t\cos(t^2)+t^4.\]

Therefore

\[\int_C\mathbf{F}\cdot d\mathbf{r} =\int_0^1\left(3t^2\sin(t^3)-2t\cos(t^2)+t^4\right)\,dt.\]

Integrate term by term:

\[\int 3t^2\sin(t^3)\,dt=-\cos(t^3),\]

\[\int -2t\cos(t^2)\,dt=-\sin(t^2),\]

and

\[\int t^4\,dt=\frac{t^5}{5}.\]

Thus

\[\int_C\mathbf{F}\cdot d\mathbf{r} =\left[-\cos(t^3)-\sin(t^2)+\frac{t^5}{5}\right]_0^1 =-\cos1-\sin1+\frac65.\]

Answer: \(\displaystyle -\cos1-\sin1+\frac65\).

4.73. Circulation Around the Unit Circle (Chapter 4, Example 9)

Find the circulation of \(\mathbf{F}=(x-y)\hat{i}+x\hat{j}\) around the circle \(\mathbf{r}(t)=\cos t\,\hat{i}+\sin t\,\hat{j}\), \(0\le t\le2\pi\).

Click to see the solution

Key Concept: Circulation is the line integral \(\oint_C\mathbf{F}\cdot d\mathbf{r}\) around a closed curve.

On the circle,

\[x=\cos t,\qquad y=\sin t,\]

\[\mathbf{F}(\mathbf{r}(t))=\langle\cos t-\sin t,\cos t\rangle.\]

Also,

\[\mathbf{r}'(t)=\langle-\sin t,\cos t\rangle.\]

Then

\[\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(\cos t-\sin t)(-\sin t)+\cos^2t.\]

Simplify:

\[-\sin t\cos t+\sin^2t+\cos^2t=1-\sin t\cos t.\]

Thus

\[\oint_C\mathbf{F}\cdot d\mathbf{r} =\int_0^{2\pi}(1-\sin t\cos t)\,dt=2\pi.\]

Answer: \(\displaystyle 2\pi\).

4.74. Flux Across the Unit Circle (Chapter 4, Example 10)

Find the outward flux of \(\mathbf{F}=(x-y)\hat{i}+x\hat{j}\) across the circle \(x^2+y^2=1\).

Click to see the solution

Key Concept: For counterclockwise orientation, outward flux is \(\displaystyle \oint_C M\,dy-N\,dx\).

Use

\[x=\cos t,\qquad y=\sin t,\qquad 0\le t\le2\pi.\]

Then

\[dx=-\sin t\,dt,\qquad dy=\cos t\,dt.\]

Here

\[M=x-y=\cos t-\sin t,\qquad N=x=\cos t.\]

Thus

\[\text{Flux}=\oint_C M\,dy-N\,dx.\]

Substitute:

\[\text{Flux}=\int_0^{2\pi}\left[(\cos t-\sin t)\cos t-\cos t(-\sin t)\right]dt.\]

The mixed terms cancel:

\[\text{Flux}=\int_0^{2\pi}\cos^2t\,dt=\pi.\]

Answer: \(\displaystyle \pi\).

4.75. Work, Circulation, and Flux Exercises (Chapter 4, Task 5)

Solve the following.

1. Find the work done by each field from \((0,0,0)\) to \((1,1,1)\) over each path:

\[\mathbf{F}_1=\sqrt z\,\hat{i}-2x\,\hat{j}+\sqrt y\,\hat{k},\]

\[\mathbf{F}_2=(3x^2-3x)\hat{i}+3z\hat{j}+\hat{k},\]

\[\mathbf{F}_3=(y+z)\hat{i}+(z+x)\hat{j}+(x+y)\hat{k}.\]

Paths:

\[C_1:\mathbf{r}(t)=t\hat{i}+t\hat{j}+t\hat{k},\]

\[C_2:\mathbf{r}(t)=t\hat{i}+t^2\hat{j}+t^4\hat{k},\]

and \(C_3\cup C_4\), where \(C_3\) goes from \((0,0,0)\) to \((1,1,0)\) and \(C_4\) goes from \((1,1,0)\) to \((1,1,1)\).

2. Find the circulation and outward flux of

\[\mathbf{F}_1=x\hat{i}+y\hat{j},\qquad \mathbf{F}_2=-y\hat{i}+x\hat{j}\]

around/across the unit circle and the ellipse \(\mathbf{r}(t)=\cos t\,\hat{i}+4\sin t\,\hat{j}\).

3. Find the outward flux of

\[\mathbf{F}_1=2x\hat{i}-3y\hat{j},\qquad \mathbf{F}_2=2x\hat{i}+(x-y)\hat{j}\]

across the circle \(\mathbf{r}(t)=a\cos t\,\hat{i}+a\sin t\,\hat{j}\), \(a>0\).

Click to see the solution

Key Concept: For work use \(\int\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt\). For counterclockwise plane curves, circulation is \(\oint M\,dx+N\,dy\) and outward flux is \(\oint M\,dy-N\,dx\).

1. Work values.

For \(C_1:\mathbf{r}(t)=\langle t,t,t\rangle\), \(0\le t\le1\):

\[\int_{C_1}\mathbf{F}_1\cdot d\mathbf{r}=\int_0^1(2\sqrt t-2t)\,dt=\frac13,\]

\[\int_{C_1}\mathbf{F}_2\cdot d\mathbf{r}=\int_0^1(3t^2+1)\,dt=2,\]

and

\[\int_{C_1}\mathbf{F}_3\cdot d\mathbf{r}=\int_0^16t\,dt=3.\]

For \(C_2:\mathbf{r}(t)=\langle t,t^2,t^4\rangle\), \(0\le t\le1\):

\[\int_{C_2}\mathbf{F}_1\cdot d\mathbf{r}=\int_0^1(4t^4-3t^2)\,dt=-\frac15,\]

\[\int_{C_2}\mathbf{F}_2\cdot d\mathbf{r}=\int_0^1(3t^2-3t+4t^3+6t^5)\,dt=\frac32,\]

and

\[\int_{C_2}\mathbf{F}_3\cdot d\mathbf{r}=\int_0^1(3t^2+5t^4+6t^5)\,dt=3.\]

For \(C_3\cup C_4\), take \(C_3:\langle t,t,0\rangle\) and \(C_4:\langle1,1,t\rangle\), \(0\le t\le1\). Then:

\[\int_{C_3\cup C_4}\mathbf{F}_1\cdot d\mathbf{r}=0,\]

\[\int_{C_3\cup C_4}\mathbf{F}_2\cdot d\mathbf{r}=\frac12,\]

and

\[\int_{C_3\cup C_4}\mathbf{F}_3\cdot d\mathbf{r}=3.\]

2. Circulation and flux.

For the unit circle \(x=\cos t\), \(y=\sin t\):

\[\operatorname{circ}(\langle x,y\rangle)=0,\qquad \operatorname{flux}(\langle x,y\rangle)=2\pi,\]

\[\operatorname{circ}(\langle-y,x\rangle)=2\pi,\qquad \operatorname{flux}(\langle-y,x\rangle)=0.\]

For the ellipse \(x=\cos t\), \(y=4\sin t\):

\[\operatorname{circ}(\langle x,y\rangle)=0,\qquad \operatorname{flux}(\langle x,y\rangle)=8\pi,\]

\[\operatorname{circ}(\langle-y,x\rangle)=8\pi,\qquad \operatorname{flux}(\langle-y,x\rangle)=0.\]

3. Flux across the circle of radius \(a\).

Use \(x=a\cos t\), \(y=a\sin t\), \(dx=-a\sin t\,dt\), and \(dy=a\cos t\,dt\).

For \(\mathbf{F}_1=\langle2x,-3y\rangle\),

\[\text{Flux}=\oint 2x\,dy-(-3y)\,dx =\int_0^{2\pi}a^2(2\cos^2t-3\sin^2t)\,dt =-\pi a^2.\]

For \(\mathbf{F}_2=\langle2x,x-y\rangle\),

\[\text{Flux}=\oint 2x\,dy-(x-y)\,dx =\int_0^{2\pi}a^2(2\cos^2t+\cos t\sin t-\sin^2t)\,dt =\pi a^2.\]

Answer: The values are listed above.

4.76. Fundamental Theorem of Line Integrals (Chapter 4, Example 11)

Calculate \(\displaystyle \int_C\mathbf{F}\cdot d\mathbf{r}\), where

\[\mathbf{F}(x,y)=(3+2xy)\hat{i}+(x^2-3y^2)\hat{j},\]

and

\[C:\mathbf{r}(t)=e^t\sin t\,\hat{i}+e^t\cos t\,\hat{j},\qquad 0\le t\le\pi.\]

Click to see the solution

Key Concept: If \(\mathbf{F}=\nabla f\), then \(\displaystyle \int_C\mathbf{F}\cdot d\mathbf{r}=f(B)-f(A)\).

Find a potential. Since

\[f_x=3+2xy,\]

integrate with respect to \(x\):

\[f=3x+x^2y+\phi(y).\]

Then

\[f_y=x^2+\phi'(y).\]

We need

\[f_y=x^2-3y^2,\]

\[\phi'(y)=-3y^2,\qquad \phi(y)=-y^3.\]

Thus

\[f(x,y)=3x+x^2y-y^3.\]

The endpoints are

\[\mathbf{r}(0)=(0,1),\qquad \mathbf{r}(\pi)=(0,-e^\pi).\]

Therefore

\[\int_C\mathbf{F}\cdot d\mathbf{r}=f(0,-e^\pi)-f(0,1).\]

Compute:

\[f(0,-e^\pi)=-(-e^\pi)^3=e^{3\pi},\qquad f(0,1)=-1.\]

\[\int_C\mathbf{F}\cdot d\mathbf{r}=e^{3\pi}+1.\]

Answer: \(\displaystyle 1+e^{3\pi}\).

4.77. Zero Curl But Not Conservative (Chapter 4, Example 12)

Let

\[\mathbf{F}=\frac{-y}{x^2+y^2}\hat{i}+\frac{x}{x^2+y^2}\hat{j}.\]

Show that \(\operatorname{curl}\mathbf{F}=\mathbf{0}\) on its natural domain, but that \(\mathbf{F}\) is not conservative.

Click to see the solution

Key Concept: Zero curl is not enough on a domain with a hole.

The natural domain is

\[\mathbb{R}^2\setminus\{(0,0)\},\]

which has a hole at the origin.

Let

\[M=\frac{-y}{x^2+y^2},\qquad N=\frac{x}{x^2+y^2}.\]

A direct derivative calculation gives

\[N_x=\frac{y^2-x^2}{(x^2+y^2)^2},\qquad M_y=\frac{y^2-x^2}{(x^2+y^2)^2},\]

\[N_x-M_y=0.\]

Thus the curl is zero away from the origin.

Now integrate around the unit circle:

\[x=\cos t,\qquad y=\sin t,\qquad 0\le t\le2\pi.\]

Then

\[\mathbf{F}(\mathbf{r}(t))=\langle-\sin t,\cos t\rangle,\]

and

\[\mathbf{r}'(t)=\langle-\sin t,\cos t\rangle.\]

Therefore

\[\oint_C\mathbf{F}\cdot d\mathbf{r} =\int_0^{2\pi}(\sin^2t+\cos^2t)\,dt =2\pi\ne0.\]

If a field were conservative, every closed-loop integral would be zero. Hence this field is not conservative on its natural domain.

Answer: \(\operatorname{curl}\mathbf{F}=0\) away from the origin, but \(\oint_C\mathbf{F}\cdot d\mathbf{r}=2\pi\), so the field is not conservative.

4.78. Use a Potential Function to Evaluate Line Integrals (Chapter 4, Task 6)

Find a function \(f\) such that \(\mathbf{F}=\nabla f\) and use it to evaluate \(\displaystyle \int_C\mathbf{F}\cdot d\mathbf{r}\).

1. \(\mathbf{F}(x, y, z) = (y^2 z + 2xz^2)\hat{i} + 2xyz\hat{j} + (xy^2 + 2x^2 z)\hat{k}\),

\[C:x=\sqrt t,\qquad y=t+1,\qquad z=t^2,\qquad 0\le t\le1.\]

2. \(\mathbf{F}(x, y, z) = yze^{xz}\hat{i} + e^{xz}\hat{j} + xye^{xz}\hat{k}\),

\[C:\mathbf{r}(t)=(t^2+1)\hat{i}+(t^2-1)\hat{j}+(t^2-2t)\hat{k},\qquad 0\le t\le2.\]

Click to see the solution

Key Concept: Once a potential function is found, only the endpoints matter.

1. Since

\[f_x=y^2z+2xz^2,\]

integrate with respect to \(x\):

\[f=xy^2z+x^2z^2+\phi(y,z).\]

Then

\[f_y=2xyz+\phi_y.\]

Matching the second component gives \(\phi_y=0\). Also,

\[f_z=xy^2+2x^2z+\phi_z.\]

Matching the third component gives \(\phi_z=0\). Therefore

\[f=xy^2z+x^2z^2+C.\]

The endpoints are

\[C(0)=(0,1,0),\qquad C(1)=(1,2,1).\]

\[\int_C\mathbf{F}\cdot d\mathbf{r}=f(1,2,1)-f(0,1,0)=(4+1)-0=5.\]

2. Since the second component is \(e^{xz}\), integrate with respect to \(y\):

\[f=ye^{xz}+\phi(x,z).\]

Then

\[f_x=yze^{xz}+\phi_x,\qquad f_z=xye^{xz}+\phi_z.\]

Matching the first and third components gives \(\phi_x=\phi_z=0\). Hence

\[f=ye^{xz}+C.\]

The endpoints are

\[C(0)=(1,-1,0),\qquad C(2)=(5,3,0).\]

Therefore

\[\int_C\mathbf{F}\cdot d\mathbf{r}=f(5,3,0)-f(1,-1,0)=3-(-1)=4.\]

Answer: 1) \(f=xy^2z+x^2z^2+C\), integral \(5\); 2) \(f=ye^{xz}+C\), integral \(4\).

4.79. Green’s Theorem on a Circle (Chapter 4, Example 13)

Calculate

\[I=\oint_C (3y-e^{\sin x})\,dx+(7x+\sqrt{y^4+1})\,dy,\]

where \(C\) is the positively oriented circle \(x^2+y^2=9\).

Click to see the solution

Key Concept: Green’s Theorem replaces a closed line integral by a double integral over the enclosed region.

Here

\[M=3y-e^{\sin x},\qquad N=7x+\sqrt{y^4+1}.\]

Green’s Theorem gives

\[I=\iint_D(N_x-M_y)\,dA,\]

where \(D\) is the disk \(x^2+y^2\le9\).

Compute:

\[N_x=7,\qquad M_y=3.\]

Therefore

\[N_x-M_y=4.\]

\[I=\iint_D4\,dA=4\operatorname{Area}(D)=4\cdot9\pi=36\pi.\]

Answer: \(\displaystyle 36\pi\).

4.80. Green’s Theorem on a Region With a Hole (Chapter 4, Example 14)

Let

\[\mathbf{F}=-\frac{y}{x^2+y^2}\hat{i}+\frac{x}{x^2+y^2}\hat{j}.\]

Calculate \(\displaystyle \int_C\mathbf{F}\cdot d\mathbf{r}\) for any positively oriented simple closed path \(C\) that encloses the origin.

Click to see the solution

Key Concept: The field is singular at the origin, so apply Green’s Theorem to the region between \(C\) and a small circle around the origin.

Let \(C'\) be a small positively oriented circle centered at the origin and lying inside \(C\). The boundary of the punctured region is \(C\cup(-C')\).

On the punctured region, the curl is zero:

\[N_x-M_y=0.\]

Green’s Theorem gives

\[\int_C\mathbf{F}\cdot d\mathbf{r}+\int_{-C'}\mathbf{F}\cdot d\mathbf{r}=0.\]

Thus

\[\int_C\mathbf{F}\cdot d\mathbf{r}=\int_{C'}\mathbf{F}\cdot d\mathbf{r}.\]

Parametrize \(C'\) by

\[x=r\cos t,\qquad y=r\sin t,\qquad 0\le t\le2\pi.\]

Then

\[\mathbf{F}(\mathbf{r}(t))=\left\langle-\frac{\sin t}{r},\frac{\cos t}{r}\right\rangle,\]

and

\[\mathbf{r}'(t)=\langle-r\sin t,r\cos t\rangle.\]

Therefore

\[\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=\sin^2t+\cos^2t=1.\]

\[\int_C\mathbf{F}\cdot d\mathbf{r}=\int_0^{2\pi}1\,dt=2\pi.\]

Answer: \(\displaystyle 2\pi\).

4.81. Green’s Theorem Exercises (Chapter 4, Task 7)

Use Green’s Theorem to evaluate the following line integrals and circulation/flux problems.

Click to see the solution

Key Concept: For positive orientation,

\[\oint_C M\,dx+N\,dy=\iint_D(N_x-M_y)\,dA.\]

For outward flux of \(\mathbf{F}=\langle M,N\rangle\),

\[\operatorname{Flux}=\iint_D(M_x+N_y)\,dA.\]

1.1 For \(\displaystyle \int_C ye^x\,dx+2e^x\,dy\) over the rectangle \(0\le x\le3\), \(0\le y\le4\):

\[M=ye^x,\qquad N=2e^x,\]

\[N_x-M_y=2e^x-e^x=e^x.\]

Thus

\[I=\int_0^3\int_0^4e^x\,dy\,dx=4(e^3-1).\]

1.2 For \(\displaystyle \int_C (x^2+y^2)\,dx+(x^2-y^2)\,dy\) over the triangle with vertices \((0,0)\), \((2,1)\), \((0,1)\):

\[N_x-M_y=2x-2y.\]

Using \(0\le x\le2\), \(x/2\le y\le1\),

\[I=\int_0^2\int_{x/2}^{1}(2x-2y)\,dy\,dx=0.\]

1.3 For \(\displaystyle \int_C y^4\,dx+2xy^3\,dy\) over the ellipse \(x^2+2y^2=2\):

\[N_x-M_y=2y^3-4y^3=-2y^3.\]

The ellipse is symmetric about the \(x\)-axis, and \(-2y^3\) is odd in \(y\), so

\[I=0.\]

1.4 For \(\displaystyle \int_C(1-y^3)\,dx+(x^3+e^{y^2})\,dy\) around the annulus between \(r=2\) and \(r=3\):

\[N_x-M_y=3x^2-(-3y^2)=3(x^2+y^2)=3r^2.\]

Therefore

\[I=\int_0^{2\pi}\int_2^3 3r^2\cdot r\,dr\,d\theta =\frac{195\pi}{2}.\]

2. For \(\mathbf{F}(x,y)=\sqrt{1+x^2}\hat{i}+\arctan x\,\hat{j}\) on the triangle \((0,0)\to(1,1)\to(0,1)\to(0,0)\):

Circulation:

\[N_x-M_y=\frac{1}{1+x^2}-0=\frac{1}{1+x^2}.\]

The region is \(0\le x\le1\), \(x\le y\le1\), so

\[\operatorname{Circ}=\int_0^1\frac{1-x}{1+x^2}\,dx =\frac{\pi}{4}-\frac12\ln2.\]

Flux:

\[M_x+N_y=\frac{x}{\sqrt{1+x^2}}+0.\]

Thus

\[\operatorname{Flux} =\int_0^1(1-x)\frac{x}{\sqrt{1+x^2}}\,dx =\frac{\sqrt2}{2}-1+\frac12\ln(1+\sqrt2).\]

3. For \(\mathbf{F}(x,y)=x(x+y)\hat{i}+xy^2\hat{j}\) around the triangle with vertices \((0,0)\), \((1,0)\), \((0,1)\):

\[M=x^2+xy,\qquad N=xy^2.\]

\[N_x-M_y=y^2-x.\]

Over \(0\le x\le1\), \(0\le y\le1-x\):

\[W=\int_0^1\int_0^{1-x}(y^2-x)\,dy\,dx=-\frac{1}{12}.\]

4. For the quarter circle of radius \(5\) and

\[\mathbf{F}=\sin x\,\hat{i}+\left(\sin y+xy^2+\frac13x^3\right)\hat{j},\]

we have

\[N_x-M_y=y^2+x^2=r^2.\]

The region is \(0\le r\le5\), \(0\le\theta\le\pi/2\). Therefore

\[W=\int_0^{\pi/2}\int_0^5 r^2\cdot r\,dr\,d\theta =\frac{625\pi}{8}.\]

5. For

\[\mathbf{F}(x,y)=\frac{2xy\hat{i}+(y^2-x^2)\hat{j}}{(x^2+y^2)^2},\]

and any positively oriented simple closed curve enclosing the origin, use a circle \(C_r\) inside the curve. On \(C_r\), \(x=r\cos t\), \(y=r\sin t\). Direct substitution gives

\[\mathbf{F}\cdot d\mathbf{r}=-\frac{\cos t}{r}\,dt,\]

\[\int_{C_r}\mathbf{F}\cdot d\mathbf{r}=\int_0^{2\pi}-\frac{\cos t}{r}\,dt=0.\]

Hence the original integral is \(0\).

6. For \(\mathbf{F}=(-\sin y)\hat{i}+(x\cos y)\hat{j}\) on the square \(0\le x\le\pi/2\), \(0\le y\le\pi/2\):

Circulation:

\[N_x-M_y=\cos y-(-\cos y)=2\cos y.\]

Thus

\[\operatorname{Circ}=\int_0^{\pi/2}\int_0^{\pi/2}2\cos y\,dy\,dx=\pi.\]

Flux:

\[M_x+N_y=0-x\sin y=-x\sin y.\]

Thus

\[\operatorname{Flux}=\int_0^{\pi/2}\int_0^{\pi/2}-x\sin y\,dy\,dx=-\frac{\pi^2}{8}.\]

7. For the cardioid \(r=a(1+\cos\theta)\) and

\[\mathbf{F}=\left(3xy-\frac{x}{1+y}\right)\hat{i}+(e^x+\arctan y)\hat{j},\]

the outward flux is

\[\operatorname{Flux}=\iint_D\left(3y-\frac{1}{1+y}+\frac{1}{1+y^2}\right)dA.\]

In polar coordinates this is

\[\int_0^{2\pi}\int_0^{a(1+\cos\theta)} \left(3r\sin\theta-\frac{1}{1+r\sin\theta}+\frac{1}{1+r^2\sin^2\theta}\right)r\,dr\,d\theta.\]

This is the Green’s-Theorem reduction of the flux to an ordinary double integral over the cardioid.

The first term integrates to zero because the cardioid is symmetric about the \(x\)-axis and \(3y\) is odd in \(y\). The remaining part has no useful elementary simplification for arbitrary \(a\):

\[\operatorname{Flux} =\iint_D\left(-\frac{1}{1+y}+\frac{1}{1+y^2}\right)dA,\]

provided the cardioid region does not cross the singular line \(y=-1\). If the region crosses that line, the flux integral is improper and Green’s Theorem cannot be applied on a region containing the singularity without splitting the domain.

8. For the region bounded above by \(y=3-x^2\) and below by \(y=x^4+1\), the intersections are at \(x=\pm1\). If the source’s \(exy\) is read literally as \(e\,xy\), then

\[M=y+e^x\ln y,\qquad N=exy,\]

and the counterclockwise circulation is

\[\int_{-1}^{1}\int_{x^4+1}^{3-x^2}\left(ey-1-\frac{e^x}{y}\right)\,dy\,dx.\]

Equivalently,

\[\int_{-1}^{1}\left[\frac{e}{2}y^2-y-e^x\ln y\right]_{y=x^4+1}^{3-x^2}\,dx.\]

Thus the exact value is

\[\int_{-1}^{1}\left(\frac{e}{2}(3-x^2)^2-(3-x^2)-e^x\ln(3-x^2)-\frac{e}{2}(x^4+1)^2+(x^4+1)+e^x\ln(x^4+1)\right)\,dx.\]

The source notation is ambiguous because \(exy\) may also mean \(e^x y\) in some lecture typography. Under the literal reading \(exy=e\,xy\), the expression above is the exact Green’s-Theorem answer.

4.82. Parametrize a Plane, Cone, and Sphere (Chapter 4, Examples 15-17)

Give standard parametrizations for a plane through a point, the cone \(z=\sqrt{x^2+y^2}\) with \(0\le z\le1\), and the sphere \(x^2+y^2+z^2=a^2\).

Click to see the solution

Key Concept: A parametrization writes every point of a surface as a vector function of two parameters.

Plane. If the plane passes through \(P_0\) with position vector \(\mathbf{r}_0\) and contains nonparallel vectors \(\mathbf{a}\) and \(\mathbf{b}\), then every point of the plane has the form

\[\mathbf{r}(u,v)=\mathbf{r}_0+u\mathbf{a}+v\mathbf{b},\qquad u,v\in\mathbb{R}.\]

If \(\mathbf{r}_0=(x_0,y_0,z_0)\), \(\mathbf{a}=(a_1,a_2,a_3)\), and \(\mathbf{b}=(b_1,b_2,b_3)\), then

\[x=x_0+ua_1+vb_1,\quad y=y_0+ua_2+vb_2,\quad z=z_0+ua_3+vb_3.\]

Cone. For \(z=\sqrt{x^2+y^2}\), cylindrical coordinates give \(z=r\). With \(0\le r\le1\) and \(0\le\theta\le2\pi\),

\[\mathbf{r}(r,\theta)=r\cos\theta\,\hat{i}+r\sin\theta\,\hat{j}+r\,\hat{k}.\]

Sphere. For \(x^2+y^2+z^2=a^2\), spherical coordinates give

\[\mathbf{r}(\phi,\theta)=a\sin\phi\cos\theta\,\hat{i}+a\sin\phi\sin\theta\,\hat{j}+a\cos\phi\,\hat{k},\]

where

\[0\le\phi\le\pi,\qquad 0\le\theta\le2\pi.\]

Answer: The parametrizations are listed above.

4.83. Surface Area of a Cone (Chapter 4, Example 18)

Find the surface area of the cone \(z=\sqrt{x^2+y^2}\) with \(0\le z\le1\).

Click to see the solution

Key Concept: For a parametric surface, \(d\sigma=\|\mathbf{r}_u\times\mathbf{r}_v\|\,du\,dv\).

Use

\[\mathbf{r}(r,\theta)=\langle r\cos\theta,r\sin\theta,r\rangle,\]

with \(0\le r\le1\), \(0\le\theta\le2\pi\).

Then

\[\mathbf{r}_r=\langle\cos\theta,\sin\theta,1\rangle,\qquad \mathbf{r}_\theta=\langle-r\sin\theta,r\cos\theta,0\rangle.\]

The cross product is

\[\mathbf{r}_r\times\mathbf{r}_\theta =\langle-r\cos\theta,-r\sin\theta,r\rangle.\]

Thus

\[\|\mathbf{r}_r\times\mathbf{r}_\theta\| =\sqrt{r^2\cos^2\theta+r^2\sin^2\theta+r^2} =\sqrt2\,r.\]

Therefore

\[A=\int_0^{2\pi}\int_0^1\sqrt2\,r\,dr\,d\theta =\sqrt2\cdot\frac12\cdot2\pi =\sqrt2\pi.\]

Answer: \(\displaystyle \sqrt2\pi\).

4.84. Surface Area of a Sphere (Chapter 4, Example 19)

Find the surface area of a sphere of radius \(a\).

Click to see the solution

Key Concept: Use the spherical parametrization and compute the surface area element.

Parametrize the sphere by

\[\mathbf{r}(\phi,\theta)=\langle a\sin\phi\cos\theta,\ a\sin\phi\sin\theta,\ a\cos\phi\rangle,\]

where \(0\le\phi\le\pi\) and \(0\le\theta\le2\pi\).

For this parametrization,

\[\|\mathbf{r}_\phi\times\mathbf{r}_\theta\|=a^2\sin\phi.\]

Therefore

\[A=\int_0^{2\pi}\int_0^\pi a^2\sin\phi\,d\phi\,d\theta.\]

Compute:

\[\int_0^\pi\sin\phi\,d\phi=2,\]

\[A=2\pi\cdot2a^2=4\pi a^2.\]

Answer: \(\displaystyle 4\pi a^2\).

4.85. Surface Area of an Explicit and an Implicit Surface (Chapter 4, Example 20)

State the surface area formulas for explicit and implicit surfaces, and apply them carefully.

Click to see the solution

Key Concept: A graph \(z=f(x,y)\) has surface element \(\sqrt{1+f_x^2+f_y^2}\,dA\). An implicit surface \(F(x,y,z)=c\) can be projected to a coordinate plane.

For \(z=f(x,y)\) over \(R\),

\[A(S)=\iint_R\sqrt{1+f_x^2+f_y^2}\,dA.\]

For an implicit surface \(F(x,y,z)=c\) projected onto a coordinate plane with unit normal \(\mathbf{p}\),

\[d\sigma=\frac{\|\nabla F\|}{|\nabla F\cdot\mathbf{p}|}\,dA.\]

Thus

\[A(S)=\iint_R\frac{\|\nabla F\|}{|\nabla F\cdot\mathbf{p}|}\,dA.\]

For the sphere \(x^2+y^2+z^2=4\), take

\[F=x^2+y^2+z^2-4.\]

Then

\[\nabla F=\langle2x,2y,2z\rangle,\qquad \|\nabla F\|=4\quad\text{on the sphere}.\]

If the surface is projected to the \(xy\)-plane, then \(\mathbf{p}=\hat{k}\) and

\[|\nabla F\cdot\hat{k}|=|2z|.\]

Since \(z^2=4-x^2-y^2\),

\[d\sigma=\frac{4}{2|z|}\,dA=\frac{2}{\sqrt{4-x^2-y^2}}\,dA.\]

This is the correct surface element over any projected region where \(z\) has one chosen sign.

For the spherical-band example from the source, the written bounds contain a typo because the sphere \(x^2+y^2+z^2=4\) has no points with \(z=3\). The projection used in the slide is the annulus

\[1\le x^2+y^2\le3.\]

Using that projected annulus on one chosen sheet of the sphere gives

\[A=\int_0^{2\pi}\int_1^{\sqrt3}\frac{2r}{\sqrt{4-r^2}}\,dr\,d\theta.\]

Compute the radial integral by the substitution \(u=4-r^2\), so \(du=-2r\,dr\):

\[\int_1^{\sqrt3}\frac{2r}{\sqrt{4-r^2}}\,dr =\left[-2\sqrt{4-r^2}\right]_1^{\sqrt3} =2(\sqrt3-1).\]

Therefore the area of one projected sheet over this annulus is

\[A=2\pi\cdot2(\sqrt3-1)=4\pi(\sqrt3-1).\]

If both the upper and lower sheets over the same annulus are included, the area is doubled.

Answer: Explicit graph: \(\displaystyle d\sigma=\sqrt{1+f_x^2+f_y^2}\,dA\); implicit surface: \(\displaystyle d\sigma=\frac{\|\nabla F\|}{|\nabla F\cdot\mathbf{p}|}\,dA\).

4.86. Parametric Surfaces and Surface Area Exercises (Chapter 4, Task 8)

Solve the parametric-surface and surface-area exercises.

Click to see the solution

Key Concept: For membership, solve for parameters. For identification, eliminate parameters. For area, compute the appropriate surface element.

1. Point membership.

For

\[\mathbf{r}(u,v)=\langle u+v,u-2v,3+u-v\rangle,\]

\(P(4,-5,1)\) lies on the surface because \(u=1\), \(v=3\) gives

\[\langle4,-5,1\rangle.\]

\(Q(0,4,6)\) does not lie on it. Solving \(u+v=0\) and \(u-2v=4\) gives \(u=4/3\), \(v=-4/3\), and then \(z=17/3\ne6\).

For

\[\mathbf{r}(u,v)=\langle1+u-v,u+v^2,u^2-v^2\rangle,\]

\(P(1,2,1)\) does not lie on the surface. The equation \(1+u-v=1\) gives \(u=v\), which forces \(z=u^2-v^2=0\), not \(1\).

\(Q(2,3,3)\) lies on it because \(u=2\), \(v=1\) gives

\[\langle2,3,3\rangle.\]

2. Identify surfaces.

For

\[\mathbf{r}(u,v)=\langle u+v,3-v,1+4u+5v\rangle,\]

we get \(v=3-y\) and \(u=x+y-3\). Hence

\[z=4+4x-y,\]

which is a plane.

For

\[\mathbf{r}(u,v)=\langle u^2,u\cos v,u\sin v\rangle,\]

we have

\[y^2+z^2=u^2=x,\]

so the surface is the paraboloid

\[x=y^2+z^2,\qquad x\ge0.\]

For

\[\mathbf{r}(s,t)=\langle3\cos t,s,\sin t\rangle,\qquad -1\le s\le1,\]

we have

\[\frac{x^2}{9}+z^2=1,\]

with \(-1\le y\le1\). This is a finite strip of an elliptic cylinder.

3. Parametric representations.

Plane through \((0,-1,5)\) containing \((2,1,4)\) and \((-3,2,5)\):

\[\mathbf{r}(u,v)=\langle0,-1,5\rangle+u\langle2,1,4\rangle+v\langle-3,2,5\rangle.\]

Part of \(4x^2-4y^2-z^2=4\) in front of the \(yz\)-plane:

\[\mathbf{r}(u,v)=\left\langle\sqrt{1+u^2+\frac{v^2}{4}},u,v\right\rangle.\]

Part of \(x^2+z^2=9\) above the \(xy\)-plane and \(-4\le y\le4\):

\[\mathbf{r}(\theta,y)=\langle3\cos\theta,y,3\sin\theta\rangle,\qquad 0\le\theta\le\pi,\quad -4\le y\le4.\]

Part of \(x^2+y^2+z^2=36\) between \(z=0\) and \(z=3\sqrt3\):

\[\mathbf{r}(\phi,\theta)=\langle6\sin\phi\cos\theta,6\sin\phi\sin\theta,6\cos\phi\rangle,\]

where

\[\frac{\pi}{6}\le\phi\le\frac{\pi}{2},\qquad 0\le\theta\le2\pi.\]

4. Surface areas.

For \(z=4-2x^2+y\) over the triangle with vertices \((0,0)\), \((1,0)\), \((1,1)\):

\[f_x=-4x,\qquad f_y=1,\]

\[d\sigma=\sqrt{16x^2+2}\,dA.\]

The triangle is \(0\le x\le1\), \(0\le y\le x\). Hence

\[A=\int_0^1\int_0^x\sqrt{16x^2+2}\,dy\,dx =\int_0^1x\sqrt{16x^2+2}\,dx =\frac{13\sqrt2}{12}.\]

For the paraboloid \(y=x^2+z^2\) inside \(x^2+z^2=16\), treat it as a graph over the \(xz\)-plane. Then

\[d\sigma=\sqrt{1+4x^2+4z^2}\,dx\,dz.\]

Use polar coordinates in the \(xz\)-plane:

\[A=\int_0^{2\pi}\int_0^4\sqrt{1+4r^2}\,r\,dr\,d\theta =\frac{\pi}{6}(65\sqrt{65}-1).\]

4.87. Surface Integral Over a Cone (Chapter 4, Example 21)

Integrate \(G(x,y,z)=x^2\) over the cone \(z=\sqrt{x^2+y^2}\), \(0\le z\le1\).

Click to see the solution

Key Concept: For the graph \(z=f(x,y)\), use \(\iint_R G(x,y,f(x,y))\sqrt{1+f_x^2+f_y^2}\,dA\).

Here

\[f(x,y)=\sqrt{x^2+y^2}.\]

Over the disk \(x^2+y^2\le1\),

\[\sqrt{1+f_x^2+f_y^2}=\sqrt2.\]

Therefore

\[\iint_S x^2\,d\sigma=\sqrt2\iint_{x^2+y^2\le1}x^2\,dA.\]

Use polar coordinates:

\[x^2=r^2\cos^2\theta,\qquad dA=r\,dr\,d\theta.\]

Thus

\[\iint_Sx^2\,d\sigma =\sqrt2\int_0^{2\pi}\int_0^1r^3\cos^2\theta\,dr\,d\theta =\sqrt2\cdot\frac14\cdot\pi =\frac{\pi\sqrt2}{4}.\]

Answer: \(\displaystyle \frac{\pi\sqrt2}{4}\).

4.88. Surface Integral Over a Slanted Cylinder (Chapter 4, Example 22)

Evaluate \(\displaystyle \iint_S z\,d\sigma\), where \(S\) is the surface whose side is the cylinder \(x^2+y^2=1\), whose bottom is the disk \(z=0\), and whose top is the plane \(z=1+x\) above the disk.

Click to see the solution

Key Concept: Split a piecewise smooth surface into simpler pieces.

Write

\[S=S_1\cup S_2\cup S_3,\]

where \(S_1\) is the cylindrical side, \(S_2\) is the bottom disk, and \(S_3\) is the top plane.

For \(S_1\), use

\[x=\cos\theta,\qquad y=\sin\theta,\qquad z=z,\]

with

\[0\le\theta\le2\pi,\qquad 0\le z\le1+\cos\theta.\]

The surface element is \(d\sigma=dz\,d\theta\). Thus

\[\iint_{S_1}z\,d\sigma =\int_0^{2\pi}\int_0^{1+\cos\theta}z\,dz\,d\theta =\frac{3\pi}{2}.\]

For \(S_2\), \(z=0\), so

\[\iint_{S_2}z\,d\sigma=0.\]

For \(S_3\), the plane is \(z=1+x\). Since \(f_x=1\) and \(f_y=0\),

\[d\sigma=\sqrt2\,dA.\]

Therefore

\[\iint_{S_3}z\,d\sigma =\sqrt2\iint_{x^2+y^2\le1}(1+x)\,dA.\]

The integral of \(x\) over the disk is zero by symmetry, so

\[\iint_{S_3}z\,d\sigma=\sqrt2\pi.\]

Thus

\[\iint_S z\,d\sigma=\frac{3\pi}{2}+\sqrt2\pi =\left(\frac32+\sqrt2\right)\pi.\]

Answer: \(\displaystyle \left(\frac32+\sqrt2\right)\pi\).

4.89. Surface Integrals of Scalar Functions (Chapter 4, Task 9)

Evaluate the scalar surface integrals from the exercises.

Click to see the solution

Key Concept: Convert each surface integral to a double integral using an appropriate parametrization or graph formula.

1. Evaluate \(\displaystyle \iint_S\sqrt{x(1+2z)}\,d\sigma\) on \(z=y^2/2\) over \(R:x\ge0\), \(y\ge0\), \(x+y\le1\).

Since \(z=y^2/2\),

\[\sqrt{x(1+2z)}=\sqrt{x(1+y^2)}.\]

Also \(f_x=0\), \(f_y=y\), so

\[d\sigma=\sqrt{1+y^2}\,dA.\]

Thus the integrand becomes \(\sqrt{x}(1+y^2)\). With \(0\le y\le1\), \(0\le x\le1-y\),

\[I=\int_0^1\int_0^{1-y}\sqrt{x}(1+y^2)\,dx\,dy=\frac{284}{945}.\]

2. Integrate \(G(x,y,z)=xyz\) over the surface of the unit cube in the first octant. On the faces \(x=0\), \(y=0\), and \(z=0\), the integrand is zero. On the faces \(x=1\), \(y=1\), and \(z=1\), the integrals are each

\[\int_0^1\int_0^1uv\,du\,dv=\frac14.\]

Therefore

\[I=\frac34.\]

3.1 For \(G=x\) over \(y=x^2\), \(0\le x\le2\), \(0\le z\le3\), parametrize by

\[\mathbf{r}(x,z)=\langle x,x^2,z\rangle.\]

Then

\[d\sigma=\sqrt{1+4x^2}\,dx\,dz.\]

Thus

\[I=\int_0^3\int_0^2x\sqrt{1+4x^2}\,dx\,dz =\frac{17\sqrt{17}-1}{4}.\]

3.2 For \(G=z\) over \(y^2+z^2=4\), \(z\ge0\), \(1\le x\le4\), use

\[\mathbf{r}(x,\theta)=\langle x,2\cos\theta,2\sin\theta\rangle,\qquad 0\le\theta\le\pi.\]

Then \(d\sigma=2\,dx\,d\theta\) and \(z=2\sin\theta\), so

\[I=\int_1^4\int_0^\pi4\sin\theta\,d\theta\,dx=24.\]

3.3 For \(G=x^2\) over the unit sphere, symmetry gives

\[\iint_Sx^2\,d\sigma=\frac13\iint_S(x^2+y^2+z^2)\,d\sigma.\]

On the unit sphere, \(x^2+y^2+z^2=1\), so

\[I=\frac13(4\pi)=\frac{4\pi}{3}.\]

3.4 For \(G=x+y+z\) over \(2x+2y+z=2\) in the first octant, write

\[z=2-2x-2y,\qquad x\ge0,\quad y\ge0,\quad x+y\le1.\]

The surface element is

\[d\sigma=\sqrt{1+(-2)^2+(-2)^2}\,dA=3\,dA.\]

Also

\[x+y+z=2-x-y.\]

Thus

\[I=3\iint_R(2-x-y)\,dA=2.\]

3.5 For \(G=yz\) over the sphere \(x^2+y^2+z^2=4\) above the cone \(z=\sqrt{x^2+y^2}\), use spherical coordinates with radius \(2\) and \(0\le\phi\le\pi/4\). The integrand contains a factor \(\sin\theta\), and the full range \(0\le\theta\le2\pi\) makes it cancel by symmetry:

\[I=0.\]

Answer: The values are 1) \(\frac{284}{945}\); 2) \(\frac34\); 3.1) \(\frac{17\sqrt{17}-1}{4}\); 3.2) \(24\); 3.3) \(\frac{4\pi}{3}\); 3.4) \(2\); 3.5) \(0\).