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

where , , and are scalar functions. The vector assigned to has components , , and .

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 , on a smaller solid region, on a surface, or even only along a curve. For example, if is a space curve, then its tangent vectors and normal vectors form vector fields along that curve. If is a scalar field, then attaching 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 is a differentiable scalar function, its gradient is the vector field

For a function of variables, the same idea becomes

The gradient is not just a list of partial derivatives. It has a geometric meaning: at each point, points in the direction where 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 represents temperature, then points in the direction of fastest temperature increase. If a particle moves perpendicular to , 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 is called a conservative vector field if it is the gradient of some scalar function:

The scalar function is called a potential function for . In coordinates, if

then is conservative when there is a function such that

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 , first integrate with respect to :

The term appears because integration with respect to treats and as constants. Then differentiate this expression with respect to and , compare with and , 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

the curl is

Expanding the determinant gives

Equivalently,

The two displayed versions are the same; the second one simply distributes the minus sign in the component.

1.3.2 Curl of a Gradient

If has continuous second-order partial derivatives, then

This follows from equality of mixed partial derivatives:

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

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

In component form, means three separate compatibility equations:

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 .

1.3.3 When Zero Curl Is Enough

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

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 , or another open simply connected region, the curl test is reliable.
Compute .
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 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

the divergence is the scalar field

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

then

The first term is zero because does not depend on .

1.4.2 Divergence of Curl

If has continuous second-order partial derivatives, then

This identity is another consequence of equality of mixed partial derivatives. It is useful as a consistency check: if some vector field is actually the curl of another field, then must be zero.

1.5 The Laplace Operator

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

This expression is called the Laplacian of and is denoted by

The equation

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:

turns a scalar field into a vector field showing fastest increase.
measures rotation of a vector field.
measures source or sink strength of a vector field.
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 behaves like a vector of partial derivatives, but products must still be interpreted carefully.

For scalar fields and vector fields with sufficiently smooth components, the standard linearity rules are

and

The product rules are

and

Two common errors should be avoided. First, do not write in the product rule for curl; the cross product is with 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 , only 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 , , and in .
Natural domain: The largest set of points where all component functions of a field are defined.
Gradient field: A vector field of the form for some differentiable scalar function .
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 such that .
Curl: The vector field that measures local rotational tendency of .
Rotation: Another name for the curl of a vector field.
Divergence: The scalar field that measures local source or sink behavior of .
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 , equal to .
Laplace’s equation: The equation .
Harmonic function: A function satisfying Laplace’s equation on its domain.

3. Formulas

Vector Field in Space: .
Gradient: .
Conservative Field Condition: .
Curl Determinant: .
Curl Expanded: .
Zero Curl Component Test: .
Divergence: .
Curl of a Gradient: , assuming continuous second-order partial derivatives.
Divergence of a Curl: , assuming continuous second-order partial derivatives.
Laplacian: .
Curl Test on Simply Connected Domains: , when is defined on an open simply connected domain and has continuous first partial derivatives.
Divergence Linearity: .
Curl Linearity: .
Divergence Product Rule: .
Curl Product Rule: .

4. Practice

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

Calculate:

a) .

b) .

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 , so the inner integral only multiplies by the length of the -interval:

Thus

For the -integral, set , so :

Therefore

b) First simplify the integrand:

The region in the -plane is

Equivalently,

This change is useful because has no elementary antiderivative, while integrating in first is easy. Then

Integrate with respect to :

Thus

Now

and, with , ,

Therefore

Answer: a) ; b) .

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

Find the volume of the solid region bounded by the planes , , , , and .

Click to see the solution

Key Concept: Translate the bounding planes into inequalities and integrate over the resulting solid.

The plane gives the bottom. For a fixed height , the plane gives , and the plane gives . Hence

This interval is nonempty when , so .

Similarly, the plane gives , and the plane gives

Therefore the volume is

Compute the inner lengths:

Thus

Answer: .

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

Set up and evaluate , where is the tetrahedron in the first octant bounded by the plane .

Click to see the solution

Key Concept: The first-octant tetrahedron is described by and .

Use the order :

Thus

A quick symmetric computation is possible. The tetrahedron has volume

Its centroid is

so the average value of over the tetrahedron is

Therefore

Answer: .

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

Let be the wedge in the first octant that is cut from the cylindrical solid by the planes and . Evaluate .

Click to see the solution

Key Concept: The cylinder is in the -variables, while the planes bound .

Because the solid lies in the first octant,

The planes and give

Thus

where is the quarter disk in the first quadrant of the -plane.

The inner integral is

Use polar coordinates in the -plane:

Then , and

Answer: .

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

Let be the solid bounded above by the plane and below by the paraboloid , with projection being the disk in the xy-plane. Evaluate .

Click to see the solution

Key Concept: Because the projection is a disk and the integrand is , cylindrical coordinates are natural.

Use

The bounds are

Therefore

Integrate with respect to :

Now

Thus

Answer: .

4.6. Double Integral by the Transformation , (Lab 12, Task 6)

Use the transformation , to evaluate where is bounded by , , , in the first quadrant.

Click to see the solution

Key Concept: Choose the transformation because it turns the four boundary curves into constant- and constant- lines.

From

we get

Since the region is in the first quadrant, and . The boundaries become

and

Compute the Jacobian:

Also, . Hence

Separate the variables:

Now

and

Therefore

Answer: .

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

Calculate the integral by changing to cylindrical coordinates:

Click to see the solution

Key Concept: The -projection is the upper half disk , .

Use cylindrical coordinates:

The bounds become

Thus

Integrate with respect to :

Now

Therefore

Answer: .

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 and below the sphere .

Click to see the solution

Key Concept: In spherical coordinates, cones have constant and spheres through the origin often have simple bounds.

The cone becomes

“Above the cone” means

The sphere is

In spherical coordinates this is

so, for ,

Also . Therefore

Integrate in :

Use , :

Thus

Answer: .

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

Describe the solid and compute its volume:

Which coordinate system is most natural?

Click to see the solution

Key Concept: The inequalities use , so cylindrical coordinates are the natural choice.

The solid lies above the plane , below the downward-opening paraboloid

and above the disk

Use cylindrical coordinates:

Then

Integrate:

Answer: The solid is a cylinder of radius capped by the paraboloid and cut below by ; the most natural coordinates are cylindrical; .

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

Click to see the solution

Key Concept: Scale the ellipsoid to the unit ball.

Let

Then

so the ellipsoid maps from the unit ball

The Jacobian is the determinant of the diagonal scaling matrix:

Therefore

Answer: .

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

Calculate:

a) over .

b) where .

c) over .

d) over .

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:

and

Therefore

b) The integrand separates:

c) Since ,

d) Separate all factors:

Compute each:

Thus

Answer: a) ; b) ; c) ; d) .

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

The solid is bounded by the planes , , and the parabolic cylinder , and by the planes and (where , ). The projection onto the xy-plane is the region , . Compute .

Click to see the solution

Key Concept: Use the given projection and place between the lower and upper planes.

The bounds are

Therefore

The -integral is

Expand:

Then

Answer: .

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

Evaluate over the region inside and above using polar coordinates.

Click to see the solution

Key Concept: The factor becomes , and .

The region above inside the unit disk is the sector

The integrand becomes

Thus

Therefore

Answer: .

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

Evaluate where is the upper half of the unit disk using polar coordinates.

Click to see the solution

Key Concept: The expression becomes .

For the upper half of the unit disk,

Therefore

Use , :

Thus

Answer: .

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

Evaluate using , .

Click to see the solution

Key Concept: The printed inner lower limit contains the same variable 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

then it implies

Since the upper limit is also

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

Therefore

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: under the literal interpretation of the printed bounds.

4.16. Transformation , Over a Square (Lab 12, Homework 6)

Use the transformation , to evaluate where is the square with vertices , , , .

Click to see the solution

Key Concept: The integrand becomes , and the square becomes a symmetric diamond in the -plane.

The inverse transformation is

The Jacobian is

The square , becomes

For each fixed , the allowed -interval is symmetric about . Since is odd, the inner integral in is zero:

Therefore the whole double integral is zero.

Answer: .

4.17. Transforming the Region Bounded by and (Lab 12, Homework 7)

Evaluate over the region in the first quadrant bounded by , , , and using an appropriate transformation.

Click to see the solution

Key Concept: Use and , because the boundary curves become and .

In the first quadrant,

Solving for and gives

The Jacobian is

Also,

Therefore

Simplify:

Now

and

Thus

Answer: .

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

Evaluate , where is the solid inside the cylinder and between the planes and .

Click to see the solution

Key Concept: Integrating first produces a function over the disk.

The bounds are

Thus

Expand:

Over the disk, by symmetry, , and

Therefore

Answer: .

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

Let be the solid enclosed by the sphere and above the cone . Compute the volume of .

Click to see the solution

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

The sphere is . The cone is

Above the cone means

Hence

Compute:

Equivalently,

Answer: .

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

Evaluate , where is the upper half of the ball (that is, ).

Click to see the solution

Key Concept: In spherical coordinates, .

The upper half ball of radius has bounds

Then

Use

Therefore

Answer: .

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

A cylindrical sensor covers the region . Assume the signal intensity is . Find the accumulated signal

Click to see the solution

Key Concept: In cylindrical coordinates, .

Since ,

Integrate in :

Thus

Compute:

Answer: .

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

Let be the solid in 3-space defined by the inequalities , , .

a) Using the coordinate transformation , , , calculate the Jacobian . Express your answer in terms of .

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

Click to see the solution

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

From

we get

Also,

The Jacobian matrix is

Expanding along the middle column,

The inequalities become

Therefore the volume is

The -interval has length , so

An antiderivative is

Thus

Answer: a) ; b) .

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

Use the change of variables , , to evaluate , where is the image of the box , , .

Click to see the solution

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

First,

The Jacobian is

Therefore

Thus

Answer: .

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

Let be the solid region in xyz-space defined by the inequalities , , . Calculate the integral by applying the transformation , , .

Click to see the solution

Key Concept: The variables , , and turn the inequalities into a rectangular box.

The inverse transformation is

The bounds become

The Jacobian is

The integrand transforms as

Therefore

First integrate with respect to :

Thus

Answer: .

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

Find the average value of on the solid . Recall that .

Click to see the solution

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

Use spherical coordinates. The upper half ball of radius has

Since and ,

Compute:

Thus

The volume of a half ball of radius is

Therefore

Answer: .

4.26. Sketch the Rotational Field (Lab 13, Example 1)

Describe the vector field by sketching some vectors.

Click to see the solution

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

At ,

so the vector points upward.

At ,

so the vector points left.

At ,

so the vector points downward.

At ,

so the vector points right.

Each vector is perpendicular to the radius vector , because

Also,

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 (Lab 13, Example 2)

Sketch the vector field on given by .

Click to see the solution

Key Concept: The vector has only a -component.

At every point,

If , the vector points upward. If , it points downward. If , the vector is zero. The magnitude is

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

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

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

Sketch the following vector fields:

a) .

b) .

c) .

d) in .

Click to see the solution

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

a) The vector is constant:

Every arrow points right and downward with the same length

b) The horizontal component is always , so every vector points left. The vertical component is :

on the line , vectors are purely horizontal and point left;
above the line , vectors point left and upward;
below the line , vectors point left and downward.

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

So all arrows have unit length. Representative values:

This field swaps the roles of and in the direction vector and normalizes the result.

d) The field is constant in space:

Every arrow points in the positive -direction with length .

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

4.29. Gradient of (Lab 13, Example 3)

Find the gradient vector field of .

Click to see the solution

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

For a two-variable scalar field,

Differentiate with respect to , treating as constant:

Therefore

Answer: .

4.30. Curl and Divergence of (Lab 13, Example 4)

Let . Find and .

Click to see the solution

Key Concept: For ,

and

Here

Compute the curl:

so the -component is

Next,

so the -component is

Finally,

so the -component is

Therefore

For divergence,

Thus

Answer: and .

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

Find the gradient vector field of the following functions:

a) .

b) .

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 ,

For , use the product rule:

Therefore

b) Let

Then

Therefore, away from the origin,

Answer: a) ; b) for .

4.32. Curl of (Lab 13, Task 3)

Compute the curl of .

Click to see the solution

Key Concept: Use .

Here

Compute:

Next,

Finally,

Therefore

Answer: .

4.33. Divergence and the Identity (Lab 13, Task 4)

Compute the divergence of and verify that for any with continuous second partials, .

Click to see the solution

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

For

we have

Now let a general field be

Then

Taking divergence,

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

All terms cancel, hence

Answer: For the given field, ; in general, when the required second partials are continuous.

4.34. Sketch (Lab 13, Homework 1)

Sketch the vector field .

Click to see the solution

Key Concept: The vector at is .

On the positive -axis, vectors point right. On the negative -axis, they point left. On the positive -axis, they point upward. On the negative -axis, they point downward.

The origin is the only zero vector:

The magnitude is

so arrows grow farther from the origin. The -component grows twice as strongly as the -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 (Lab 13, Homework 2)

Use a Computer Algebra System (CAS) to plot the vector field and describe its appearance.

Click to see the solution

Key Concept: The components are always nonnegative because .

The horizontal component is

so it depends only on the distance from the -axis. It is zero when and grows as grows.

The vertical component is

so it depends only on the distance from the -axis. It is zero when and grows as grows.

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

on the -axis, vectors are vertical upward;
on the -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 (Lab 13, Homework 3)

Find the gradient vector field of .

Click to see the solution

Key Concept: Apply the chain rule to .

Write

Then

and

Therefore

where .

Answer: .

4.37. Curl and Divergence of (Lab 13, Homework 4)

Compute both the curl and divergence of .

Click to see the solution

Key Concept: Use the component formulas directly.

Here

The curl is

Compute:

Thus

The divergence is

Answer: and .

4.38. Verify (Lab 13, Homework 5)

Verify that for .

Click to see the solution

Key Concept: By definition, , which is the Laplacian.

Let

First,

Next,

Differentiate again with respect to :

Finally,

Differentiate again with respect to :

Therefore

equals

This is exactly .

Answer: , with the explicit value above.

4.39. Show That Is Conservative (Lab 13, Homework 6)

Show that is conservative by computing its curl.

Click to see the solution

Key Concept: On a simply connected domain such as , a continuously differentiable field with zero curl is conservative.

Write

Compute the curl components:

Next,

Finally,

Therefore

Since the field is defined and continuously differentiable on all of , it is conservative.

A potential function is visible by integrating with respect to :

Indeed,

Answer: , so is conservative; one potential is .

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

Explain why the gravitational field is conservative.

Click to see the solution

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

Let

The scalar function

is defined on the natural domain . Since

we have

Thus is a gradient field on its natural domain, so it is conservative there.

Answer: on , 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) .

b) .

Click to see the solution

Key Concept: For ,

and

a) Here

Compute the derivatives needed for the curl:

Therefore

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

Hence

b) Now

The curl derivatives are

Thus

For the divergence,

Therefore

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 such that .

a) .

b) .

c) .

d) .

Click to see the solution

Key Concept: In the plane, a continuously differentiable field on a simply connected domain is conservative exactly when . In space, use the curl test: a nonzero curl means the field is not conservative.

a) Here and . Then

The domain is all of , so the field is conservative. Find from

Integrating with respect to gives

Then

Thus

b) Let

Then

So the field is conservative on . Integrate with respect to :

Differentiate with respect to :

Compare with :

Therefore

c) Let

Already one compatibility condition fails:

Since in general, , and the field is not conservative.

d) Let

Again, check one curl component:

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

Answer: a) conservative, ; b) conservative, ; c) not conservative; d) not conservative.

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

Calculate

where is the straight-line segment from to .

Click to see the solution

Key Concept: For a scalar line integral, parametrize the curve and use .

A convenient parametrization of the line segment is

Then

Along the curve,

Therefore

Compute:

Answer: .

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

Calculate

along the curve

Click to see the solution

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

Here

Also

and therefore

Thus

Answer: .

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

Find the line integral of

from to over each path:

a) , .

b) , .

c) , where is the segment from to and is the segment from to .

Click to see the solution

Key Concept: For a vector line integral,

a) For , and . Then

Hence

b) For , and . Then

The dot product is

Therefore

Compute term by term:

c) First parametrize by , . Then , and

Thus

Next parametrize by , . Then , and

Therefore

Answer: a) ; b) ; c) .

4.46. Circulation and Flux of (Lab 14, Task 6)

Find the circulation and flux of around and across each curve:

a) The circle , .

b) The ellipse , .

Click to see the solution

Key Concept: Green’s theorem gives circulation from and outward flux from .

Here and . Therefore

so the circulation around any positively oriented simple closed curve is .

For flux,

Thus the outward flux equals

a) The unit circle has area , so

b) The ellipse has semiaxes and , so its area is . Therefore

Answer: a) circulation , flux ; b) circulation , flux .

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

Find a function such that , and use it to evaluate for

where

Click to see the solution

Key Concept: If , then .

Let

Start with :

Differentiate this expression with respect to :

Since must equal , we get . Differentiate with respect to :

Since must equal , we get . Thus one potential is

The curve starts at

and ends at

Therefore

Compute:

Answer: and .

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

Evaluate

where is the positively oriented rectangle with vertices , , , and .

Click to see the solution

Key Concept: Green’s theorem converts into a double integral of over the enclosed region.

Here

Then

The rectangle is , . Hence

Integrating first in gives

Answer: .

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

Evaluate

where is the positively oriented ellipse .

Click to see the solution

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

Let

Then

Therefore

The ellipse is symmetric about the -axis. The function is odd in , so its integral over this symmetric region is zero:

Answer: .

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

moving a particle counterclockwise along the closed triangle

Click to see the solution

Key Concept: Work around a closed plane curve is the circulation .

Here

By Green’s theorem,

Compute the derivatives:

So the integrand is

The triangle is

Thus

Separate the two pieces:

and

Therefore

Answer: .

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

Integrate over the parabolic cylinder

Click to see the solution

Key Concept: Parametrize the surface and use .

Use the parametrization

Then

Their cross product has magnitude

Since on the surface,

Use , so :

Multiplying by the -length gives

Answer: .

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

Integrate over the unit sphere

Click to see the solution

Key Concept: Use symmetry on the sphere: the integrals of , , and over the full sphere are equal.

On the unit sphere,

By symmetry,

Adding these three equal integrals gives

The surface area of the unit sphere is , so

Therefore

Answer: .

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

Integrate over the portion of the plane

that lies in the first octant.

Click to see the solution

Key Concept: Write the plane as a graph over its triangular projection.

Solve for :

In the first octant, , so

The surface element for is

Here and , so

On the plane,

Therefore

where is the triangle , . Now

Thus

Answer: .

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

Find the flux of

through the surface cut from the parabolic cylinder by the planes , , and , directed away from the -axis.

Click to see the solution

Key Concept: For an oriented parametrized surface, flux is with the cross product chosen in the required direction.

The surface can be parametrized by

Since , we have

Compute

Choose

This normal points away from the -axis because its projection in the -plane points outward from the axis.

On the surface,

Therefore

The flux is

Integrate in :

The odd term integrates to zero over , so

Answer: .

4.55. Flux of Through a First-Octant Sphere (Lab 15, Task 8)

Find the flux of across the portion of the sphere

in the first octant, directed away from the origin.

Click to see the solution

Key Concept: On a sphere of radius , the outward unit normal is .

Since ,

Use spherical coordinates on the sphere:

The first octant corresponds to

Therefore

This simplifies to

The angular integrals are

Hence

Answer: .

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

Find the flux of

outward through the closed surface consisting of the paraboloid

and the disk

Click to see the solution

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

The divergence is

The enclosed solid is

In cylindrical coordinates,

Its volume is

Compute:

By the divergence theorem,

Answer: .

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

Find the counterclockwise circulation and the outward flux of

around and across the boundary of the square

Click to see the solution

Key Concept: For , Green’s theorem gives

and the flux form gives

Here

For circulation,

Thus

For flux,

Therefore

Separate the variables:

Answer: circulation ; outward flux .

4.58. Surface Integral Over a Spherical Cap Above a Cone (Lab 15, Homework 2)

Integrate over the part of the sphere

that lies above the cone

Click to see the solution

Key Concept: Symmetry can make a surface integral zero even before doing the full calculation.

The sphere has radius . The cone corresponds to in spherical coordinates. The part above the cone has

On the sphere,

Thus

The surface element on the sphere is

So the integrand contains a factor :

Since

the whole surface integral is zero.

Answer: .

4.59. Upward Flux Through a Plane Above a Square (Lab 15, Homework 3)

Find the flux of

upward across the portion of the plane

that lies above the square

Click to see the solution

Key Concept: For a graph with upward orientation, use the vector surface element .

Solve the plane equation for :

Then

so the upward vector surface element is

On the plane,

Therefore

Simplify:

The flux is

Compute the terms:

and

Therefore

Answer: .

4.60. Interpret Basic Vector Fields (Chapter 4, Illustrative Examples)

The source gives three basic examples of vector fields:

The tangent vectors and normal vectors along a space curve .
The gradient field attached to points of a level surface of a scalar function .
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 , then its tangent direction is determined by . Where , the unit tangent vector is

As 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 , the unit normal vector

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 is differentiable, then

At a point on a level surface , the gradient is perpendicular to that level surface, provided . Therefore attaching 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

is conservative over its natural domain, and

is a potential function for it. Check why.

Click to see the solution

Key Concept: To check that is a potential for , compute and compare it with .

Differentiate with respect to :

Thus

Since , the field is conservative.

Answer: is conservative, with potential .

4.62. Conservative Polynomial Vector Field (Chapter 4, Example 2)

Show that 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 . Integrating with respect to gives

where is independent of .

Differentiate this expression with respect to :

The field requires , so

Therefore depends only on : .

Now differentiate with respect to :

The field requires , so

Thus is constant.

Therefore a potential function is

Answer: is conservative, and is a potential.

4.63. Divergence of a Polynomial Vector Field (Chapter 4, Example 3)

For the field , compute .

Click to see the solution

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

Here

Therefore

Compute:

because does not contain ;

and

Thus

Answer: .

4.64. Curl and Divergence of Four Vector Fields (Chapter 4, Task 1)

Find the curl and the divergence of each vector field:

Click to see the solution

Key Concept: For ,

and

1. Let

Then

and

The divergence is

2. Let

Then

and

Hence

The divergence is

3. Let

Then

and

Since each component is independent of its own matching variable,

4. Let

Then

and

Therefore

The divergence is

4.65. Determine Whether Vector Fields Are Conservative (Chapter 4, Task 2)

Determine whether each vector field is conservative. If it is conservative, find such that .

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

Check one compatibility condition:

These are not equal in general. Therefore the field is not conservative.

2. Here

Check:

These are not equal in general. Therefore the field is not conservative.

3. Here

Integrate with respect to :

Then

Since this must equal , we get . Also,

Since this must equal , we get . Hence is constant, and

So the field is conservative.

4. Here

Integrate with respect to :

Then

Matching gives . Also,

Matching gives . Hence

So the field is conservative.

Answer: 1) not conservative; 2) not conservative; 3) conservative with ; 4) conservative with .

4.66. Prove Basic Vector Calculus Identities (Chapter 4, Task 3)

Let be a scalar field, and let and 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 terms in the first two formulas and writes the curl product rule incorrectly. The corrected standard identities are:

and

Let

For divergence linearity,

Using linearity of partial derivatives,

Curl linearity is proved the same way by applying linearity to each component of

For the divergence product rule,

Apply the product rule:

Group terms:

Thus

For the curl product rule, compute

Expanding each component by the product rule gives one group equal to and the remaining group equal to . Hence

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 , where is the upper half of the unit circle .

Click to see the solution

Key Concept: For a scalar line integral, parametrize the curve and use .

The upper half of the unit circle is

Then

Substitute into the integral:

Compute the two terms separately:

For the second term, let , so :

Therefore

Answer: .

4.68. Scalar Line Integral Along Two Different Paths (Chapter 4, Example 5)

Calculate , where:

is the line segment joining the origin to .
, where joins the origin to and joins to .

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

Then

The integrand becomes

Hence

2. Piecewise path. For use

Thus

For use

Thus

Therefore

Answer: Direct segment: ; piecewise path: .

4.69. Line Integral with Respect to , , and (Chapter 4, Example 6)

Calculate , where consists of the line segment from to , followed by the line segment from to .

Click to see the solution

Key Concept: For integrals with , , and , substitute the parametrization and use , , .

For , use

Then

This simplifies to

Therefore

For , use

Then

Thus

Hence

Answer: .

4.70. Scalar Line Integral Exercises (Chapter 4, Task 4)

Compute the following scalar line integrals:

, where is the straight-line segment from to .
along , .
Integrate over the path followed by followed by from to :

Calculate , where goes along the parabola from to , then returns to the origin along .
Calculate , where is the unit square traversed counterclockwise.

Click to see the solution

Key Concept: Parametrize each curve piece and replace by speed times .

1. The segment is

Then , and

Hence

2. Here

Also,

Therefore

3. On , and , so

On , and , so

Adding,

4. For the parabola , use , , . Then

For the line , scalar line integrals do not depend on orientation, so use , , . Then

Thus

Therefore

5. Break the square into four sides. On the bottom side , :

On the right side , :

On the top side , the same scalar integral gives

On the left side :

Answer: 1) ; 2) ; 3) ; 4) ; 5) .

4.71. Work Along a Space Curve (Chapter 4, Example 7)

Find the work done by along , .

Click to see the solution

Key Concept: Work is .

We have

Along the curve,

Thus

Dot with :

Therefore

Compute:

Answer: .

4.72. Vector Line Integral Along a Parametric Curve (Chapter 4, Example 8)

Calculate , where

and

Click to see the solution

Key Concept: Substitute the parametrization into the field and dot with the velocity vector.

We have

Along the curve,

Then

Therefore

Integrate term by term:

and

Thus

Answer: .

4.73. Circulation Around the Unit Circle (Chapter 4, Example 9)

Find the circulation of around the circle , .

Click to see the solution

Key Concept: Circulation is the line integral around a closed curve.

On the circle,

Also,

Then

Simplify:

Thus

Answer: .

4.74. Flux Across the Unit Circle (Chapter 4, Example 10)

Find the outward flux of across the circle .

Click to see the solution

Key Concept: For counterclockwise orientation, outward flux is .

Use

Then

Here

Thus

Substitute:

The mixed terms cancel:

Answer: .

4.75. Work, Circulation, and Flux Exercises (Chapter 4, Task 5)

Solve the following.

1. Find the work done by each field from to over each path:

Paths:

and , where goes from to and goes from to .

2. Find the circulation and outward flux of

around/across the unit circle and the ellipse .

3. Find the outward flux of

across the circle , .

Click to see the solution

Key Concept: For work use . For counterclockwise plane curves, circulation is and outward flux is .

1. Work values.

For , :

and

For , :

and

For , take and , . Then:

and

2. Circulation and flux.

For the unit circle , :

For the ellipse , :

3. Flux across the circle of radius .

Use , , , and .

For ,

Answer: The values are listed above.

4.76. Fundamental Theorem of Line Integrals (Chapter 4, Example 11)

Calculate , where

and

Click to see the solution

Key Concept: If , then .

Find a potential. Since

integrate with respect to :

Then

We need

Thus

The endpoints are

Therefore

Compute:

Answer: .

4.77. Zero Curl But Not Conservative (Chapter 4, Example 12)

Let

Show that on its natural domain, but that 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

which has a hole at the origin.

Let

A direct derivative calculation gives

Thus the curl is zero away from the origin.

Now integrate around the unit circle:

Then

and

Therefore

If a field were conservative, every closed-loop integral would be zero. Hence this field is not conservative on its natural domain.

Answer: away from the origin, but , so the field is not conservative.

4.78. Use a Potential Function to Evaluate Line Integrals (Chapter 4, Task 6)

Find a function such that and use it to evaluate .

1. ,

2. ,

Click to see the solution

Key Concept: Once a potential function is found, only the endpoints matter.

1. Since

integrate with respect to :

Then

Matching the second component gives . Also,

Matching the third component gives . Therefore

The endpoints are

2. Since the second component is , integrate with respect to :

Then

Matching the first and third components gives . Hence

The endpoints are

Therefore

Answer: 1) , integral ; 2) , integral .

4.79. Green’s Theorem on a Circle (Chapter 4, Example 13)

Calculate

where is the positively oriented circle .

Click to see the solution

Key Concept: Green’s Theorem replaces a closed line integral by a double integral over the enclosed region.

Here

Green’s Theorem gives

where is the disk .

Compute:

Therefore

Answer: .

4.80. Green’s Theorem on a Region With a Hole (Chapter 4, Example 14)

Let

Calculate for any positively oriented simple closed path 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 and a small circle around the origin.

Let be a small positively oriented circle centered at the origin and lying inside . The boundary of the punctured region is .

On the punctured region, the curl is zero:

Green’s Theorem gives

Thus

Parametrize by

Then

and

Therefore

Answer: .

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,

For outward flux of ,

1.1 For over the rectangle , :

Thus

1.2 For over the triangle with vertices , , :

Using , ,

1.3 For over the ellipse :

The ellipse is symmetric about the -axis, and is odd in , so

1.4 For around the annulus between and :

Therefore

2. For on the triangle :

Circulation:

The region is , , so

Flux:

Thus

3. For around the triangle with vertices , , :

Over , :

4. For the quarter circle of radius and

we have

The region is , . Therefore

5. For

and any positively oriented simple closed curve enclosing the origin, use a circle inside the curve. On , , . Direct substitution gives

Hence the original integral is .

6. For on the square , :

Circulation:

Thus

Flux:

Thus

7. For the cardioid and

the outward flux is

In polar coordinates this is

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 -axis and is odd in . The remaining part has no useful elementary simplification for arbitrary :

provided the cardioid region does not cross the singular line . 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 and below by , the intersections are at . If the source’s is read literally as , then

and the counterclockwise circulation is

Equivalently,

Thus the exact value is

The source notation is ambiguous because may also mean in some lecture typography. Under the literal reading , 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 with , and the sphere .

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 with position vector and contains nonparallel vectors and , then every point of the plane has the form

If , , and , then

Cone. For , cylindrical coordinates give . With and ,

Sphere. For , spherical coordinates give

where

Answer: The parametrizations are listed above.

4.83. Surface Area of a Cone (Chapter 4, Example 18)

Find the surface area of the cone with .

Click to see the solution

Key Concept: For a parametric surface, .

Use

with , .

Then

The cross product is

Thus

Therefore

Answer: .

4.84. Surface Area of a Sphere (Chapter 4, Example 19)

Find the surface area of a sphere of radius .

Click to see the solution

Key Concept: Use the spherical parametrization and compute the surface area element.

Parametrize the sphere by

where and .

For this parametrization,

Therefore

Compute:

Answer: .

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 has surface element . An implicit surface can be projected to a coordinate plane.

For over ,

For an implicit surface projected onto a coordinate plane with unit normal ,

Thus

For the sphere , take

Then

If the surface is projected to the -plane, then and

Since ,

This is the correct surface element over any projected region where has one chosen sign.

For the spherical-band example from the source, the written bounds contain a typo because the sphere has no points with . The projection used in the slide is the annulus

Using that projected annulus on one chosen sheet of the sphere gives

Compute the radial integral by the substitution , so :

Therefore the area of one projected sheet over this annulus is

If both the upper and lower sheets over the same annulus are included, the area is doubled.

Answer: Explicit graph: ; implicit surface: .

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

lies on the surface because , gives

does not lie on it. Solving and gives , , and then .

For

does not lie on the surface. The equation gives , which forces , not .

lies on it because , gives

2. Identify surfaces.

For

we get and . Hence

which is a plane.

For

we have

so the surface is the paraboloid

For

we have

with . This is a finite strip of an elliptic cylinder.

3. Parametric representations.

Plane through containing and :

Part of in front of the -plane:

Part of above the -plane and :

Part of between and :

where

4. Surface areas.

For over the triangle with vertices , , :

The triangle is , . Hence

For the paraboloid inside , treat it as a graph over the -plane. Then

Use polar coordinates in the -plane:

4.87. Surface Integral Over a Cone (Chapter 4, Example 21)

Integrate over the cone , .

Click to see the solution

Key Concept: For the graph , use .

Here

Over the disk ,

Therefore

Use polar coordinates:

Thus

Answer: .

4.88. Surface Integral Over a Slanted Cylinder (Chapter 4, Example 22)

Evaluate , where is the surface whose side is the cylinder , whose bottom is the disk , and whose top is the plane above the disk.

Click to see the solution

Key Concept: Split a piecewise smooth surface into simpler pieces.

Write

where is the cylindrical side, is the bottom disk, and is the top plane.

For , use

with

The surface element is . Thus

For , , so

For , the plane is . Since and ,

Therefore

The integral of over the disk is zero by symmetry, so

Thus

Answer: .

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 on over , , .

Since ,

Also , , so

Thus the integrand becomes . With , ,

2. Integrate over the surface of the unit cube in the first octant. On the faces , , and , the integrand is zero. On the faces , , and , the integrals are each

Therefore

3.1 For over , , , parametrize by

Then

Thus

3.2 For over , , , use

Then and , so

3.3 For over the unit sphere, symmetry gives

On the unit sphere, , so

3.4 For over in the first octant, write

The surface element is

Also

Thus

3.5 For over the sphere above the cone , use spherical coordinates with radius and . The integrand contains a factor , and the full range makes it cancel by symmetry:

Answer: The values are 1) ; 2) ; 3.1) ; 3.2) ; 3.3) ; 3.4) ; 3.5) .