EXAMPLE 1 Evaluation of a Line Integral in the Plane

Find the value of the line integral (3) when F(r) = [−y, −xy] = −yi − xyj and C is the circular arc in Fig. 220 from A to B.

Solution. We may represent C by r(t) = [cos t, sin t] = cos t i + sin t j, where 0 t π/2. Then x(t) = cos t, y(t) = sin t, and

By differentiation, r′(t) = [−sin t, cos t] = −sin t i + cos t j, so that by (3) [use (10) in App. 3.1; set cos t = u in the second term]

EXAMPLE 2 Line Integral in Space

The evaluation of line integrals in space is practically the same as it is in the plane. To see this, find the value of (3) when F(r) = [z, x, y] = zi + xj + yk and C is the helix (Fig. 221)

Solution. From (4) we have x(t) = cos t, y(t) = sin t, z(t) = 3t. Thus

The dot product is 3t(−sin t) + cos² t + 3 sin t. Hence (3) gives

THEOREM 1 Direction-Preserving Parametric Transformations

Any representations of C that give the same positive direction on C also yield the same value of the line integral (3).

PROOF

The proof follows by the chain rule. Let r(t) be the given representation with a t b as in (3). Consider the transformation t = φ(t*) which transforms the t interval to a* t* b* and has a positive derivative dt/dt*. We write r(t) = r(φ(t*)) = r*(t*). Then dt = (dt/dt*) dt* and

EXAMPLE 3 Work Done by a Variable Force

If F in Example 1 is a force, the work done by F in the displacement along the quarter-circle is 0.4521, measured in suitable units, say, newton-meters (nt · m, also called joules, abbreviation J; see also inside front cover). Similarly in Example 2.

EXAMPLE 4 Work Done Equals the Gain in Kinetic Energy

Let F be a force, so that (3) is work. Let t be time, so that dr/dt = v, velocity. Then we can write (3) as

Now by Newton's second law, that is, force = mass × acceleration, we get

where m is the mass of the body displaced. Substitution into (5) gives [see (11), Sec. 9.4]

On the right, m|v|²/2 is the kinetic energy. Hence the work done equals the gain in kinetic energy. This is a basic law in mechanics.

EXAMPLE 5 A Line Integral of the Form (8)

Integrate F(r) = [xy, yz, z] along the helix in Example 2.

Solution. F(r(t)) = [cos t sin t, 3t sin t, 3t] integrated with respect to t from 0 to 2π gives

THEOREM 2 Path Dependence

The line integral (3) generally depends not only on F and on the endpoints A and B of the path, but also on the path itself along which the integral is taken.

PROOF

Almost any example will show this. Take, for instance, the straight segment C₁: r₁(t) = [t, t, 0] and the parabola C₂: r₂(t) = [t, t², 0] with 0 t 1 (Fig. 223) and integrate F = [0, xy, 0]. Then , so that integration gives 1/3 and 2/5, respectively.

PROBLEM SET 10.1

1. WRITING PROJECT. From Definite Integrals to Line Integrals. Write a short report (1–2 pages) with examples on line integrals as generalizations of definite integrals. The latter give the area under a curve. Explain the corresponding geometric interpretation of a line integral.

2–11 LINE INTEGRAL. WORK

Calculate for the given data. If F is a force, this gives the work done by the force in the displacement along C. Show the details.

• 2. F = [y², −x²], C: y = 4x² from (0, 0) to (1, 4)
• 3. F as in Prob. 2, C from (0, 0) straight to (1, 4). Compare.
• 4. F = [xy, x²y²], C from (2, 0) straight to (0, 2)
• 5. F as in Prob. 4, C the quarter-circle from (2, 0) to (0, 2) with center (0, 0)
• 6. F = [x − y, y − z, z − x], C: r = [2 cos t, t, 2 sin t] from (2, 0, 0) to (2, 2π, 0)
• 7. F = [x², y², z²], C: r = [cos t, sin t, e^t] from (1, 0, 1) to (1, 0, e^2π). Sketch C.
• 8. F = [e^x, cosh y, sinh z], C: r = [t, t², t³] from (0, 0, 0) to . Sketch C.
• 9. F = [x + y, y + z, z + x], C: r = [2t, 5t, t] from t = 0 to 1. Also from t = −1 to 1.
• 10. F = [x, −z, 2y] from (0, 0, 0) straight to (1, 1, 0), then to (1, 1, 1), back to (0, 0, 0)
• 11. F = [e^−x, e^−y, e^−z], C: r = [t, t², t] from (0, 0, 0) to (2, 4, 2). Sketch C.
• 12. PROJECT. Change of Parameter. Path Dependence.

  Consider the integral .

  (a) One path, several representations. Find the value of the integral when r = [cos t, sin t], 0 t π/2. Show that the value remains the same if you set t = −p or t = p² or apply two other parametric transformations of your own choice.

  (b) Several paths. Evaluate the integral when C: y = xⁿ, thus r = [t, tⁿ], 0 t 1, where n = 1, 2, 3, …. Note that these infinitely many paths have the same endpoints.

  (c) Limit. What is the limit in (b) as n → ∞? Can you confirm your result by direct integration without referring to (b)?

  (d) Show path dependence with a simple example of your choice involving two paths.

• 13. ML-Inequality, Estimation of Line Integrals. Let F be a vector function defined on a curve C. Let |F| be bounded, say, |F| M on C, where M is some positive number. Show that

  

• 14. Using (9), find a bound for the absolute value of the work W done by the force F = [x², y] in the displacement from (0, 0) straight to (3, 4). Integrate exactly and compare.

15–20 INTEGRALS (8) AND (8*)

Evaluate them with F or f and C as follows.

• 15. F = [y², z², x²], C: r = [3 cos t, 3 sin t, 2t], 0 t 4π
• 16. f = 3x + y + 5z, C: r = [t, cosh t, sinh t], 0 t 1. Sketch C.
• 17. F = [x + y, y + z, z + x], C; r = [4 cos t, sin t, 0], 0 t π
• 18. F = [y^1/3, x^1/3, 0], C the hypocycloid r = [cos³ t, sin³ t, 0], 0 t π/4
• 19. f = xyz, C: r = [4t, 3t², 12t], −2 t 2. Sketch C.
• 20. F = [xz, yz, x²y²], C: r = [t, t, e^t], 0 t 5. Sketch C.

THEOREM 1 Path Independence

A line integral (1) with continuous F₁, F₂, F₃ in a domain D in space is path independent in D if and only if F = [F₁, F₂, F₃] is the gradient of some function f in D,

PROOF

(a) We assume that (2) holds for some function f in D and show that this implies path independence. Let C be any path in D from any point A to any point B in D, given by r(t) = [x(t), y(t), z(t)], where a t b. Then from (2), the chain rule in Sec. 9.6, and (3′) in the last section we obtain

(b) The more complicated proof of the converse, that path independence implies (2) for some f, is given in App. 4.

EXAMPLE 1 Path Independence

Show that the integral is path independent in any domain in space and find its value in the integration from A: (0, 0, 0) to B: (2, 2, 2).

Solution. F = [2x, 2y, 4z] = grad f, where f = x² + y² + 2z² because ∂f/∂x = 2x = F₁, ∂f/∂y = 2y = F₂, ∂f/∂z = 4z = F₃. Hence the integral is independent of path according to Theorem 1, and (3) gives f(B) − f(A) = f(2, 2, 2) − f(0, 0, 0) = 4 + 4 + 8 = 16.

If you want to check this, use the most convenient path C: r(t) = [t, t, t], 0 t 2, on which F(r(t) = [2t, 2t, 4t], so that F(r(t)) • r′(t) = 2t + 2t + 4t = 8t, and integration from 0 to 2 gives 8 · 2²/2 = 16.

If you did not see the potential by inspection, use the method in the next example.

EXAMPLE 2 Path Independence. Determination of a Potential

Evaluate the integral from A: (0, 1, 2) to B: (1, −1, 7) by showing that F has a potential and applying (3).

Solution. If F has a potential f, we should have

We show that we can satisfy these conditions. By integration of f_x and differentiation,

This gives f(x, y, z) = x³ + y³z and by (3),

THEOREM 2 Path Independence

The integral (1) is path independent in a domain D if and only if its value around every closed path in D is zero.

PROOF

If we have path independence, then integration from A to B along C₁ and along C₂ in Fig. 225 gives the same value. Now C₁ and C₂ together make up a closed curve C, and if we integrate from A along C₁ to B as before, but then in the opposite sense along C₂ back to A (so that this second integral is multiplied by −1), the sum of the two integrals is zero, but this is the integral around the closed curve C.

Fig. 225. Proof of Theorem 2

Conversely, assume that the integral around any closed path C in D is zero. Given any points A and B and any two curves C₁ and C₂ from A to B in D, we see that C₁ with the orientation reversed and C₂ together form a closed path C. By assumption, the integral over C is zero. Hence the integrals over C₁ and C₂, both taken from A to B, must be equal. This proves the theorem.

THEOREM 3* Path Independence

The integral (1) is path independent in a domain D in space if and only if the differential form (4) has continuous coefficient functions F₁, F₂, F₃ and is exact in D.

THEOREM 3 Criterion for Exactness and Path Independence

Let F₁, F₂, F₃ in the line integral (1),

be continuous and have continuous first partial derivatives in a domain D in space. Then:

(a) If the differential form (4) is exact in D—and thus (1) is path independent by Theorem 3*—, then in D,

in components (see Sec. 9.9)

(b) If (6) holds in D and D is simply connected, then (4) is exact in D—and thus (1) is path independent by Theorem 3*.

PROOF

(a) If (4) is exact in D, then F = grad f in D by Theorem 3*, and, furthermore, curl F = curl (grad f) = 0 by (2) in Sec. 9.9, so that (6) holds.

(b) The proof needs “Stokes's theorem” and will be given in Sec. 10.9.

EXAMPLE 3 Exactness and Independence of Path. Determination of a Potential

Using (6′), show that the differential form under the integral sign of

is exact, so that we have independence of path in any domain, and find the value of I from A: (0, 0, 1) to B: (1, π/4, 2).

Solution. Exactness follows from (6′), which gives

To find f, we integrate F₂ (which is “long,” so that we save work) and then differentiate to compare with F₁ and F₃,

h′ = 0 implies h = const and we can take h = 0, so that g = 0 in the first line. This gives, by (3),

EXAMPLE 4 On the Assumption of Simple Connectedness in Theorem 3

Let

Differentiation shows that (6′) is satisfied in any domain of the xy-plane not containing the origin, for example, in the domain shown in Fig. 226. Indeed, F₁ and F₂ do not depend on z, and F₃ = 0, so that the first two relations in (6′) are trivially true, and the third is verified by differentiation:

Clearly, D in Fig. 226 is not simply connected. If the integral

were independent of path in D, then I = 0 on any closed curve in D, for example, on the circle x² + y² = 1. But setting x = r cos θ, y = r sin θ and noting that the circle is represented by r = 1, we have

so that −y dx + x dy = sin² θ dθ + cos² θ dθ = dθ and counterclockwise integration gives

Since D is not simply connected, we cannot apply Theorem 3 and cannot conclude that I is independent of path in D.

Although F = grad f, where f = arctan (y/x) (verify!), we cannot apply Theorem 1 either because the polar angle f = θ = arctan (y/x) is not single-valued, as it is required for a function in calculus.

PROBLEM SET 10.2

1. WRITING PROJECT. Report on Path Independence. Make a list of the main ideas and facts on path independence and dependence in this section. Then work this list into a report. Explain the definitions and the practical usefulness of the theorems, with illustrative examples of your own. No proofs.
2. On Example 4. Does the situation in Example 4 of the text change if you take the domain 3/2?

3–9 PATH INDEPENDENT INTEGRALS

Show that the form under the integral sign is exact in the plane (Probs. 3–4) or in space (Probs. 5–9) and evaluate the integral. Show the details of your work.

• 3.
• 4.
• 5.
• 6.
• 7.
• 8.
• 9.
• 10. PROJECT. Path Dependence. (a) Show that is path dependent in the xy-plane.

  (b) Integrate from (0, 0) along the straight-line segment to (1, b), 0 b 1, and then vertically up to (1, 1); see the figure. For which b is I maximum? What is its maximum value?

  (c) Integrate I from (0, 0) along the straight-line segment to (c, 1), 0 c 1, and then horizontally to (1, 1). For c = 1, do you get the same value as for b = 1 in (b)? For which c is I maximum? What is its maximum value?

  

  Project 10. Path Dependence

• 11. On Example 4. Show that in Example 4 of the text, F = grad (arctan (y/x)). Give examples of domains in which the integral is path independent.
• 12. CAS EXPERIMENT. Extension of Project 10. Integrate x²y dx + 2xy² dy over various circles through the points (0, 0) and (1, 1). Find experimentally the smallest value of the integral and the approximate location of the center of the circle.

13–19 PATH INDEPENDENCE?

Check, and if independent, integrate from (0, 0, 0) to (a, b, c).

• 13.
• 14. (sinh xy) (z dx − x dz)
• 15. x²y dx − 4xy² dy + 8z²x dz
• 16. e^y dx + (xe^y − e^z) dy − ye^z dz
• 17. 4y dx + z dy + (y − 2z)dz
• 18. (cos xy)(yz dx + xz dy) − 2 sin xy dz
• 19. (cos (x² + 2y² + z²))(2x dx + 4y dy + 2z dz)
• 20. Path Dependence. Construct three simple examples in each of which two equations (6′) are satisfied, but the third is not.

EXAMPLE 1 Change of Variables in a Double Integral

Evaluate the following double integral over the square R in Fig. 232.

Solution. The shape of R suggests the transformation x + y = u, x − y = υ. Then , . The Jacobian is

R corresponds to the square 0 u 2, 0 υ 2. Therefore,

EXAMPLE 2 Double Integrals in Polar Coordinates. Center of Gravity. Moments of Inertia

Let f(x, y) = 1 be the mass density in the region in Fig. 233. Find the total mass, the center of gravity, and the moments of inertia I_x, I_y, I₀.

Solution. We use the polar coordinates just defined and formula (8). This gives the total mass

Fig. 233. Example 2

The center of gravity has the coordinates

The moments of inertia are

Why are and less than ?

PROBLEM SET 10.3

1. Mean value theorem. Illustrate (2) with an example.

2–8 DOUBLE INTEGRALS

Describe the region of integration and evaluate.

• 2.
• 3.
• 4. Prob. 3, order reversed.
• 5.
• 6.
• 7. Prob. 6, order reversed.
• 8.

9–11 VOLUME

Find the volume of the given region in space.

• 9. The region beneath z = 4x² + 9y² and above the rectangle with vertices (0, 0), (3, 0), (3, 2), (0, 2) in the xy-plane.
• 10. The first octant region bounded by the coordinate planes and the surfaces y = 1 − x², z = 1 − x². Sketch it.
• 11. The region above the xy-plane and below the paraboloid z = 1 − (x² + y²).

12–16 CENTER OF GRAVITY

Find the center of gravity of a mass of density f(x, y) = 1 in the given region R.

• 12.
• 13.
• 14.
• 15.
• 16.

17–20 MOMENTS OF INERTIA

Find I_x, I_y, I₀ of a mass of density f(x, y) = 1 in the region R in the figures, which the engineer is likely to need, along with other profiles listed in engineering handbooks.

• 17. R as in Prob. 13.
• 18. R as in Prob. 12.
• 19.
• 20.

THEOREM 1 Green's Theorem in the Plane⁴ (Transformation between Double Integrals and Line Integrals)

Let R be a closed bounded region (see Sec. 10.3) in the xy-plane whose boundary C consists of finitely many smooth curves (see Sec. 10.1). Let F₁(x, y) and F₂(x, y) be functions that are continuous and have continuous partial derivatives ∂F₁/∂y and ∂F₂/∂x everywhere in some domain containing R. Then

Here we integrate along the entire boundary C of R in such a sense that R is on the left as we advance in the direction of integration (see Fig. 234).

EXAMPLE 1 Verification of Green's Theorem in the Plane

Green's theorem in the plane will be quite important in our further work. Before proving it, let us get used to it by verifying it for F₁ = y² − 7y, F₂ = 2xy + 2x and C the circle x² + y² = 1.

Solution. In (1) on the left we get

since the circular disk R has area π.

We now show that the line integral in (1) on the right gives the same value, 9π. We must orient C counterclockwise, say, r(t) = [cos t, sin t]. Then r′(t) = [−sin t, cos t], and on C,

Hence the line integral in (1) becomes, verifying Green's theorem,

PROOF

We prove Green's theorem in the plane, first for a special region R that can be represented in both forms

and

EXAMPLE 2 Area of a Plane Region as a Line Integral Over the Boundary

In (1) we first choose F₁ = 0, F₂ = x and then F₁ = −y, F₂ = 0. This gives

respectively. The double integral is the area A of R. By addition we have

where we integrate as indicated in Green's theorem. This interesting formula expresses the area of R in terms of a line integral over the boundary. It is used, for instance, in the theory of certain planimeters (mechanical instruments for measuring area). See also Prob. 11.

For an ellipse x²/a² + y²/b² = 1 or x = a cos t, y = b sin t we get x′ = −a sin t, y′ = b cos t; thus from (4) we obtain the familiar formula for the area of the region bounded by an ellipse,

EXAMPLE 3 Area of a Plane Region in Polar Coordinates

Let r and θ be polar coordinates defined by x = r cos θ, y = r sin θ. Then

and (4) becomes a formula that is well known from calculus, namely,

As an application of (5), we consider the cardioid r = a(1 − cos θ), where 0 θ 2π (Fig. 239). We find

EXAMPLE 4 Transformation of a Double Integral of the Laplacian of a Function into a Line Integral of Its Normal Derivative

The Laplacian plays an important role in physics and engineering. A first impression of this was obtained in Sec. 9.7, and we shall discuss this further in Chap. 12. At present, let us use Green's theorem for deriving a basic integral formula involving the Laplacian.

We take a function w(x, y) that is continuous and has continuous first and second partial derivatives in a domain of the xy-plane containing a region R of the type indicated in Green's theorem. We set F₁ = −∂w/∂y and F₂ = ∂w/∂x. Then ∂F₁/∂y and ∂F₂/∂x are continuous in R, and in (1) on the left we obtain

the Laplacian of w (see Sec. 9.7). Furthermore, using those expressions for F₁ and F₂, we get in (1) on the right

where s is the arc length of C, and C is oriented as shown in Fig. 240. The integrand of the last integral may be written as the dot product

The vector n is a unit normal vector to C, because the vector r′ (s) = dr/ds = [dx/ds, dy/ds] is the unit tangent vector of C, and r′ • n = 0, so that n is perpendicular to r′. Also, n is directed to the exterior of C because in Fig. 240 the positive x-component dx/ds of r′ is the negative y-component of n, and similarly at other points. From this and (4) in Sec. 9.7 we see that the left side of (8) is the derivative of w in the direction of the outward normal of C. This derivative is called the normal derivative of w and is denoted by ∂w/∂n; that is, ∂w/∂n = (grad w) • n. Because of (6), (7), and (8), Green's theorem gives the desired formula relating the Laplacian to the normal derivative,

For instance, w = x² − y² satisfies Laplace's equation ∇²w = 0. Hence its normal derivative integrated over a closed curve must give 0. Can you verify this directly by integration, say, for the square 0 x 1, 0 y 1?

PROBLEM SET 10.4

1–10 LINE INTEGRALS: EVALUATION BY GREEN'S THEOREM

Evaluate counterclockwise around the boundary C of the region R by Green's theorem, where

1. F = [y, −x], C the circle x² + y² = 1/4
2. F = [6y², 2x − 2y⁴], R the square with vertices ±(2, 2), ±(2, −2)
3. F = [x²e^y, y², e^x], R the rectangle with vertices (0, 0), (2, 0), (2, 3), (0, 3)
4. F = [x cosh 2y, 2x² sinh 2y], R: x² y x
5. F = [x² + y², x² − y²], R: 1 y 2 − x²
6. F = [cosh y, −sinh x], R: 1 x 3, x y 3x
7. F = grad (x³ cos² (xy)), R as in Prob. 5
8. F = [−e^−x cos y, −e^−x sin y], R the semidisk x² + y² 16, x 0
9. F = [e^y/x, e^y ln x + 2x], R: 1 + x⁴ y 2
10. F = [x²y, −x/y²], R: 1 x² + y² 4, x 0, y x. Sketch R.
11. CAS EXPERIMENT. Apply (4) to figures of your choice whose area can also be obtained by another method and compare the results.
12. PROJECT. Other Forms of Green's Theorem in the Plane. Let R and C be as in Green's theorem, r′ a unit tangent vector, and n the outer unit normal vector of C (Fig. 240 in Example 4). Show that (1) may be written

    

    or

    

    where k is a unit vector perpendicular to the xy-plane. Verify (10) and (11) for F = [7x, −3y] and C the circle x² + y² = 4 as well as for an example of your own choice.

13–17 INTEGRAL OF THE NORMAL DERIVATIVE

Using (9), find the value of taken counterclockwise over the boundary C of the region R.

• 13. w = cosh x, R the triangle with vertices (0, 0), (4, 2), (0, 2).
• 14. w = x²y + xy², R: x² + y² 1, x 0, y 0
• 15. w = e^x cos y + xy³, R: 1 y 10 − x², x 0
• 16. W = x² + y², C; x² + y² = 4. Confirm the answer by direct integration.
• 17. w = x³ − y³, 0 y x², |x| 2
• 18. Laplace's equation. Show that for a solution w(x, y) of Laplace's equation ∇²w = 0 in a region R with boundary curve C and outer unit normal vector n,

  

• 19. Show that w = e^x sin y satisfies Laplace's equation ∇²w = 0 and, using (12), integrate w(∂w/∂n) counterclockwise around the boundary curve C of the rectangle 0 x 2, 0 y 5.
• 20. Same task as in Prob. 19 when w = x² + y² and C the boundary curve of the triangle with vertices (0, 0), (1, 0), (0, 1).

EXAMPLE 1 Parametric Representation of a Cylinder

The circular cylinder x² + y² = a², −1 z 1, has radius a, height 2, and the z-axis as axis. A parametric representation is

The components of r are x = a cos u, y = a sin u, z = υ. The parameters u, υ vary in the rectangle R; 0 u 2π, −1 υ 1 in the uυ-plane. The curves u = const are vertical straight lines. The curves υ = const are parallel circles. The point P in Fig. 242 corresponds to u = π/3 = 60°, υ = 0.7.

EXAMPLE 2 Parametric Representation of a Sphere

A sphere x² + y² + z² = a² can be represented in the form

where the parameters u, υ vary in the rectangle R in the uυ-plane given by the inequalities 0 u 2π, −π/2 υ π/2. The components of r are

The curves u = const and υ = const are the “meridians” and “parallels” on S (see Fig. 243). This representation is used ingeography for measuring the latitude and longitude of points on the globe.

Another parametric representation of the sphere also used in mathematics is

where 0 u 2π, 0 υ π.

EXAMPLE 3 Parametric Representation of a Cone

A circular cone can be represented by

in components x = u cos υ, y = u sin υ, z = u. The parameters vary in the rectangle R: 0 u H, 0 υ 2π. Check that x² + y² = z², as it should be. What are the curves u = const and υ = const?

THEOREM 1 Tangent Plane and Surface Normal

If a surface S is given by (2) with continuous r_u = ∂r/∂u and r_υ = ∂r/∂υ satisfying (4) at every point of S, then S has, at every point P, a unique tangent plane passing through P and spanned by r_u and r_υ, and a unique normal whose direction depends continuously on the points of S. A normal vector is given by (4) and the corresponding unit normal vector by (5). (See Fig. 244.)

EXAMPLE 4 Unit Normal Vector of a Sphere

From (5*) we find that the sphere g(x, y, z) = x² + y² + z² − a² = 0 has the unit normal vector

We see that n has the direction of the position vector [x, y, z] of the corresponding point. Is it obvious that this must be the case?

EXAMPLE 5 Unit Normal Vector of a Cone

At the apex of the cone in Example 3, the unit normal vector n becomes undetermined because from (5*) we get

We are now ready to discuss surface integrals and their applications, beginning in the next section.

PROBLEM SET 10.5

1–8 PARAMETRIC SURFACE REPRESENTATION

Familiarize yourself with parametric representations of important surfaces by deriving a representation (1), by finding the parameter curves (curves u = const and υ = const) of the surface and a normal vector N = r_u × r_υ of the surface. Show the details of your work.

1. xy-plane r(u, υ) = (u, υ) (thus ui + υj; similarly in Probs. 2–8).
2. xy-plane in polar coordinates r(u, υ) = [u cos υ, u sin υ] (thus u = r, υ = θ)
3. Cone r(u, υ) = [u cos υ, u sin υ, cu]
4. Elliptic cylinder r(u, υ) = [a cos υ, b sin υ, u]
5. Paraboloid of revolution r(u, υ) = [u cos υ, u sin υ, υ²]
6. Helicoid r(u, υ) = [u cos υ, u sin υ, υ]. Explain the name.
7. Ellipsoid r(u, υ) = [a cos υ cos u, b cos υ sin u, c sin υ]
8. Hyperbolic paraboloid r(u, υ) = [au cosh υ, bu sinh υ, u²]
9. CAS EXPERIMENT. Graphing Surfaces, Dependence on a, b, c. Graph the surfaces in Probs. 3–8. In Prob. 6 generalize the surface by introducing parameters a, b. Then find out in Probs. 4 and 6–8 how the shape of the surfaces depends on a, b, c.
10. Orthogonal parameter curves u = const and υ = const on r(u, υ) occur if and only if r_u • r_υ = 0. Give examples. Prove it.
11. Satisfying (4). Represent the paraboloid in Prob. 5 so that and show .
12. Condition (4). Find the points in Probs. 1–8 at which (4) N ≠ 0 does not hold. Indicate whether this results from the shape of the surface or from the choice of the representation.
13. Representation z = f(x, y). Show that z = f(x, y) or g = z − f(x, y) = 0 can be written (f_u = ∂f/∂u etc.)

    

14–19 DERIVE A PARAMETRIC REPRESENTATION

Find a normal vector. The answer gives one representation; there are many. Sketch the surface and parameter curves.

• 14. Plane 4x + 3y + 2z = 12
• 15. Cylinder of revolution (x − 2)² + (y + 1)² = 25
• 16. Ellipsoid
• 17. Sphere x² + (y + 2.8)² + (z − 3.2)² = 2.25
• 18. Elliptic cone
• 19. Hyperbolic cylinder x² − y² = 1
• 20. PROJECT. Tangent Planes T(P) will be less important in our work, but you should know how to represent them.

  (a) If S: r(u, υ), then T(P): (r* − r r_u r_υ) = 0 (a scalar triple product) or

  

  (b) If S: g(x, y, z) = 0, then

  

  (c) If S: z = f(x, y), then

  

  Interpret (a)−(c) geometrically. Give two examples for (a), two for (b), and two for (c).

EXAMPLE 1 Flux Through a Surface

Compute the flux of water through the parabolic cylinder S: y = x², 0 x 2, 0 z 3 (Fig. 245) if the velocity vector is v = F = [3z², 6, 6xy], speed being measured in meters/sec. (Generally, F = ρv, but water has the density ρ = 1 g/cm³ = 1 ton/m³.)

Fig. 245. Surface S in Example 1

Solution. Writing x = u and z = υ, we have y = x² = u². Hence a representation of S is

By differentiation and by the definition of the cross product,

On S, writing simply F(S) for F[r(u, υ)], we have F(S) = [3υ², 6, 6uυ]. Hence F(S) • N = 6uυ² − 6. By integration we thus get from (3) the flux

or 72,000 liters/sec. Note that the y-component of F is positive (equal to 6), so that in Fig. 245 the flow goes from left to right.

Let us confirm this result by (5). Since

we see that cos α > 0, cos β < 0, and cos γ. Hence the second term of (5) on the right gets a minus sign, and the last term is absent. This gives, in agreement with the previous result,

EXAMPLE 2 Surface Integral

Evaluate (3) when F = [x², 0, 3y²] and S is the portion of the plane x + y + z = 1 in the first octant (Fig. 246).

Solution. Writing x = u, and y = υ, we have z = 1 − x − y = 1 − u − υ. Hence we can represent the plane x + y + z = 1 in the form r(u, υ) = [u, υ, 1, − u − υ]. We obtain the first-octant portion S of this plane by restricting x = u and y = υ to the projection R of S in the xy-plane. R is the triangle bounded by the two coordinate axes and the straight line x + y = 1, obtained from x + y + z = 1 by setting z = 0. Thus 0 x 1 − y, 0 y 1.

Fig. 246. Portion of a plane in Example 2

By inspection or by differentiation,

Hence F(S) • N = [u², 0, 3υ²] • [1, 1, 1] = u² + 3υ². By (3),

THEOREM 1 Change of Orientation in a Surface Integral

The replacement of n by −n (hence of N by −N) corresponds to the multiplication of the integral in (3) or (4) by −1.

EXAMPLE 3 Change of Orientation in a Surface Integral

In Example 1 we now represent S by , 0 υ 2, 0 u 3. Then

For F = [3z², 6, 6xz] we now get . Hence and integration gives the old result times −1,

EXAMPLE 4 Area of a Sphere

For a sphere r(u, υ) = [a cos υ cos u, a cos υ sin u, a sin υ], 0 u 2π, −π/2 u π/2 [see (3) in Sec. 10.5], we obtain by direct calculation (verify!)

Using cos² u + sin² u = 1 and then cos² υ + sin² υ = 1, we obtain

With this, (8) gives the familiar formula (note that |cos υ| = cos υ when −π/2 υ π/2)

EXAMPLE 5 Torus Surface (Doughnut Surface): Representation and Area

A torus surface S is obtained by rotating a circle C about a straight line L in space so that C does not intersect or touch L but its plane always passes through L. If L is the z-axis and C has radius b and its center has distance a (> b) from L, as in Fig. 249, then S can be represented by

where 0 u 2π, 0 υ 2π. Thus

Hence |r_u × r_υ| = b(a × b cos υ), and (8) gives the total area of the torus,

EXAMPLE 6 Moment of Inertia of a Surface

Find the moment of inertia I of a spherical lamina S: x² + y² + z² = a² of constant mass density and total mass M about the z-axis.

Solution. If a mass is distributed over a surface S and μ(x, y, z) is the density of the mass (= mass per unit area), then the moment of inertia I of the mass with respect to a given axis L is defined by the surface integral

where D(x, y, z) is the distance of the point (x, y, z) from L. Since, in the present example, μ is constant and S has the area A = 4πa², we have μ = M/A = M/(4πa²).

For S we use the same representation as in Example 4. Then D² = x² + y² = a² cos² υ. Also, as in that example, dA = a² cos υ du dυ. This gives the following result. [In the integration, use cos³ υ = cos υ (1 − sin² υ).]

PROBLEM SET 10.6

1–10 FLUX INTEGRALS (3)

Evaluate the integral for the given data. Describe the kind of surface. Show the details of your work.

1. F = [−x², y², 0], S: r = [u, υ, 3u − 2υ], 0 u 1.5, −2 υ 2
2. F = [e^y, e^x, 1], S: x + y + z = 1, x 0, y 0, z 0
3. F = [0, x, 0], S: x² + y² + z² = 1, x 0, y 0, z 0
4. F = [e^y, −e^z, e^x], S: x² + y² = 25, x 0, y 0, 0 z 2
5. F = [x, y, z], S: r = [u cos υ, u sin υ, u²], 0 u 4, −π υ π
6. F = [cosh y, 0, sinh x], S; z = x + y², 0 y x, 0 x 1
7. F = [0, sin y, cos z], S the cylinder x = y², where 0 y π/4 and 0 z y
8. F = [tan xy, x, y], S: y² + z² = 1, 2 x 5, y 0, z 0
9. F = [0, sinh z, cosh x], S: x² + z² = 4, , 0 y 5, z 0
10. F = [y², x², z⁴], , 0 z 8, y 0
11. CAS EXPERIMENT. Flux Integral. Write a program for evaluating surface integrals (3) that prints intermediate results (F, F • N, the integral over one of the two variables). Can you obtain experimentally some rules on functions and surfaces giving integrals that can be evaluated by the usual methods of calculus? Make a list of positive and negative results.

12–16 SURFACE INTEGRALS (6)

Evaluate these integrals for the following data. Indicate the kind of surface. Show the details.

• 12. G = cos x + sin x, S the portion of x + y + z = 1 in the first octant
• 13. G = x + y + z, z = x + 2y, 0 x π, 0 y x
• 14. G = ax + by + cz, S: x² + y² + z² = 1, y = 0, z = 0
• 15. G = (1 + 9xz)^3/2, S: r = [u, υ, u³], 0 u 1, −2 υ 2
• 16. G = arctan (y/x), S: z = x² + y², 1 z 9, x 0, y 0
• 17. Fun with Möbius. Make Möbius strips from long slim rectangles R of grid paper (graph paper) by pasting the short sides together after giving the paper a half-twist. In each case count the number of parts obtained by cutting along lines parallel to the edge. (a) Make R three squares wide and cut until you reach the beginning. (b) Make R four squares wide. Begin cutting one square away from the edge until you reach the beginning. Then cut the portion that is still two squares wide. (c) Make R five squares wide and cut similarly. (d) Make R six squares wide and cut. Formulate a conjecture about the number of parts obtained.
• 18. Gauss “Double Ring” (See Möbius, Works 2, 518–559). Make a paper cross (Fig. 251) into a “double ring” by joining opposite arms along their outer edges (without twist), one ring below the plane of the cross and the other above. Show experimentally that one can choose any four boundary points A, B, C, D and join A and C as well as B and D by two nonintersecting curves. What happens if you cut along the two curves? If you make a half-twist in each of the two rings and then cut? (Cf. E. Kreyszig, Proc. CSHPM 13 (2000), 23–43.)

  

  Fig. 251. Problem 18. Gauss “Double Ring”

APPLICATIONS

• 19. Center of gravity. Justify the following formulas for the mass M and the center of gravity of a lamina S of density (mass per unit area) σ(x, y, z) in space:

  

• 20. Moments of inertia. Justify the following formulas for the moments of inertia of the lamina in Prob. 19 about the x-, y-, and z-axes, respectively:

  

• 21. Find a formula for the moment of inertia of the lamina in Prob. 20 about the line y = x, z = 0.

22–23 Find the moment of inertia of a lamina S of density 1 about an axis B, where

• 22. S: x² + y² = 1, 0 z, h, B: the line z = h/2 in the xz- plane
• 23. S: x² + y² = z², 0 z h, B: the z-axis
• 24. Steiner's theorem.⁶ If I_B is the moment of inertia of a mass distribution of total mass M with respect to a line B through the center of gravity, show that its moment of inertia I_K with respect to a line K, which is parallel to B and has the distance k from it is

  

• 25. Using Steiner's theorem, find the moment of inertia of a mass of density 1 on the sphere S: x² + y² + z² = 1 about the line K: x = 1, y = 0 from the moment of inertia of the mass about a suitable line B, which you must first calculate.
• 26. TEAM PROJECT. First Fundamental Form ofS. Given a surface S: r(u, υ), the differential form

  

  with coefficients (in standard notation, unrelated to F, G elsewhere in this chapter)

  

  is called the first fundamental form of S. This form is basic because it permits us to calculate lengths, angles, and areas on S. To show this prove (a)–(c):

  (a) For a curve C: u = u(t), υ = υ(t), a t b, on S, formulas (10), Sec. 9.5, and (14) give the length

  

  (b) The angle γ between two intersecting curves C₁: u = g(t), υ = h(t) and C₂: u = p(t), υ = q(t) on S: r(u, v) is obtained from

  

  where a = r_ug′ + r_υh′ and b = r_up′ + r_υq′ are tangent vectors of C₁ and C₂.

  (c) The square of the length of the normal vector N can be written

  

  so that formula (8) for the area A(S) of S becomes

  

  (d) For polar coordinates u (= r) and υ (= θ) defined by x = u cos υ, y = u sin υ we have E = 1, F = 0, G = u², so that

  

  Calculate from this and (18) the area of a disk of radius a.

  (e) Find the first fundamental form of the torus in Example 5. Use it to calculate the area A of the torus. Show that A can also be obtained by the theorem of Pappus,⁷ which states that the area of a surface of revolution equals the product of the length of a meridian C and the length of the path of the center of gravity of C when C is rotated through the angle 2π.

  (f) Calculate the first fundamental form for the usual representations of important surfaces of your own choice (cylinder, cone, etc.) and apply them to the calculation of lengths and areas on these surfaces.

THEOREM 1 Divergence Theorem of Gauss (Transformation Between Triple and Surface Integrals)

Let T be a closed bounded region in space whose boundary is a piecewise smooth orientable surface S. Let F(x, y, z) be a vector function that is continuous and has continuous first partial derivatives in some domain containing T. Then

In components of F = [F₁, F₂, F₃] and of the outer unit normal vector n = [cos α, cos β, cos γ] of S (as in Fig. 253), formula (2) becomes

EXAMPLE 1 Evaluation of a Surface Integral by the Divergence Theorem

Before we prove the theorem, let us show a typical application. Evaluate

where S is the closed surface in Fig. 252 consisting of the cylinder x² + y² = a²(0 z b) and the circular disks z = 0 and z = b(x² + y² a²).

Fig. 252. Surface S in Example 1

Solution.. F₁ = x³, F₂ = x²y, F₃ = x²z. Hence div F = 3x² + x² + x² = 5x². The form of the surface suggests that we introduce polar coordinates r, θ defined by x = r cos θ, y = r sin θ (thus cylindrical coordinates r, θ, z). Then the volume element is dx dy dz = r dr dθ dz, and we obtain

PROOF

We prove the divergence theorem, beginning with the first equation in (2*). This equation is true if and only if the integrals of each component on both sides are equal; that is,

EXAMPLE 2 Verification of the Divergence Theorem

Evaluate over the sphere S: x² + y² + z² = 4 (a) by (2), (b) directly.

Solution. (a) div F = div [7x, 0, −z] = div [7xi − zk] = 7 − 1 = 6. Answer:6 · .

(b) We can represent S by (3), Sec. 10.5 (with a = 2), and we shall use n dA = N du dυ [see (3*), Sec. 10.6]. Accordingly,

Then

Now on S we have x = 2 cos υ cos u, z = 2 sin υ, so that F = [7x, 0, −z] becomes on S

and

On S we have to integrate over u from 0 to 2π. This gives

The integral of cos υ sin² υ equals (sin³ υ)/3, and that of cos³ υ = cos υ (1 − sin² υ) equals sin υ − (sin³ υ)/3. On S we have −π/2 υ π/2, so that by substituting these limits we get

as hoped for. To see the point of Gauss's theorem, compare the amounts of work.

THEOREM 2 Invariance of the Divergence

The divergence of a vector function F with continuous first partial derivatives in a region T is independent of the particular choice of Cartesian coordinates. For any P in T it is given by (11).

PROBLEM SET 10.7

1–8 APPLICATION: MASS DISTRIBUTION

Find the total mass of a mass distribution of density σ in a region T in space.

1. σ = x² + y² + z², T the box |x| 4, |y| 1, 0 z 2
2. σ = xyz, T the box 0 x a, 0 y b, 0 z c
3. σ = e^−x−y−z, T: 0 x 1 − y, 0 y 1, 0 z 2
4. σ as in Prob. 3, T the tetrahedron with vertices (0, 0, 0), (3, 0, 0), (0, 3, 0), (0, 0, 3)
5. σ = sin 2x cos 2y, , , 0 z 6
6. σ x²y²z², T the cylindrical region x² + z² 16, |y| 4
7. σ = arctan (y/x), T: x² + y² + z² a², z 0
8. σ = x² + y², T as in Prob. 7

9–18 APPLICATION OF THE DIVERGENCE THEOREM

Evaluate the surface integral by the divergence theorem. Show the details.

• 9. F = [x², 0, z²], S the surface of the box |x| 1, |y| 3, 0 z 2
• 10. Solve Prob. 9 by direct integration.
• 11. F = [e^x, e^y, e^z], S the surface of the cube |x| 1, |y| 1, |z| 1
• 12. F = [x³ − y³, y³ − z³, z³ − x³], S the surface of x² + y² + z² 25, z 0
• 13. F = [sin y, cos x, cos z], S, the surface of x² + y² 4, |z| 2 (a cylinder and two disks!)
• 14. F as in Prob. 13, S the surface of x² + y² 9, 0 z 2
• 15. , S the surface of the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)
• 16. F = [cosh x, z, y], S as in Prob. 15
• 17. F = [x², y², z²], S the surface of the cone x² + y² z², 0 z h
• 18. F = [xy, yz, zx], S the surface of the cone x² + y² 4z², 0 z 2

19–23 APPLICATION: MOMENT OF INERTIA

Given a mass of density 1 in a region T of space, find the moment of intertia about the x-axis

• 19. The box −a x a, −b y b, −c z c
• 20. The ball x² + y² + z² a²
• 21. The cylinder y² + z² a², 0 x h
• 22. The paraboloid y² + z² x, 0 x h
• 23. The cone y² + z² x², 0 x h
• 24. Why is I_x in Prob. 23 for large h larger than I_x in Prob. 22 (and the same h)? Why is it smaller for h = 1? Give physical reason.
• 25. Show that for a solid of revolution, . Solve Probs. 20–23 by this formula.

EXAMPLE 1 Fluid Flow. Physical Interpretation of the Divergence

From the divergence theorem we may obtain an intuitive interpretation of the divergence of a vector. For this purpose we consider the flow of an incompressible fluid (see Sec. 9.8) of constant density ρ = 1 which is steady, that is, does not vary with time. Such a flow is determined by the field of its velocity vector v(P) at any point P.

Let S be the boundary surface of a region T in space, and let n be the outer unit normal vector of S. Then v • n is the normal component of v in the direction of n, and |v • n dA| is the mass of fluid leaving T (if v • n > 0 at some P) or entering T (if v • n ≤ 0 at P) per unit time at some point P of S through a small portion ΔS of S area ΔA. Hence the total mass of fluid that flows across S from T to the outside per unit time is given by the surface integral

Division by the volume V of T gives the average flow out of T:

Since the flow is steady and the fluid is incompressible, the amount of fluid flowing outward must be continuously supplied. Hence, if the value of the integral (1) is different from zero, there must be sources (positive sources and negative sources, called sinks) in T, that is, points where fluid is produced or disappears.

If we let T shrink down to a fixed point P in T, we obtain from (1) the source intensity at P given by the right side of (11) in the last section with F • n replaced by v • n, that is,

Hence the divergence of the velocity vector v of a steady incompressible flow is the source intensity of the flow at the corresponding point.

There are no sources in T if and only if div v is zero everywhere in T. Then for any closed surface S in T we have

EXAMPLE 2 Modeling of Heat Flow. Heat or Diffusion Equation

Physical experiments show that in a body, heat flows in the direction of decreasing temperature, and the rate of flow is proportional to the gradient of the temperature. This means that the velocity v of the heat flow in a body is of the form

where U(x, y, z, t) is temperature, t is time, and K is called the thermal conductivity of the body; in ordinary physical circumstances K is a constant. Using this information, set up the mathematical model of heat flow, the so-called heat equation or diffusion equation.

Solution. Let T be a region in the body bounded by a surface S with outer unit normal vector n such that the divergence theorem applies. Then v · n is the component of v in the direction of n, and the amount of heat leaving T per unit time is

This expression is obtained similarly to the corresponding surface integral in the last example. Using

(the Laplacian; see (3) in Sec. 9.8), we have by the divergence theorem and (3)

On the other hand, the total amount of heat H in T is

where the constant σ is the specific heat of the material of the body and ρ is the density (= mass per unit volume) of the material. Hence the time rate of decrease of H is

and this must be equal to the above amount of heat leaving T. From (4) we thus have

or

Since this holds for any region T in the body, the integrand (if continuous) must be zero everywhere; that is,

where c² is called the thermal diffusivity of the material. This partial differential equation is called the heat equation. It is the fundamental equation for heat conduction. And our derivation is another impressive demonstration of the great importance of the divergence theorem. Methods for solving heat problems will be shown in Chap. 12.

The heat equation is also called the diffusion equation because it also models diffusion processes of motions of molecules tending to level off differences in density or pressure in gases or liquids.

If heat flow does not depend on time, it is called steady-state heat flow. Then ∂U/∂t = 0, so that (5) reduces to Laplace's equation ∇²U = 0. We met this equation in Secs. 9.7 and 9.8, and we shall now see that the divergence theorem adds basic insights into the nature of solutions of this equation.

EXAMPLE 3 A Basic Property of Solutions of Laplace's Equation

The integrands in the divergence theorem are div F and F • n (Sec. 10.7). If F is the gradient of a scalar function, say, F = grad f, then div F = div (grad f) = ∇²f; see (3), Sec. 9.8. Also, F • n = n • F = n • grad f. This is the directional derivative of f in the outer normal direction of S, the boundary surface of the region T in the theorem. This derivative is called the (outer) normal derivative of f and is denoted by ∂f/∂n. Thus the formula in the divergence theorem becomes

This is the three-dimensional analog of (9) in Sec. 10.4. Because of the assumptions in the divergence theorem this gives the following result.

THEOREM 1 A Basic Property of Harmonic Functions

Let f(x, y, z) be a harmonic function in some domain D is space. Let S be any piecewise smooth closed orientable surface in D whose entire region it encloses belongs to D. Then the integral of the normal derivative of f taken over S is zero. (For “piecewise smooth” see Sec. 10.5.)

EXAMPLE 4 Green's Theorems

Let f and g be scalar functions such that F = f grad g satisfies the assumptions of the divergence theorem in some region T. Then

Also, since f is a scalar function,

Now n • grad g is the directional derivative ∂g/∂n of g in the outer normal direction of S. Hence the formula in the divergence theorem becomes “Green's first formula”

Formula (8) together with the assumptions is known as the first form of Green's theorem.

Interchanging f and g we obtain a similar formula. Subtracting this formula from (8) we find

This formula is called Green's second formula or (together with the assumptions) the second form of Green's theorem.

EXAMPLE 5 Uniqueness of Solutions of Laplace's Equation

Let f be harmonic in a domain D and let f be zero everywhere on a piecewise smooth closed orientable surface S in D whose entire region T it encloses belongs to D. Then ∇²g is zero in T, and the surface integral in (8) is zero, so that (8) with g = f gives

Since f is harmonic, grad f and thus |grad f| are continuous in T and on S, and since |grad f| is nonnegative, to make the integral over T zero, grad f must be the zero vector everywhere in T. Hence f_x = f_y = f_z = 0, and f is constant in T and, because of continuity, it is equal to its value 0 on S. This proves the following theorem.

THEOREM 2 Harmonic Functions

Let f(x, y, z) be harmonic in some domain D and zero at every point of a piecewise smooth closed orientable surface S in D whose entire region T it encloses belongs to D. Then f is identically zero in T.

THEOREM 3 Uniqueness Theorem for Laplace's Equation

Let T be a region that satisfies the assumptions of the divergence theorem, and let f(x, y, z) be a harmonic function in a domain D that contains T and its boundary surface S. Then f is uniquely determined in T by its values on S.

THEOREM 3* Uniqueness Theorem for the Dirichlet Problem

If the assumptions in Theorem 3 are satisfied and the Dirichlet problem for the Laplace equation has a solution in T, then this solution is unique.

PROBLEM SET 10.8

1–6 VERIFICATIONS

1. Harmonic functions. Verify Theorem 1 for f = 2z² − x² − y² and S the surface of the box 0 x a, 0 y b, 0 z c.
2. Harmonic functions. Verify Theorem 1 for f = x² − y² and the surface of the cylinder x² + y² = 4, 0 z h.
3. Green's first identity. Verify (8) for f = 4y², g = x², S the surface of the “unit cube” 0 x 1, 0 y 1, 0 z 1. What are the assumptions on f and g in (8)? Must f and g be harmonic?
4. Green's first identity. Verify (8) for f = x, g = y² + z², S the surface of the box 0 x 1, 0 y 2, 0 z 3.
5. Green's second identity. Verify (9) for f = 6y², g = 2x², S the unit cube in Prob. 3.
6. Green's second identity. Verify (9) for f = x², g = y⁴, S the unit cube in Prob. 3.

7–11 VOLUME

Use the divergence theorem, assuming that the assumptions on T and S are satisfied.

• 7. Show that a region T with boundary surface S has the volume

  

• 8. Cone. Using the third expression for υ in Prob. 7, verify V = πa² h/3 for the volume of a circular cone of height h and radius of base a.
• 9. Ball. Find the volume under a hemisphere of radius a from in Prob. 7.
• 10. Volume. Show that a region T with boundary surface S has the volume

  

  where r is the distance of a variable point P: (x, y, z) on S from the origin O and φ is the angle between the directed line OP and the outer normal of S at P. Make a sketch. Hint. Use (2) in Sec. 10.7 with F = [x, y, z].

• 11. Ball. Find the volume of a ball of radius a from Prob. 10.
• 12. TEAM PROJECT. Divergence Theorem and Potential Theory. The importance of the divergence theorem in potential theory is obvious from (7)–(9) and Theorems 1–3. To emphasize it further, consider functions f and g that are harmonic in some domain D containing a region T with boundary surface S such that T satisfies the assumptions in the divergence theorem. Prove, and illustrate by examples, that then:

  (a)

  (b) If ∂g/∂n = 0 on S, then g is constant in T.

  (c)

  (d) If ∂f/∂n = ∂g/∂n on S, then f = g + c in T, where c is a constant.

  (e) The Laplacian can be represented independently of coordinate systems in the form

  

  where d(T) is the maximum distance of the points of a region T bounded by S(T) from the point at which the Laplacian is evaluated and V(T) is the volume of T.

THEOREM 1 Stokes's Theorem⁹ (Transformation Between Surface and Line Integrals)

Let S be a piecewise smooth⁹ oriented surface in space and let the boundary of S be a piecewise smooth simple closed curve C. Let F(x, y, z) be a continuous vector function that has continuous first partial derivatives in a domain in space containing S. Then

Here n is a unit normal vector of S and, depending on n, the integration around C is taken in the sense shown in Fig. 254. Furthermore, r′ = dr/ds is the unit tangent vector and s the arc length of C.

In components, formula (2) becomes

Here,, F = [F₁, F₂, F₃], N = [N₁, N₂, N₃], n dA = N du dυ, r′ ds = [dx, dy, dz], and R is the region with boundary curve in the uυ-plane corresponding to S represented by r(u, υ).

EXAMPLE 1 Verification of Stokes's Theorem

Before we prove Stokes's theorem, let us first get used to it by verifying it for F = [y, z, x] and S the paraboloid (Fig. 255)

Solution. The curve C, oriented as in Fig. 255, is the circle r(s) = [cos s, sin s, 0]. Its unit tangent vector is r′(s) = [−sin s, cos s, 0]. The function F = [y, z, x] on C is F (r(s)) = [sin s, 0, cos s]. Hence

We now consider the surface integral. We have F₁ = y, F₂ = z, F₃ = x, so that in (2*) we obtain

A normal vector of S is N = grad (z − f(x, y)) = [2x, 2y, 1]. Hence (curl F) • N = −2x − 2y − 1. Now n dA = N dx dy (see (3*) in Sec. 10.6 with x, y instead of u, υ). Using polar coordinates r, θ defined by x = r cos θ, y = r sin θ and denoting the projection of S into the xy-plane by R, we thus obtain

PROOF

We prove Stokes's theorem. Obviously, (2) holds if the integrals of each component on both sides of (2*) are equal; that is,

EXAMPLE 2 Green's Theorem in the Plane as a Special Case of Stokes's Theorem

Let F = [F₁, F₂] = F₁i + F₂j be a vector function that is continuously differentiable in a domain in the xy-plane containing a simply connected bounded closed region S whose boundary C is a piecewise smooth simple closed curve. Then, according to (1),

Hence the formula in Stokes's theorem now takes the form

This shows that Green's theorem in the plane (Sec. 10.4) is a special case of Stokes's theorem (which we needed in the proof of the latter!).

EXAMPLE 3 Evaluation of a Line Integral by Stokes's Theorem

Evaluate ∫_C F • r′ ds, where C is the circle x² + y² = 4, z = −3, oriented counterclockwise as seen by a person standing at the origin, and, with respect to right-handed Cartesian coordinates,

Solution. As a surface S bounded by C we can take the plane circular disk x² + y² 4 in the plane z = −3. Then n in Stokes's theorem points in the positive z-direction; thus n = k. Hence (curl F) • n is simply the component of curl F in the positive z-direction. Since F with z = −3 has the components F₁ = y, F₂ = −27x, F₃ = 3y³, we thus obtain

Hence the integral over S in Stokes's theorem equals −28 times the area 4π of the disk S. This yields the answer −28 · 4π = −112π ≈ −352. Confirm this by direct calculation, which involves somewhat more work.

EXAMPLE 4 Physical Meaning of the Curl in Fluid Motion. Circulation

Let S_ro be a circular disk of radius r₀ and center P bounded by the circle C_r0 (Fig. 257), and let F(Q) ≡ F(x, y, z) be a continuously differentiable vector function in a domain containing S_r0. Then by Stokes's theorem and the mean value theorem for surface integrals (see Sec. 10.6),

where A_r0 is the area of S_r0 and P* is a suitable point of S_r0. This may be written in the form

Fig. 257. Example 4

In the case of a fluid motion with velocity vector F = v, the integral

is called the circulation of the flow around C_r0. It measures the extent to which the corresponding fluid motion is a rotation around the circle C_r0. If we now let r₀ approach zero, we find

that is, the component of the curl in the positive normal direction can be regarded as the specific circulation (circulation per unit area) of the flow in the surface at the corresponding point.

EXAMPLE 5 Work Done in the Displacement around a Closed Curve

Find the work done by the force F = 2xy³ sin zi + 3x²y² sin zj + x²y³ cos z k in the displacement around the curve of intersection of the paraboloid z = x² + y² and the cylinder (x − 1)² + y² = 1.

Solution. This work is given by the line integral in Stokes's theorem. Now F = grad f, where f = x²y³ sin z and curl (grad f) = 0 (see (2) in Sec. 9.9), so that (curl F) • n = 0 and the work is 0 by Stokes's theorem. This agrees with the fact that the present field is conservative (definition in Sec. 9.7).

PROBLEM SET 10.9

1–10 DIRECT INTEGRATION OF SURFACE INTEGRALS

Evaluate the surface integral directly for the given F and S.

1. F = [z², −x², 0], S the rectangle with vertices (0, 0, 0), (1, 0, 0), (0, 4, 4), (1, 4, 4)
2. F = [−13 sin y, 3 sinh z, x], S the rectangle with vertices (0, 0, 2), (4, 0, 2), (4, π/2, 2), (0, π/2, 2)
3. F = [e^−z, e^−z cos y, e^−z sin y], S: z = y²/2, −1 x 1, 0 y 1
4. F as in Prob. 1, z = xy (0 x 1, 0 y 4). Compare with Prob. 1.
5. , S: 0 x a, 0 y a, z = 1
6. F = [y³, −x³, 0], S: x² + y² 1, z = 0
7. F = [e^y, e^z, e^x], S: z = x² (0 x 2, 0 y 1)
8. F = [z², x², y²], , y 0, 0 z h
9. Verify Stokes's theorem for F and S in Prob. 5.
10. Verify Stokes's theorem for F and S in Prob. 6.
11. Stokes's theorem not applicable. Evaluate , F = (x² + y²)⁻¹ [−y, x], C: x² + y² = 1, z = 0, oriented clockwise. Why can Stokes's theorem not be applied? What (false) result would it give?
12. WRITING PROJECT. Grad, Div, Curl in Connection with Integrals. Make a list of ideas and results on this topic in this chapter. See whether you can rearrange or combine parts of your material. Then subdivide the material into 3–5 portions and work out the details of each portion. Include no proofs but simple typical examples of your own that lead to a better understanding of the material.

13–20 EVALUATION OF

Calculate this line integral by Stokes's theorem for the given F and C. Assume the Cartesian coordinates to be right-handed and the z-component of the surface normal to be nonnegative.

• 13. F = [−5y, 4x, z], C the circle x² + y² = 16, z = 4
• 14. F = [z³, x³, y³], C the circle x = 2, y² + z² = 9
• 15. F = [y², x², z + x] around the triangle with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0)
• 16. F = [e^y, 0, e^x], C as in Prob. 15
• 17. F = [0, z³, 0], C the boundary curve of the cylinder x² + y² = 1, x 0, y 0, 0 z 1
• 18. F = [−y, 2z, 0], C the boundary curve of y² + z² = 4, z 0, 0 x h
• 19. F = [z, e^z, 0], C the boundary curve of the portion of the cone , x 0, y 0, 0 z 1
• 20. F = [0, cos x, 0], C the boundary curve of y² + z² = 4, y 0, z 0, 0 x π

Chapter 9 extended differential calculus to vectors, that is, to vector functions v(x, y, z) or v(t). Similarly, Chapter 10 extends integral calculus to vector functions. This involves line integrals (Sec. 10.1), double integrals (Sec. 10.3), surface integrals (Sec. 10.6), and triple integrals (Sec. 10.7) and the three “big” theorems for transforming these integrals into one another, the theorems of Green (Sec. 10.4), Gauss (Sec. 10.7), and Stokes (Sec. 10.9).

The analog of the definite integral of calculus is the line integral (Sec. 10.1)

where C: r(t) = [x(t), y(t), z(t)] = x(t)i + y(t)j + z(t)k (a t b) is a curve in space (or in the plane). Physically, (1) may represent the work done by a (variable) force in a displacement. Other kinds of line integrals and their applications are also discussed in Sec. 10.1.

Independence of path of a line integral in a domain D means that the integral of a given function over any path C with endpoints P and Q has the same value for all paths from P to Q that lie in D; here P and Q are fixed. An integral (1) is independent of path in D if and only if the differential form F₁ dx + F₂ dy + F₃ dz with continuous F₁, F₂, F₃ is exact in D (Sec. 10.2). Also, if curl F = 0, where F = [F₁, F₂, F₃], has continuous first partial derivatives in a simply connected domain D, then the integral (1) is independent of path in D (Sec. 10.2).

Integral Theorems. The formula of Green's theorem in the plane (Sec. 10.4)

transforms double integrals over a region R in the xy-plane into line integrals over the boundary curve C of R and conversely. For other forms of (2) see Sec. 10.4.

Similarly, the formula of the divergence theorem of Gauss (Sec. 10.7)

transforms triple integrals over a region T in space into surface integrals over the boundary surface S of T, and conversely. Formula (3) implies Green's formulas

Finally, the formula of Stokes's theorem (Sec. 10.9)

transforms surface integrals over a surface S into line integrals over the boundary curve C of S and conversely.