9.14 Stokes’ Theorem

INTRODUCTION

Green’s theorem of the preceding section has two vector forms. In this and in Section 9.16 we shall generalize these forms to three dimensions.

Vector Form of Green’s Theorem

If F(x, y) = P(x, y)i + Q(x, y)j is a two-dimensional vector field, then

From (12) and (13) of Section 9.8, Green’s theorem can be written in vector notation as

(1)

that is, the line integral of the tangential component of F is the double integral of the normal component of curl F.

Green’s Theorem in 3-Space

The vector form of Green’s theorem given in (1) relates a line integral around a piecewise-smooth simple closed curve C forming the boundary of a plane region R to a double integral over R. Green’s theorem in 3-space relates a line integral around a piecewise-smooth simple closed curve C forming the boundary of a surface S with a surface integral over S. Suppose z = f(x, y) is a continuous function whose graph is a piecewise-smooth orientable surface over a region R on the xy-plane. Let C form the boundary of S and let the projection of C onto the xy-plane form the boundary of R. The positive direction on C is induced by the orientation of the surface S; the positive direction on C corresponds to the direction a person would have to walk on C to have his or her head point in the direction of the orientation of the surface while keeping the surface to the left. See FIGURE 9.14.1. More precisely, the positive orientation of C is in accordance with the right-hand rule: If the thumb of the right hand points in the direction of the orientation of the surface, then roughly the fingers of the right hand wrap around the surface in the positive direction. Finally, let T be a unit tangent vector to C that points in the positive direction. The three-dimensional form of Green’s theorem, which we now give, is called Stokes’ theorem after the Irish mathematical physicist George G. Stokes (1819–1903).

A graph. A surface is graphed on a three dimensional x y z coordinate system. A closed curve labeled C goes forms the boundary of the surface. A vector labeled n is normal to the surface at a marked point. A vector labeled T is tangent to the curve C at another marked point. A person walks along the curve C with the head pointing in the direction of the vector n while keeping the graphed surface to the left. The curve C is projected vertically down on the x y plane. The region inside the closed projected curve in the x y plane is labeled R. Several arrows placed along the closed projected curve are oriented counter clockwise.

FIGURE 9.14.1 Boundary C of surface S has positive orientation

THEOREM 9.14.1 Stokes’ Theorem

Let S be a piecewise-smooth orientable surface bounded by a piecewise-smooth simple closed curve C. Let F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a vector field for which P, Q, and R are continuous and have continuous first partial derivatives in a region of 3-space containing S. If C is traversed in the positive direction, then

(2)

where n is a unit normal to S in the direction of the orientation of S.

PARTIAL PROOF:

Suppose the surface S is oriented upward and is defined by a function z = f(x, y) that has continuous second partial derivatives. From Definition 9.7.1 we have

Furthermore, if we write g(x, y, z) = z – f(x, y) = 0, then

Hence,

(3)

Our goal is now to show that F · dr reduces to (3).

If Cxy is the projection of C onto the xy-plane and has the parametric equations x = x(t), y = y(t), a ≤ t ≤ b, then parametric equations for C are x = x(t), y = y(t), z = f(x(t), y(t)), a ≤ t ≤ b. Thus,

(4)

Now,

(5)

Similarly,

(6)

Subtracting (6) from (5) and using the fact that ∂2f/∂x∂y = ∂2f/∂y∂x, we see that (4) becomes, after rearranging,

This last expression is the same as the right side of (3), which was to be shown. ≡

EXAMPLE 1 Verifying Stokes’ Theorem

Let S be the part of the cylinder z = 1 – x2 for 0 ≤ x ≤ 1, –2 ≤ y ≤ 2. Verify Stokes’ theorem for the vector field F = xyi + yzj + xzk. Assume S is oriented upward.

SOLUTION

The surface S, the curve C (which is composed of the union of C1, C2, C3, and C4), and the region R are shown in FIGURE 9.14.2.

2 graphs. Graph a. A quarter cylinder is graphed on a three dimensional x y z coordinate system. The quarter cylinder, labeled S: z = 1 minus x^2, 0 <= x <= 1, negative 2 <= y <= 2, is positioned horizontally. The central axis of the quarter cylinder is on the y axis. The length of the cylinder ranges from y = negative 2 to y = 2. The shaded lateral surface of the cylinder is bound by 2 lines and 2 curves. The first line, labeled C subscript 1, begins at the point (1, negative 2, 0), goes horizontally to the right and ends at the point (2, 2, 0). The first curve, labeled C subscript 2, begins at the point (2, 2, 0), goes upward following a circular path, and ends at the point (0, 2, 1). The second line, labeled c subscript 3, begins at the point (0, 2, 1), goes horizontally to the left, and ends at the point (0, negative 2, 1). The second curve, labeled C subscript 4, begins at the point (0, negative 2, 1), goes downward following a circular path, and ends at the point (1, negative 2, 0). The vertical projection of the surface on the x y plane is a rectangular region labeled R and bound by vertices at the following points: (2, negative 2, 0), (2, 2, 0), (0, 2, 0), and (0, negative 2, 0). Graph b. A rectangular region is graphed on an x y coordinate plane. The region is bound by the following vertices at the following points: (1, negative 2), (1, 2), (0, 2), and (0, negative 2). Several arrows on the rectangle are oriented counter clockwise. The region inside the rectangle is shaded and labeled R.

FIGURE 9.14.2 Surface S and region R in Example 1

Surface Integral: From F = xyi + yzj + xzk, we find

Now, if g(x, y, z) = z + x2 – 1 = 0 defines the cylinder, then the upper normal is

Therefore,

To evaluate the latter surface integral, we use (5) of Section 9.13:

(7)

Line Integral: We write =

On C1: = x = 1, z = 0, dx = 0, dz = 0, so

On C2: = y = 2, z = 1 – x2, dy = 0, dz = –2x dx, so

On C3: x = 0, z = 1, dx = 0, dz = 0, so

+ y dy + 0 = dy = 0.

On C4: y = –2, z = 1 – x2, dy = 0, dz = –2x dx, so

Hence,

which, of course, agrees with (7). ≡

EXAMPLE 2 Using Stokes’ Theorem

Evaluate z dx + x dy + y dz, where C is the trace of the cylinder x2 + y2 = 1 in the plane y + z = 2. Orient C counterclockwise as viewed from above. See FIGURE 9.14.3.

A graph. A cylinder and a plane is graphed on a three dimensional x y z coordinate system. The central axis of the cylinder is on the z axis, and the circular base of the cylinder on the x y plane is labeled x^2 + y^2 = 1. An inclined plane, labeled y + z = 2, intersects the cylinder. The intersection of the plane and the cylinder is a closed curve labeled C. The region inside the closed curve C is shaded and labeled S. Several arrows on the closed curve are oriented counter clockwise.

FIGURE 9.14.3 Curve C in Example 2

SOLUTION

If F = zi + xj + yk, then

curl F = = i + j + k.

The given orientation of C corresponds to an upward orientation of the surface S. Thus, if g(x, y, z) = y + z – 2 = 0 defines the plane, then the upper normal is

Hence, from (2),

≡

Note that if F is the gradient of a scalar function, then, in view of (5) in Section 9.7, (2) implies that the circulation F · dr is zero. Conversely, it can be shown that if the circulation is zero for every simple closed curve, then F is the gradient of a scalar function. In other words, F is irrotational if and only if F = ∇ϕ, where ϕ is a potential for F. Equivalently, this gives a test for a conservative vector field:

F is a conservative vector field if and only if curl F = 0.

Physical Interpretation of Curl

In Section 9.8 we saw that if F is a velocity field of a fluid, then the circulation F · dr of F around C is a measure of the amount by which the fluid tends to turn the curve C by circulating around it. The circulation of F is closely related to the curl of F. To see this, suppose P0(x 0, y0, z0) is any point in the fluid and Cr is a small circle of radius r centered at P0. See FIGURE 9.14.4. Then by Stokes’ theorem,

· dr = F) · n dS.(8)

A circular closed curve is labeled C subscript r. The center of the circle is labeled P subscript 0. The region inside the circle is shaded and labeled S subscript r. A vector labeled n(P subscript 0) begins at the point P subscript 0 and is normal to the surface S subscript r. 2 arrows on the circle are oriented counter clockwise.

FIGURE 9.14.4 Curve Cr and surface Sr in (8)

Now at all points P(x, y, z) within the small circle Cr, if we take curl F(P) ≈ curl F(P0), then (8) gives the approximation

(9)

where Ar is the area (πr2) of the circular surface Sr. As we let r → 0, the approximation curl F(P) ≈ curl F(P0) becomes better, and so (9) yields

(10)

Thus we see that the normal component of curl F is the limiting value of the ratio of the circulation of F to the area of the circular surface. For a small but fixed value of r, we have

(11)

Roughly then, the curl of F is the circulation of F per unit area. If curl F(P0) ≠ 0, then the left-hand side of (11) is a maximum when the circle Cr is situated in a manner so that n(P0) points in the same direction as curl F(P0). In this case, the circulation on the right side of (11) will also be a maximum. Thus, a paddle wheel inserted into the fluid at P0 will rotate fastest when its axis points in the direction of curl F(P0). See FIGURE 9.14.5. Note, too, that the paddle wheel will not rotate if its axis is perpendicular to curl F(P0).

A paddle wheel, consisting of 2 paddles, rotates counter clockwise around an axis at a point labeled P subscript 0. The circular surface covered by the rotating paddles is shaded. The following information is given beside the surface: curl F(P subscript 0).

FIGURE 9.14.5 Paddle wheel

REMARKS

The value of the surface integral in (2) is determined solely by the integral around its boundary C. This basically means that the shape of the surface S is irrelevant. Assuming that the hypotheses of Theorem 9.14.1 are satisfied, then for two different surfaces S1 and S2 with the same orientation and with the same boundary C, we have

See FIGURE 9.14.6 and Problems 17 and 18 in Exercises 9.14.

2 graphs. Graph a. A surface with 2 humps is graphed on a three dimensional x y z coordinate system. The surface is labeled S subscript 1. The surface is bound at the bottom by a closed curve labeled C. 2 arrows on the curve are oriented counter clockwise. A vector labeled n on the surface points upward to the left. Graph b. A surface with its curve side facing down is graphed on a three dimensional x y z coordinate system. The surface is labeled S subscript 1. The surface is bound at the top by a closed curve labeled C similar to the curve C in graph a. An arrow on the curve is oriented counter clockwise. A vector labeled n on the surface at the bottom points upward to the left.

FIGURE 9.14.6 Two surfaces with same boundary C

9.13 Exercises Answers to selected odd-numbered problems begin on page ANS-26.

In Problems 1–4, verify Stokes’ theorem. Assume that the surface S is oriented upward.

  1. F = 5yi – 5xj + 3k; S that portion of the plane z = 1 within the cylinder x2 + y2 = 4
  2. F = 2zi – 3xj + 4yk; S that portion of the paraboloid z = 16 – x2 – y2 for z ≥ 0
  3. F = zi + xj + yk; S that portion of the plane 2x + y + 2 z = 6 in the first octant
  4. F = xi + yj + zk; S that portion of the sphere x2 + y2 + z2 = 1 for z ≥ 0

In Problems 5–12, use Stokes’ theorem to evaluate F · dr. Assume C is oriented counterclockwise as viewed from above.

  1. F = (2z + x)i + (y – z)j + (x + y)k; C the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1)
  2. F = z2y cos xyi + z2x(1 + cos xy)j + 2z sin xyk; C the boundary of the plane z = 1 – y shown in FIGURE 9.14.7 on page 571.
    A graph. A surface given by a portion of a plane is graphed on a three dimensional x y z coordinate system. The portion of the plane is formed by a closed curve labeled C, consisting of 4 lines. The first line, graphed on the x y plane, begins at the point (0, 0, 1), and ends at the point (2, 0, 1). The second line begins at the point (2, 0, 1) and ends at the point (2, 1, 0). The third line begins at the point (2, 1, 0) and ends at the point (0, 1, 0). The fourth line, labeled z = 1 minus y, begins at the point (0, 1, 0) and ends at the point (0, 0, 1). The surface is shaded. 4 arrows on the closed curve C are oriented counter clockwise.

    FIGURE 9.14.7 Curve C for Problem 6

  3. F = xyi + 2yzj + xzk; C the boundary given in Problem 6
  4. F = (x + 2 z)i + (3x + y)j + (2y – z)k; C the curve of intersection of the plane x + 2y + z = 4 with the coordinate planes
  5. F = y3i – x3j + z3k; C the trace of the cylinder x2 + y2 = 1 in the plane x + y + z = 1
  6. F = x2yi + (x + y2)j + xy2zk; C the boundary of the surface shown in FIGURE 9.14.8
    A graph. A surface is graphed on a three dimensional x y z coordinate system. The surface is bound by a closed curve labeled C, consisting of 3 curves. The first curve, labeled z = 9 minus y^2, begins at the point (0, 3, 0), curves upward, and ends at the point (0, 0, 9). The second curve begins at the point (0, 0, 9), curves downward similar to the first curve, and ends at a point on the x y plane, on the line labeled y = 2 x. The third curve begins at the point the second curve ends, follows a circular path, and ends at the point (0, 3, 0). The surface is shaded. 3 arrows on the closed curve C are oriented counter clockwise.

    FIGURE 9.14.8 Curve C for Problem 10

  7. F = xi + x3y2j + zk; C the boundary of the semi-ellipsoid z = in the plane z = 0
  8. F = zi + xj + yk; C the curve of intersection of the plane x + y + z = 0 and the sphere x2 + y2 + z2 = 1 [Hint: Recall that the area of an ellipse x/a2 + y2/b2 = 1 is πab.]

In Problems 13–16, use Stokes’ theorem to evaluate (curl F) · n dS. Assume that the surface S is oriented upward.

  1. F = 6yzi + 5xj + yzk; S that portion of the paraboloid z = x2 + y2 for 0 ≤ z ≤ 4
  2. F = yi + (y – x)j + z2k; S that portion of the sphere x2 + y2 + (z – 4)2 = 25 for z ≥ 0
  3. F = 3x2i + 8x3yj + 3x2yk; S that portion of the plane z = x that lies inside the rectangular cylinder defined by the planes x = 0, y = 0, x = 2, y = 2
  4. F = 2xy2zi + 2x2yzj + (x2y2 – 6x)k; S that portion of the plane z = y that lies inside the cylinder x2 + y2 = 1
  5. Use Stokes’ theorem to evaluate

    z2dx + xy2 dy + tan−1 y dz,

    where C is the circle x2 + y2 = 9, by finding a surface S with C as its boundary and such that the orientation of C is counterclockwise as viewed from above.

  6. Consider the surface integral (curl F) · n dS, where F = xyzk and S is that portion of the paraboloid z = 1 – x2 – y2 for z ≥ 0 oriented upward.
    1. Evaluate the surface integral by the method of Section 9.13; that is, do not use Stokes’ theorem.
    2. Evaluate the surface integral by finding a simpler surface that is oriented upward and has the same boundary as the paraboloid.
    3. Use Stokes’ theorem to verify the result in part (b).