CalculusVolume3-OP_mktoy8b-161

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conservative, the curl of F is zero, so ∬
S
curl F · dS = 0. Since the boundary of S is a closed curve, ∫
C
F · dr is also zero.
Example 6.73
Verifying Stokes’ Theorem for a Specific Case
Verify that Stokes’ theorem is true for vector field F(x, y, z) = 〈 y, 2z, x2 〉 and surface S, where S is the
paravbolid z = 4 - x2 - y2 .
Figure 6.83 Verifying Stokes’ theorem for a hemisphere in a
vector field.
Solution
As a surface integral, you have g⎛
⎝x, y⎞
⎠ = 4 - x2 - y2, gx = −2y and
curl F = | i j k
∂
∂ x
∂
∂ y
∂
∂z
y 2z x2 | = ⟨−2, −2x, −1⟩.
By Equation 6.19,
∬
S
curl F · dS = ∬
D
curl F⎛
⎝r⎛
⎝ϕ, θ⎞
⎠
⎞
⎠ · ⎛
⎝tϕ × tθ
⎞
⎠dA
= ∬
D
〈 −2, −2x, −1 〉 · 〈 2x, 2y, 1 〉 dA
= ∫
−2
2
∫
4 - x2
4 − x2
⎛
⎝−4x−4xy−1⎞
⎠dydx
= ∫
−2
2 ⎛
⎝−8x 4 - x2−2 4 - x2⎞
⎠dx
= −4π
Chapter 6 | Vector Calculus 793
(6.22)
6.61
As a line integral, you can parameterize C by r(t) = 〈 2 cos t, 2 sin t, 0 〉 0 ≤ t ≤ 2π . By Equation 6.19,
∫
C
F · dr = ∫
0
2π
〈 2sin t, 0, 4cos2 t 〉 · 〈 −2sin t, 2cos t, 0 〉 dt
= ∫
0
2π
−4sin2 tdt = −4π
Therefore, we have verified Stokes' theorem for this example.
Verify that Stokes’ theorem is true for vector field F(x, y, z) = 〈 y, x, −z 〉 and surface S, where S is
the upwardly oriented portion of the graph of f (x, y) = x2 y over a triangle in the xy-plane with vertices
(0, 0), (2, 0), and (0, 2).
Applying Stokes’ Theorem
Stokes’ theorem translates between the flux integral of surface S to a line integral around the boundary of S. Therefore, the
theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the
line integral into a surface integral or vice versa. We now study some examples of each kind of translation.
Example 6.74
Calculating a Surface Integral
Calculate surface integral ∬
S
curl F · dS, where S is the surface, oriented outward, in Figure 6.84 and
F = 〈 z, 2xy, x + y 〉 .
794 Chapter 6 | Vector Calculus
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Figure 6.84 A complicated surface in a vector field.
Solution
Note that to calculate ∬
S
curl F · dS without using Stokes’ theorem, we would need to use Equation 6.19. Use
of this equation requires a parameterization of S. Surface S is complicated enough that it would be extremely
difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful
for this problem. Instead, we use Stokes’ theorem, noting that the boundary C of the surface is merely a single
circle with radius 1.
The curl of F is 〈 1, 1, 2y 〉 . By Stokes’ theorem,
∬
S
curl F · dS = ∫
C
F · dr,
where C has parameterization r (t) = 〈 −sin t, 0, 1 − cos t 〉 , 0 ≤ t < 2π. By Equation 6.9,
Chapter 6 | Vector Calculus 795
∬
S
curl F · dS = ∫
C
F · dr
= ∫
0
2π
〈 1 − cos t , 0, − sin t 〉 · 〈 − cos t, 0, sin t 〉 dt
= ∫
0
2π
⎛
⎝−cos t + cos2 t − sin2 t⎞
⎠dt
= ⎡
⎣−sin t + 1
2sin(2t)⎤
⎦0
2π
= ( − sin(2π) + 1
2 sin(4 π ))−( − sin 0 + 1
2sin 0)
= 0.
An amazing consequence of Stokes’ theorem is that if S′ is any other smooth surface with boundary C and the same
orientation as S, then ∬
S
curl F · dS = ∫
C
F · dr = 0 because Stokes’ theorem says the surface integral depends on the line
integral around the boundary only.
In Example 6.74, we calculated a surface integral simply by using information about the boundary of the surface. In
general, let S1 and S2 be smooth surfaces with the same boundary C and the same orientation. By Stokes’ theorem,
(6.23)∬
S1
curl F · dS = ∫
C
F · dr = ∬
S2
curl F · dS.
Therefore, if ∬
S1
curl F · dS is difficult to calculate but ∬
S2
curl F · dS is easy to calculate, Stokes’ theorem allows
us to calculate the easier surface integral. In Example 6.74, we could have calculated ∬
S
curl F · dS by calculating
∬
S′
curl F · dS, where S′ is the disk enclosed by boundary curve C (a much more simple surface with which to work).
Equation 6.23 shows that flux integrals of curl vector fields are surface independent in the same way that line integrals
of gradient fields are path independent. Recall that if F is a two-dimensional conservative vector field defined on a simply
connected domain, f is a potential function for F, and C is a curve in the domain of F, then ∫
C
F · dr depends only on the
endpoints of C. Therefore if C′ is any other curve with the same starting point and endpoint as C (that is, C′ has the same
orientation as C), then ∫
C
F · dr = ∫
C′
F · dr. In other words, the value of the integral depends on the boundary of the path
only; it does not really depend on the path itself.
Analogously, suppose that S and S′ are surfaces with the same boundary and same orientation, and suppose that G is a three-
dimensional vector field that can be written as the curl of another vector field F (so that F is like a “potential field” of G).
By Equation 6.23,
∬
S
G · dS = ∬
S
curl F · dS = ∫
C
F · dr = ∬
S′
curl F · dS = ∬
S′
G · dS.
Therefore, the flux integral of G does not depend on the surface, only on the boundary of the surface. Flux integrals of
vector fields that can be written as the curl of a vector field are surface independent in the same way that line integrals of
vector fields that can be written as the gradient of a scalar function are path independent.
796 Chapter 6 | Vector Calculus
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6.62
6.63
Use Stokes’ theorem to calculate surface integral ∬
S
curl F · dS, where F = 〈 z, x, y 〉 and S is the
surface as shown in the following figure. The boundary curve, C, is oriented clockwise.
Example 6.75
Calculating a Line Integral
Calculate the line integral ∫
C
F · dr, where F = 〈 xy, x2 + y2 + z2, yz 〉 and C is the boundary of the
parallelogram with vertices (0, 0, 1), (0, 1, 0), (2, 0, −1), and (2, 1, −2).
Solution
To calculate the line integral directly, we need to parameterize each side of the parallelogram separately, calculate
four separate line integrals, and add the result. This is not overly complicated, but it is time-consuming.
By contrast, let’s calculate the line integral using Stokes’ theorem. Let S denote the surface of the parallelogram.
Note that S is the portion of the graph of z = 1 − x − y for (x, y) varying over the rectangular region
with vertices (0, 0), (0, 1), (2, 0), and (2, 1) in the xy-plane. Therefore, a parameterization of S is
〈 x, y, 1 − x − y 〉 , 0 ≤ x ≤ 2, 0 ≤ y ≤ 1. The curl of F is − 〈 z, 0, x 〉 , and Stokes’ theorem and
Equation 6.19 give
∫
C
F · dr = ∬
S
curl F · dS
= ∫
0
2
∫
0
1
curl F(x, y) · ⎛
⎝tx×ty
⎞
⎠dydx
= ∫
0
2
∫
0
1
〈 −⎛
⎝1 − x − y⎞
⎠,0,x 〉 · ⎛
⎝ 〈 1, 0, −1 〉 × 〈 0, 1, −1 〉 ⎞
⎠dydx
= ∫
0
2
∫
0
1
〈 x + y − 1, 0, x 〉 · 〈 1, 1, 1 〉 dydx
∫
0
2
∫
0
1
2x + y − 1dydx
= 3.
Use Stokes’ theorem to calculate line integral ∫
C
F · dr, where F = 〈 z, x, y 〉 and C is oriented
clockwise and is the boundary of a triangle with vertices (0, 0, 1), (3, 0, −2), and (0, 1, 2).
Chapter 6 | Vector Calculus 797