Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 17.8 STOKES’ THEOREM ET SECTION 16.8 ¤ 297
43. S consists of the hemisphere S 1 given by z = a 2 − x 2 − y 2 and the disk S 2 given by 0 ≤ x 2 + y 2 ≤ a 2 , z =0.
On S 1 : E = a sin φ cos θ i + a sin φ sin θ j +2a cos φ k,
T φ × T θ = a 2 sin 2 φ cos θ i + a 2 sin 2 φ sin θ j + a 2 sin φ cos φ k. Thus
S 1
E · dS = 2π π/2
(a 3 sin 3 φ +2a 3 sin φ cos 2 φ) dφ dθ
0 0
= 2π π/2
(a 3 sin φ + a 3 sin φ cos 2 φ) dφ dθ =(2π)a 3
1+ 1 0 0 3 =
8
3 πa3
On S 2: E = x i + y j,andr y × r x = −k so S 2
E · dS =0. Hence the total charge is q = ε 0
S E · dS = 8 3 πa3 ε 0.
45. K∇u =6.5(4y j +4z k). S is given by r(x, θ) =x i + √ 6cosθ j + √ 6sinθ k and since we want the inward heat flow, we
use r x × r θ = − √ 6cosθ j − √ 6sinθ k. Then the rate of heat flow inward is given by
S (−K ∇u) · dS = 2π 4
−(6.5)(−24) dx dθ =(2π)(156)(4) = 1248π.
0 0
47. Let S be a sphere of radius a centered at the origin. Then |r| = a and F(r) =cr/ |r| 3 = c/a 3 (x i + y j + z k). A
parametric representation for S is r(φ, θ) =a sin φ cos θ i + a sin φ sin θ j + a cos φ k, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π. Then
r φ = a cos φ cos θ i + a cos φ sin θ j − a sin φ k, r θ = −a sin φ sin θ i + a sin φ cos θ j, and the outward orientation is given
by r φ × r θ = a 2 sin 2 φ cos θ i + a 2 sin 2 φ sin θ j + a 2 sin φ cos φ k. Theflux of F across S is
S F · dS = π
2π c
(a sin φ cos θ i + a sin φ sin θ j + a cos φ k)
0 0
a3 = c
a 3 π
0
Thus the flux does not depend on the radius a.
· a 2 sin 2 φ cos θ i + a 2 sin 2 φ sin θ j + a 2 sin φ cos φ k dθ dφ
2π
a 3 sin 3 φ +sinφ cos 2 φ dθ dφ = c π
2π
sin φdθdφ=4πc
0 0 0
17.8 Stokes' Theorem ET 16.8
1. Both H and P are oriented piecewise-smooth surfaces that are bounded by the simple, closed, smooth curve x 2 + y 2 =4,
z =0(whichwecantaketobeorientedpositivelyforbothsurfaces).ThenH and P satisfy the hypotheses of Stokes’
Theorem, so by (3) we know curl F · dS = F · dr = curl F · dS (where C is the boundary curve).
H C P
3. The paraboloid z = x 2 + y 2 intersects the cylinder x 2 + y 2 =4in the circle x 2 + y 2 =4, z =4. This boundary curve C
should be oriented in the counterclockwise direction when viewed from above, so a vector equation of C is
r(t) =2cost i +2sint j +4k, 0 ≤ t ≤ 2π. Thenr 0 (t) =−2sint i +2cost j,
F(r(t)) = (4 cos 2 t)(16) i +(4sin 2 t)(16) j +(2cost)(2 sin t)(4) k =64cos 2 t i +64sin 2 t j +16sint cos t k,
andbyStokes’Theorem,
S curl F · dS = C F · dr = 2π
0
F(r(t)) · r 0 (t) dt = 2π
0 (−128 cos2 t sin t +128sin 2 t cos t +0)dt
=128 1
3 cos3 t + 1 3 sin3 t 2π
=0
0