Solução_Calculo_Stewart_6e

F.
F(r(t)) = − sin t j +cost k, F(r(t)) · r 0 (t) =− cos 2 t,and C
TX.10 SECTION 17.9 THE DIVERGENCE THEOREM ET SECTION 16.9 ¤ 299
Now curl F = −i − j − k, andS can be parametrized (see Example 17.6.10 [ ET 16.6.10]) by
r(φ, θ) =sinφ cos θ i +sinφ sin θ j +cosφ k, 0 ≤ θ ≤ π, 0 ≤ φ ≤ π. Then
r φ × r θ =sin 2 φ cos θ i +sin 2 φ sin θ j +sinφ cos φ k and
curl F · dS = curl F · (r
S φ × r θ ) dA = π
0
x 2 +z 2 ≤1
F · dr = 2π
0
− cos 2 tdt= − 1 2 t − 1 4 sin 2t 2π
0
= −π.
π
0 (− sin2 φ cos θ − sin 2 φ sin θ − sin φ cos φ) dθ dφ
= π
0 (−2sin2 φ − π sin φ cos φ) dφ = 1
2 sin 2φ − φ − π 2 sin2 φ π
0 = −π
17. It is easier to use Stokes’ Theorem than to compute the work directly. Let S be the planar region enclosed by the path of the
particle, so S is the portion of the plane z = 1 y for 0 ≤ x ≤ 1, 0 ≤ y ≤ 2,withupwardorientation.
2
curl F =8y i +2z j +2y k and
F · dr = curl F · dS = C S D
= 1
0
2
0
3
ydydx= 1
2 0
−8y (0) − 2z
1
1
2 +2y dA =
0
3
4 y2 y=2
y=0 dx = 1
0
3 dx =3
2
0
2y −
1
2 y dy dx
19. Assume S is centered at the origin with radius a and let H 1 and H 2 be the upper and lower hemispheres, respectively, of S.
Then S curl F · dS = H 1
curl F · dS + H 2
curl F · dS = C 1
F · dr + C 2
F · dr by Stokes’ Theorem. But C 1 is the
circle x 2 + y 2 = a 2 oriented in the counterclockwise direction while C 2 is the same circle oriented in the clockwise direction.
Hence C 2
F · dr = − C 1
F · dr so curl F · dS =0as desired.
S
17.9 The Divergence Theorem ET 16.9
1. div F =3+x +2x =3+3x,so
div F dV = 1 1
1 (3x +3)dx dy dz = 9 (notice the triple integral is
E 0 0 0 2
three times the volume of the cube plus three times x).
To compute S F · dS,on
S 1: n = i, F =3i + y j +2z k,and S 1
F · dS = S 1
3 dS =3;
S 2 : F =3x i + x j +2xz k, n = j and S 2
F · dS = S 2
xdS = 1 2 ;
S 3: F =3x i + xy j +2x k, n = k and S 3
F · dS = S 3
2xdS =1;
S 4 : F = 0, S 4
F · dS =0; S 5 : F =3x i +2x k, n = −j and S 5
F · dS = S 5
0 dS =0;
S 6 : F =3x i + xy j, n = −k and S 6
F · dS = S 6
0 dS =0.Thus S F · dS = 9 2 .
3. div F = x + y + z,so
E div F dV = 2π
0
= 2π
0
1
1 (r cos θ + r sin θ + z) rdzdrdθ= 2π
1
0 0 0 0 r 2 cos θ + r 2 sin θ + 1 r 2
dr dθ
1 cos θ + 1 sin θ + 1
3 3 4 dθ =
1
(2π) = π 4 2
Let S 1 be the top of the cylinder, S 2 the bottom, and S 3 the vertical edge. On S 1, z =1, n = k,andF = xy i + y j + x k,so