Section 16.9 The Divergence Theorem
EX.2
divF = 2x + x + 1 = 3x + 1.
Let S1 = {(x, y, z) | z = 4 − x2− y2 and x2+ y2 ≤ 4} and S2 = {(x, y, 0) | x2+ y2 ≤ 4}. Then Z Z Z
E
divFdV = Z Z Z
E
(3x + 1)dV = Z 2π
0
Z 2 0
Z 4−r2 0
(3r cos θ + 1)rdzdrdθ
= Z 2
0
Z 2π 0
r(3r cos θ + 1)(4 − r2)dθdr = 8π.
On S1, let D = {(x, y) | x2+ y2 ≤ 4}.
Z Z
S1
F · dS = Z Z
D
[−(x2)(−2x) − (xy)(−2y) + (4 − x2− y2)]dA
= Z 2π
0
Z 2 0
(2r cos θ · r2+ 4 − r2)rdrdθ = 8π.
On S2, since z = 0 and n = −k, we have F · n = 0 which impliesRR
S2F · dS = 0. Therefore Z Z
S
F · dS = Z Z
S1
F · dS + Z Z
S2
F · dS = 8π.
EX.3
divF = 1.
Let S = {(x, y, z) | x2+ y2+ z2 = 16} and D = {(x, y) | x2+ y2 ≤ 16}. Then Z Z Z
E
divFdV = Z Z Z
E
1dV = colume of E = 4π
3 43 = 356 3 π.
Next, we compute RR
SF · dS. We use parametric representation
γ(φ, θ) = (2 cos φ, 2 sin φ sin θ, 2 sin φ cos θ), 0 ≤ φ ≤ π, 0 ≤ θ ≤ θ.
Then
F(γ(φ, θ)) = (2 cos φ, 2 sin φ sin θ, 2 sin φ cos θ), and
γφ× γθ = (42sin2φ cos θ, 42sin2φ sin θ, 42sin φ cos φ).
Therefore,
F(γ(φ, θ)) · γφ× γθ = cos φ sin2φ cos θ + sin3φ sin2θ + sin2φ cos φ cos θ.
1
Z Z
S
F · dS = Z Z
D
F(γ(φ, θ)) · γφ× γθdA
= Z 2π
0
Z π 0
cos φ sin2φ cos θ + sin3φ sin2θ + sin2φ cos φ cos θdφdθ
= 43( Z π
0
sin3φdφ)(
Z 2π 0
sin2θdθ) = 256 3 π.
EX.5
divF = yez+ 2xyz3− yez = 2xyz3. Let E = {(x, y, z) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1}.
Z Z
S
F · dS = Z Z Z
E
divFdV = Z 3
0
Z 2 0
Z 1 0
2xyz3dzdydx
= 2(
Z 3 0
xdx)(
Z 2 0
ydy)(
Z 1 0
z3dz) = 2 · 32 2 · 22
2 · 1 4 = 9
2.
EX.7
divF = 3y2+ 3z2.
Let E = {(x, y, z) | y2+z2 = 1, −1 ≤ x ≤ 2}. Use cylindrical coordinates y = r cos θ, z = r sin θ, x = x.
Z Z
S
F · dS = Z Z Z
E
divFdV = Z Z Z
E
(3y2+ 3z2)dV
= Z 2π
0
Z 1 0
Z 2
−1
(3r2cos2θ + 3r2sin2θ)rdxdrdθ
= 3(
Z 2π 0
1dθ)(
Z 1 0
r3dr)(
Z 2
−1
1dx) = 9 2π.
EX.13
F = r
|r| = x
(x2+ y2+ z2)12i + y
(x2+ y2+ z2)12j + z
(x2+ y2+ z2)12k.
Then
∂
∂x
x
(x2+ y2+ z2)12 = (x2+ y2+ z2) − x2
(x2+ y2+ z2)1+12 = |r|2− x2
|r|3 ,
∂
∂y
y
(x2+ y2+ z2)12 = (x2+ y2+ z2) − y2
(x2+ y2+ z2)1+12 = |r|2− y2
|r|3 ,
∂
∂z
z
(x2+ y2+ z2)12 = (x2+ y2+ z2) − z2
(x2+ y2+ z2)1+12 = |r|2− z2
|r|p+2 .
2
Therefore, divF = 3|r|2− |r|2
|r|3 = 2
|r|. Z Z
S
F · dS = Z Z Z
E
divFdV = Z Z Z
E
2
px2+ y2+ z2dV
= Z π2
0
Z 2π 0
Z 1 0
2
ρρ2sin φdρdθdφ = 2π.
EX.15
Z Z
S
F · dS = Z Z Z
E
divFdV = Z Z Z
E
√
3 − x2dV
= Z 1
−1
Z 1
−1
Z 2−x4−y4 0
√
3 − x2dzdydx = 341 60
√ 2 + 81
20sin−1( 1
√3).
EX.17
For S1, we have n = (0, 0, −1), so F · n = −x2z − y2 = −y2 (since z = 0 on S1). So if D is the unit disk, we get
Z Z
S1
F · dS = Z Z
S1
F · ndS = Z Z
D
(−y2)dA = − Z 2π
0
Z 1 0
r2(sin2θ)rdrdθ = −π 4.
Now, since S2 is closed, we can use the Divergence theorem. Since divF = z2+ y2+ x2, we use spherical coordinate to get
Z Z
S2
F · dS = Z Z Z
E
divFdV = Z 2π
0
Z π2
0
Z 1 0
ρ2· ρ2sin φdρdφdθ = 2 5π.
Finally,
Z Z
S
F · dS = Z Z
S2
F · dS − Z Z
S1
F · dS = 13 20π.
EX.18
We create a closed surface S2 = S1 ∪ S, where S is the part of paraboloid x2 + y2 + z = 2 that lies above the plane z = 1, and S1 is the disk x2 + y2 = 1 on the plane z = 1 oriented downward, and we then apply Divergence Theorem. Since the disk S1 is oriented downward, its unit normal vector is n = (0, 0, −1) and F · n = −z = −1 on S1. So
Z Z
S1
F · dS = Z Z
S1
F · ndS = Z Z
S1
(−1)dS = −π.
3
Let E be the region bounded by S2. Then Z Z
S2
F · dS = Z Z Z
E
divFdV = Z Z Z
E
1dV
= Z 1
0
Z 2π 0
Z 2−r3 0
rdzdθdr = Z 1
0
Z 2π 0
(r − r3)dθdr = π 2. Thus the flux of F across S is
Z Z
S
F · dS = Z Z
S2
F · dS − Z Z
S1
F · dS = 3 2π.
EX.23
Let
F = r
|r|p = x
(x2+ y2+ z2)p2i + y
(x2+ y2+ z2)p2j + z
(x2+ y2+ z2)p2k.
Then
∂
∂x
x
(x2+ y2+ z2)p2 = (x2+ y2 + z2) − px2
(x2+ y2+ z2)1+p2 = |r|2 − px2
|r|p+2 ,
∂
∂y
y
(x2+ y2+ z2)p2 = (x2+ y2 + z2) − py2
(x2+ y2+ z2)1+p2 = |r|2− py2
|r|p+2 ,
∂
∂z
z
(x2+ y2+ z2)p2 = (x2+ y2 + z2) − pz2
(x2+ y2+ z2)1+p2 = |r|2 − pz2
|r|p+2 . Therefore, divF = 3|r||r|2−p|r|p+2 2 = 3−p|r|p. Consequently, divF = 0 if and only if p = 3.
Hence, divE(x) = 0.
EX.24
We need to find F so that RR
SF · dS = RR
S(2x + 2y + z2)dS, so F · n = 2x + 2y + z2. But for S, n = √(x,y,z)
x2+y2+z2 = (x, y, z). Take F = (2, 2, z) and divF = 1. If B = {(x, y, z) : x2+ y2+ z2 ≤ 1}, then Z Z
S
(2x + 2y + z2)dS = Z Z Z
B
1dV = 4 3π.
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