Calculo

Section 15.6
C15S06.001: The right-hand side in the divergence theorem (Eq. (1)) is ∇· F dV =
B B

3 dV = 3 ·

4 π · 13 = 4π 3

and the left-hand side in the divergence theorem is F · n dS =
S S

x, y, z · x, y, z dS =
S

(x2 + y 2 + z 2 ) dS =
S

1 dS = 1 · 4π · 12 = 4π.

Note that we integrate a constant function by multiplying its value by the size (length, area, or volume)of the domain of the integral. We will continue to do so without further comment. C15S06.002: Here we have F(x, y, z) = (x2 + y 2 + z 2 )1/2 x, y, z , is a unit vector normal to the surface S. Because F · n = value 9 on S. Therefore
1 2 3 (x

and

n=

1 x, y, z 3

+ y 2 + z 2 )3/2 , F · n takes on the constant

F · n dS = 9 · area(S) = 9 · 4π · 32 = 324π ≈ 1017.8760197630930093.
SNext, let B denote the solid ball bounded by S. Then ∇· F = and thus ∇· F dV =
B 2π B π 0 0 3 π π

x2 x2 + y 2 + z 2

+

y2 x2 + y 2 + z 2

+

z2 x2 + y 2 + z 2

+ 3 x2 + y 2 + z 2 = 4 x2 + y 2 + z 2 ,

4

x2 + y 2 + z 2 dV

=
0

4ρ3 sin φ dρ dφ dθ = 2π
0

81 sin φ dφ = 2π −81 cos φ
0

= 324π.

C15S06.003: On the face F of the cube in the plane x = 2, a unit vectornormal to F is n = i, and F · i = x = 2. Hence F · n dS = 2 · area(F ) = 8.
F

By symmetry, the same result obtains on the faces in the planes y = 2 and z = 2. On the face G of the cube in the plane x = 0, a unit vector normal to G is n = −i, and F · (−i) = −x = 0. Hence F · n dS = 0.
G

By symmetry, the same result holds on the faces in the other two coordinate planes. Hence 1

F · n dS = 3· 8 + 3 · 0 = 24.
S

Let B denote the solid cube bounded by S and let V denote the volume of B. Because ∇· F = 3, we also have ∇· F dV = 3 · V = 3 · 8 = 24.
B

C15S06.004: On the face F of the cube in the plane x = 2, a unit vector normal to F is n = i, and F · i = xy = 2y. Hence
2 2 2

F · n dS =
F 0 0

2y dy dz =
0

4 dz = 8.

By symmetry, the same result holds on the faces inthe planes y = 2 and z = 2. On the face G of the cube in the plane x = 0, a unit vector normal to G is n = −i, and F · (−i) = −xy = 0. Hence F · n dS = 0.
G

By symmetry, the same result holds on the faces in the other two coordinate planes. Hence F · n dS = 3 · 8 + 3 · 0 = 24.
S

Let B denote the solid cube bounded by S. Because ∇· F = y + z + x, we see that
2 2 0 2 2 0 2 2 2

∇· F dV =B 0

(x + y + z) dx dy dz

=
0 0

(2 + 2y + 2z) dy dz =
0

(8 + 4z) dz = 8z + 2z 2
0

= 24.

C15S06.005: On the face F of the tetrahedron that lies in the plane x = 0, a unit vector normal to F is n = −i, and F · n = −x − y = −y. Hence
1 1−z 1

F · n dS =
F 0 0

(−y) dy dz =
0

1 1 − (1 − z)2 dz = (1 − z)3 2 6

1 0

1 =− . 6

By symmetry the same result holds on thefaces in the other two coordinate planes. On the fourth face G of the tetrahedron, a unit vector normal to G is √ n= 3 3 1, 1, 1 ,

and G is part of the graph of z = h(x, y) = 1 − x − y, so that dS = Therefore 2 1 + (hx )2 + (hy )2 dA = √ 3 dA.

1

1−x

1

1−x

1

F · n dS =
G G

(2x + 2y + 2z) dA =
G

2 dA =
0 0

2 dy dx =
0

2y
0

dx = 2x − x2
0

= 1,

andtherefore F · n dS = 1 − 3 ·
S

1 1 = . 6 2

Let B denote the solid tetrahedron itself, with volume V . Then ∇· F dV =
B B

3 dV = 3 · V = 3 ·

1 1 ·1·1= . 6 2

C15S06.006: Let B denote the cube bounded by the surface S. Then
2 2 0 0 2 2

F · n dS =
S 2 B

∇· F dV =
B 2

(2x + 2y + 2z) dV =
0 2

(2x + 2y + 2z) dz dy dx

=
0 0

(4 + 4x + 4y) dy dx =
0

(16 + 8x) dx = 16x+ 4x2
0

= 48.

C15S06.007: Let B denote the solid cylinder bounded by the surface S. Then
2π 3 0 4 −1

F · n dS =
S B 3

∇· F dV =
B 4

3(x2 + y 2 + z 2 ) dV =
0 3

3(r2 + z 2 ) · r dz dr dθ
3 0

= 2π
0

3r3 z + rz 3
−1

dr = 2π
0

(65r + 15r3 ) dr = 2π

65 2 15 4 r + r 2 4

= 2π ·

2385 2385 = π ≈ 3746.3492394058284367. 4 2

C15S06.008: Denote by B the...