17.3 Divergence Theorem
Lukas Geyer
Montana State University
M273, Fall 2011
Lukas Geyer (MSU)
17.3 Divergence Theorem
M273, Fall 2011
1 / 11
Fundamental Theorems of Vector Analysis
I
ZZ
F · ds =
Green’s Theorem
curl F dA
D
∂D
I
I
D plane domain, F = hP, Qi
∂Q
∂P
curlhP, Qi =
−
∂x
dy
I
ZZ
F · ds =
Stokes’ Theorem
I
I
curl F · dS
S
∂S
S surface in space,
F = hP, Q, Ri
∂R
∂Q ∂P
∂R ∂Q
∂P
curlhP, Q, Ri =
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x
∂y
ZZ
ZZZ
F · dS =
Divergence Theorem
∂W
I
I
div F dV
W
W region in space, F = hP, Q, Ri
∂P
∂Q
∂R
divhP, Q, Ri =
+
+
∂x
∂y
∂z
Lukas Geyer (MSU)
17.3 Divergence Theorem
M273, Fall 2011
2 / 11
Divergence Theorem
Theorem
ZZ
ZZZ
F · dS =
div F dV
W
∂W
Remarks
W is a region in space.
∂W is oriented with outward-pointing normal vector
F = hF1 , F2 , F3 i is a smooth vector field.
∂ ∂ ∂
,
,
· hF1 , F2 , F3 i
div F = ∇ · F =
∂x ∂y ∂z
=
∂F1 ∂F2 ∂F3
+
+
∂x
∂y
∂z
Lukas Geyer (MSU)
17.3 Divergence Theorem
M273, Fall 2011
3 / 11
Example I
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the surface integral
The boundary of W consists of the upper hemisphere of radius 2 and the
disk of radius 2 in the xy -plane. The upper hemisphere is parametrized by
G (φ, θ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ),
0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.
Tφ = h2 cos φ cos θ, 2 cos φ sin θ, −2 sin φi,
Tθ = h−2 sin φ sin θ, 2 sin φ cos θ, 0i,
n = Tφ × Tθ = h4 sin2 φ cos θ, 4 sin2 φ sin θ, 4 cos φ sin φi
n points up, so the orientation is correct.
Lukas Geyer (MSU)
17.3 Divergence Theorem
M273, Fall 2011
4 / 11
Example II
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the surface integral
G (φ, θ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ),
0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.
n = h4 sin2 φ cos θ, 4 sin2 φ sin θ, 4 cos φ sin φi
F · n = h2 sin φ sin θ, 2 sin φ cos θ, 1 + 2 cos φi
·
h4 sin2 φ cos θ, 4 sin2 φ sin θ, 4 cos φ sin φi
= 16 sin3 φ cos θ sin θ + 4 cos φ sin φ(1 + 2 cos φ)
Lukas Geyer (MSU)
17.3 Divergence Theorem
M273, Fall 2011
5 / 11
Example III
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the surface integral
G (φ, θ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ),
0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.
F · n = 16 sin3 φ cos θ sin θ + 4 cos φ sin φ(1 + 2 cos φ)
ZZ
Z
π/2 Z 2π
F · dS =
S1
0
Z
π/2 Z 2π
+
4 cos φ sin φ(1 + 2 cos φ) dθ dφ
0
Lukas Geyer (MSU)
16 sin3 φ cos θ sin θ dθ dφ
0
0
17.3 Divergence Theorem
M273, Fall 2011
6 / 11
Example IV
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the surface integral
ZZ
Z
π/2 Z 2π
F · dS =
S1
0
Z
16 sin3 φ cos θ sin θ dθ dφ
0
π/2 Z 2π
+
4 cos φ sin φ(1 + 2 cos φ) dθ dφ
0
0
Z
= 0 + 8π
π/2
(cos φ + 2 cos2 φ) sin φ dφ
0
1
2
= 8π − cos2 φ − cos3 φ
2
3
Lukas Geyer (MSU)
φ=π/2
= 8π
φ=0
17.3 Divergence Theorem
1 2
+
2 3
=
M273, Fall 2011
28π
.
3
7 / 11
Example V
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the surface integral
The disk of radius 2 in the xy -plane is parametrized by
G (r , θ) = (r cos θ, r sin θ, 0),
Tr
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.
= hcos θ, sin θ, 0i,
Tθ = h−r sin θ, r cos θ, 0i,
n = Tr × Tθ = h0, 0, r i
n points up, so the orientation is incorrect, and we have to change the sign
on n.
Lukas Geyer (MSU)
17.3 Divergence Theorem
M273, Fall 2011
8 / 11
Example VI
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the surface integral
G (r , θ) = (r cos θ, r sin θ, 0),
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.
n = h0, 0, −r i
ZZ
Z
2 Z 2π
F · dS =
S2
hr sin θ, r cos θ, 1i · h0, 0, −r i dθ dr
0
Z
0
2 Z 2π
(−r ) dθ dr = −2π
=
0
Lukas Geyer (MSU)
Z
0
17.3 Divergence Theorem
2
r dr = −4π
0
M273, Fall 2011
9 / 11
Example VII
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the surface integral
Adding up the results gives
ZZ
ZZ
ZZ
F · dS +
F · dS =
∂W
Lukas Geyer (MSU)
S1
F · dS =
S2
17.3 Divergence Theorem
28π
16π
− 4π =
.
3
3
M273, Fall 2011
10 / 11
Example VIII
Example
Verify the Divergence Theorem for the region given by x 2 + y 2 + z 2 ≤ 4,
z ≥ 0, and for the vector field F = hy , x, 1 + zi.
Computing the volume integral
∂F1 ∂F2 ∂F3
+
+
= 0 + 0 + 1 = 1.
∂x
∂y
∂z
ZZZ
ZZZ
16π
1 4π · 23
=
div F dV =
1 dV = vol(W) = ·
2
3
3
W
W
div F =
Lukas Geyer (MSU)
17.3 Divergence Theorem
M273, Fall 2011
11 / 11