Spring 2004 Math 253/501–503
14 Vector Calculus
14.9 The Divergence Theorem
c
Tue, 19/Apr
2004,
Art Belmonte
• The respective compositions w (sk (r, θ )) are
i
h
w (s1 (r, θ )) = r 3 cos θ, r 3 sin θ, 3r 4
w (s2 (r, θ ))
Let E be a simple solid region whose closed boundary surface S
has positive (outward) oriention. Let w be a vector field whose
components have continuous partial derivatives on an open region
containing E. Then
ZZZ
ZZ
w · dS =
div w dV.
• Now set up the surface integral, which consists of two pieces
(one for each subsurface).
ZZ
ZZ
w · dS =
w (s (r, θ )) · (sr × sθ ) dr dθ
E
Hand / MATLAB Examples
S
The Divergence Theorem is always used in the forward direction.
Instead of computing a difficult surface integral (or several of
them), we compute just one triple integral. That said, we’ll start
the party off by computing both sides of the Divergence Theorem
for our first problem.
D
Z
=
2π
−
0
=
8π
3
Z
1
Z
2π
r 5 dr dθ +
0
Z
3r dr dθ
0
0
≈ 8.38
→
−
• The divergence is div w = ∇ · w or
h
i
∂ ∂ ∂
,
· x z, yz, 3z 2 = z + z + 6z = 8z.
∂ x ∂y , ∂z
Verify that the Divergence
Theorem is true for the vector field
w = x z, yz, 3z 2 on the solid region E bounded by the
paraboloid z = x 2 + y 2 and the plane z = 1.
Stewart 936/2
• Therefore,
RRR
R 2π R 1 R 1
8
E div w dV = 0
0 r 2 8z · r dz dr dθ = 3 π ≈ 8.38.
1
0.5
MATLAB diary file
Herewith a transcript of the needful.
1
0
1
1
Triple Integral Well, that was a world ‘o‘ hurt (and we didn’t
even show all the computations). Instead, apply the Divergence
Theorem and compute one triple integral.
936/2
z
[r cos θ, r sin θ, 3]
• To apply the Divergence Theorem, we must use an outward
orientation. As we see in the MATLAB diary file below,
∂s2 /∂r × ∂s2 /∂θ = [0, 0, r ] gives an upward (and hence
outward) orientation for the circular disk on top. However,
∂s1 /∂r × ∂s1 /∂θ = [−2r 2 cos θ, −2r 2 sin θ, r ] gives an
upward (and hence inward) orientation for the paraboloid.
Therefore, we negate the result returned by siv when this
piece of the surface integral is computed.
Summary
S
=
0.5
%
% Stewart 936/2
%
syms r t x y z
w = [x*z y*z 3*zˆ2];
s1 = [r*cos(t) r*sin(t) rˆ2];
s2 = [r*cos(t) r*sin(t) 1];
%
s1r = diff(s1,r); s1t = diff(s1,t);
c1 = simple(cross(s1r,s1t))
0
0
−0.5
y
−1
−1
x
Solution
Surface Integral
c1 =
• Parameterize the two pieces of the closed surface using
cylindrical coordinates. The first is the parboloid, the second
the circular disk on top. For both parameterizations, the
ranges of the parameters are 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π. This
ought to be old hat to you by now.
i
h
s1 (r, θ) = r cos θ, r sin θ, r 2 ,
[ -2*rˆ2*cos(t), -2*rˆ2*sin(t),
s2r = diff(s2,r); s2t = diff(s2,t);
c2 = simple(cross(s2r,s2t))
c2 =
[ 0, 0, r]
s2 (r, θ) = [r cos θ, r sin θ, 1]
%
1
r]
937/6
M = [r 0 1; t 0 2*pi];
S = -siv(w,s1,M) + siv(w,s2,M);
pretty(S)
Use
RR the Divergence Theorem to evaluate the surface integral
S w · dS, where S is the surface of the tetrahedron with vertices
8/3 pi
floated = double(S)
floated =
8.3776
%
div w = div(w)
(0, 0, 0) ,
(1, 0, 0) , (0, 1, 0) ,
The vector field is w = 3x y, y 2 , −x 2 y 4 .
(0, 0, 1) .
div w =
Stewart 937/6
8*z
via DT = int(int(int(div w * r, z,rˆ2,1), r,0,1), t,0,2*pi);
pretty(via DT)
1
0.8
8/3 pi
0.6
z
%
0.4
echo off; diary off
0.2
0
0
0
0.5
0.5
937/4
1 1
x
y
Use
RR the Divergence Theorem to evaluate the surface integral
S w · dS, where S is the surface of the rectangular box bounded
by the planes
x = 0,
x = 3,
y = 2,
The vector field is w = x 2 y, −x 2 z, z 2 y .
y = 0,
z = 0,
Solution
z = 1.
Via plane3pt, an equation for the plane containing the slanted
upper plane is x + y + z = 1 or z = 1 − x − y. Now apply the
Divergence Theorem.
Stewart 937/4
div w dV =
1.5
E
1
z
1 Z 1−x
Z
ZZZ
0.5
−0.5
5y dz d y d x =
0
5
24
≈ 0.21
3
2
1
y
0
1−x−y
%
% Stewart 937/6
%
our plane = plane3pt([1 0 0], [0 1 0], [0 0 1])
0
2
0
Z
our plane =
1
0
0
x
x+y+z = 1
%
syms x y z
w = [3*x*y yˆ2
div w = div(w)
Solution
RR
RRR
Via the Divergence Theorem, S w · dS =
E div w dV or
Z 1Z 2Z 3
2x y + 0 + 2yz d x d y dz = 24.
0
0
div w =
5*y
0
%
% Stewart 937/4
%
syms x y z
w = [xˆ2*y -xˆ2*z
div w = div(w)
-xˆ2*yˆ4];
via DT = int(int(int(div w, z,0,1-x-y), y,0,1-x), x,0,1);
pretty(via DT)
5/24
floated = double(via DT)
floated =
0.2083
%
zˆ2*y];
echo off; diary off
div w =
2*y*(x+z)
937/8
via DT = int(int(int(div w, x,0,3), y,0,2), z,0,1);
pretty(via DT)
Use
integral
RR the Divergence Theorem to evaluate4the surface
4 + z 4 = 1. The
w
·
dS,
where
S
is
the
“fat
sphere”
x
+
y
S
vector field is w = x + e y tan z , 3xe xz , cos y − z .
24
%
echo off; diary off
2
937/12
Stewart 937/8: "Fat sphere"
Use
RR the Divergence Theorem to evaluate the surface integral
S is the surface of the solid bounded by the
S w · dS, wherep
p
hemispheres z = 4 − x 2 − y 2 , z = 1 − x 2 − y 2 , and the
plane z = 0. The vector field is
h
i
w = x 3 + y sin z, y 3 + z sin x, 3z .
1
0.5
z
0
−0.5
−1
1
1
0
0
−1
y
x
−1
Stewart 937/12: Bottom view
Stewart 937/12: View from the top
E
%
% Stewart 937/8
%
syms x y z
w = [x + exp(y*tan(z))
div w = div(w)
2
1
1
0
−2
0
2
The divergence is 0. Therefore,
ZZZ
ZZZ
ZZ
w · dS =
div w · dV =
0 dV = 0.
S
2
z
z
Solution
1
2
−1
2
0
0
0
−1
y
E
0
−2
1
y
x
−2
2
−2
x
Solution
3*x*exp(x*z)
cos(y)-z];
Apply the Divergence Theorem, then switch to spherical
coordinates! [Recall that x 2 + y 2 = r 2 = (ρ sin φ)2 .]
ZZZ
ZZ
w · dS =
div w dV
S
Z Z ZE
3x 2 + 3y 2 + 3 dV
=
div w =
0
via DT = 0;
%
echo off; diary off
E
Z
=
937/11
=
Use
RR the Divergence Theorem to evaluate the surface integral
S w · dS, where S is the surface of the solid bounded by the
circular cylinder x 2 + yh2 = 9 and the iplanes z = 0 and z = y − 3.
2
The vector field is w = ye z , y 2 , e xy .
2π
0
Z
π/2 Z 2
0
1
3 1 + (ρ sin φ)2 · ρ 2 sin φ dρ dφ dθ
194
π ≈ 121.89
5
%
% Stewart 937/12
%
syms p r t x y z
w = [xˆ3 + y*sin(z)
yˆ3 + z*sin(x)
div w = div(w); pretty(div w)
Stewart 937/11
3*z];
2
2
3 x + 3 y + 3
% Switch to spherical coordinates!
via DT = int(int(int(3 * (1 + (r*sin(p))ˆ2) * rˆ2*sin(p),...
r,1,2), p,0,pi/2), t,0,2*pi);
pretty(via DT)
0
z
−2
−4
194/5 pi
−6
−2
0
y
2
2
0
floated = double(via DT)
floated =
121.8938
%
−2
echo off; diary off
x
Solution
937/16
This is a homework problem. Verify that you obtain the answer
− 81
2 π ≈ −127.23. Apply the Divergence Theorem, then use
cylindrical coordinates. Ask if you have questions!
Compute the flux of the vector field
h
w = sin x cos2 y, sin3 y cos4 z,
3
sin5 z cos6 x
i
across the surface of the cubeRR0 ≤ x, y, z ≤ π2 . In other words,
compute the surface integral S w · dS by applying the
Divergence Theorem. Here is a plot of the 3-D vector field.
Stewart 937/16: Vector field in space
z
1.5
1
0.5
0
0
0
0.5
0.5
y
1
1
1.5
1.5 x
Solution
The last target: compute the divergence and render the needful.
%
% Stewart 937/16s
%
syms x y z
w = [sin(x)ˆ1 * cos(y)ˆ2 ...
sin(y)ˆ3 * cos(z)ˆ4 ...
sin(z)ˆ5 * cos(x)ˆ6];
div w = div(w); pretty(div w)
2
2
4
cos(x) cos(y) + 3 sin(y) cos(z) cos(y)
4
6
+ 5 sin(z) cos(x) cos(z)
%
via DT = int(int(int(div w, ...
x,0,pi/2), y,0,pi/2), z,0,pi/2);
pretty(via DT)
19
2
-- pi
64
floated = double(via DT)
floated =
2.9300
%
echo off; diary off
4