Download Lesson 26: The Divergence Theorem

Lesson 26: The Divergence Theorem
August 3rd, 2015
Section 16.9
At the end of our lecture on Green’s Theorem, we derived two
vector versions of it. One involved the curl and the other
divergence. The divergence version is
Z
C
F · n ds =
ZZ
div F(x , y ) dA
D
where C is the positively oriented boundary curve of the plane
region D.
What we will see today is that this integral essentially extends
to vector fields on R3 .
Section 16.9
Theorem (The Divergence Theorem)
Let E be a simple solid region and let S be the boundary
surface of E , given with positive (outward) orientation. Let F
be a vector field whose component functions have continuous
partial derivatives on an open region that contains E . Then
ZZ
F · dS =
S
ZZZ
div F dV
E
Here E is a simple region (i.e a region of type I, II, and III
such as a sphere, ellipsoid, or cube) and S is a closed surface
whose unit normal points outward from E . The theorem is
true for more general regions E as well.
Section 16.9
Example
Verify that the Divergence Theorem is true for the vector field
F(x , y , z) = 3x i + xy j + 2xz k on the region E which is the
cube bounded by the planes x = 0, x = 1, y = 0, y = 1,
z = 0, and z = 1.
Section 16.9
Example
Find the (outward) flux of the vector field
F(x , y , z) = z i + y j + x k
over the unit sphere x 2 + y 2 + z 2 = 1.
Section 16.9
Example
Evaluate
RR
S
F · dS where
2
F(x , y , z) = xy i + (y 2 + e xz ) j + sin(xy ) k
and S is the surface of the region E bounded by the parabolic
cylinder z = 1 − x 2 and the planes z = 0, y = 0, and
y +z =2
Section 16.9
When E is a region lying between two surfaces S1 and S2 with
S1 lying inside S2 , we can still apply the Divergence Theorem.
Consider the picture below
So S1 is the sphere, S2 is the cube (ignore the fact that it’s
not smooth), and E is the region between them. Let n1 be the
unit normal of S1 and n2 the unit normal of S2 . Then the
boundary of E is S1 ∪ S2 , and the unit normal n of E is
n = n2 on S2 and n = −n1 on S1 .
Section 16.9
Applying the Divergence Theorem, this gives us
ZZZ
div F dV =
ZZ
E
F · dS =
ZZ
S
=
ZZ
S1
=−
F · n dS
S
F · (−n1 ) dS +
ZZ
F · dS +
ZZ
S1
S2
F · n2 dS
F · dS
S2
Section 16.9
ZZ
Example
Recall that when an electric charge Q is positioned at the
origin, the electric field created is
E(x) =
εQ
x
|x|3
where x = hx , y , zi is a position vector. Use the Divergence
Theorem to show that the electric flux of E through any
closed surface S that encloses the origin is
ZZ
E · dS = 4πεQ
S
Section 16.9
As we’ve seen, divergence (like curl) also relates to fluid flow.
Let v(x , y , z) be the velocity field of a fluid with constant
density ρ. Then F = ρv is the rate of flow per unit area. Take
a point P0 (x0 , y0 , z0 ) in the fluid and consider a ball Ba
centered at P0 of radius a (with a very small).
Since div F is continuous, div F(P) ≈ div F(P0 ) in Ba . The flux
over the boundary sphere Sa is then approximately
ZZ
ZZZ
ZZZ
F · dS =
Sa
div F dV ≈
Ba
div F(P0 ) dV = div F(P0 )V (Ba )
Ba
Letting a → 0 gives a better approximation and suggests that
1 ZZ
div F(P0 ) = lim
F · dS
a→0 V (Ba )
Sa
Section 16.9
This last equation tells us that div F(P0 ) is the net rate of
outward flux per unit volume at P0 (hence the name
divergence).
If div F(P) > 0, the net flow is outward near P. In this case
we call P a source. If div F(P) < 0, the net flow is inward
near P and P is called a sink.
Section 16.9
Example
Use the Divergence Theorem to calculate S F · dS, where
F(x , y , z) = x 2 sin y i + x cos y j − xz sin y k and S is the “fat
sphere” x 8 + y 8 + z 8 = 8.
RR
Section 16.9
Example
Compute the flux of F(x , y , z) = |r|r across S, where
r = x√i + y j + z k and S consists of the hemisphere
z = 1 − x 2 − y 2 and the disc x 2 + y 2 ≤ 1 in the xy -plane.
Section 16.9