Static Vector Fields Homework 3
Due 10/24/12 @ 4pm
Practice Solutions
PRACTICE:
1. Calculate the divergence of each of the following vector fields.
~ = z 2 x̂ + x2 ŷ − y 2 ẑ
(a) F
~ = e−x x̂ + e−y ŷ + e−z ẑ
(b) G
~ = yz x̂ + zx ŷ + xy ẑ
(c) H
(d) ~I = x2 x̂ + z 2 ŷ + y 2 ẑ
~ = xy x̂ + xz ŷ + yz ẑ
(e) J
Solution:
~ =0
~ ·F
∇
~ = −e−x − e−y − e−z
~ ·G
∇
~ =0
~ ·H
∇
~ · ~I = 2x
∇
~ = 2y
~ ·J
∇
2. Choose some simple vector fields of your own and find the divergence of them both
by hand and using Mathematica or Maple. Choose some that are written in terms of
rectangular coordinates and others in cylindrical and/or spherical.
3. If you need more practice visualizing divergence, go through the Mathematica notebook
on the course website.
4. Consider the vector field F~ = (x + 2)x̂ + (z + 2)ẑ.
(a) Calculate the divergence of F~ .
Solution:
∂
∂
∂
Fx +
Fy + Fz
∂x
∂y
∂z
∂
∂
∂
=
(x + 2) +
(0) + (z + 2)
∂x
∂y
∂z
= 2
~ · F~ =
∇
(b) In which direction does the vector field F~ point on the plane z = x? What is the
value of F~ · n̂ on this plane where n̂ is the unit normal to the plane?
1
Solution:
On the plane z = x, the x and z components of F~ are equal, therefore F~ points
in the ı̂ + k̂ direction, i.e. parallel to the plane. Therefore we see, without any
further calculation, that F~ · n̂, the flux of F~ through this plane is zero.
(c) Verify the divergence theorem for this vector field where the volume involved is
drawn below.
Solution:
The divergence theorem states that the flux of any (sufficiently smooth) vector
field through any closed surface is the same as the integral of the divergence of
that vector field over the interior of the surface.
I
Z
~
~ · F~ dτ
F · n̂ dA = ∇
Calculating the left-hand side: We showed in part (b) that there is no flux through
the angled plane. There is also no flux through the triangular end pieces because F~
has no ̂ component (perpendicular to those planes). The (outward) flux through
the square boundary in the x, y-plane is:
Z 1Z 1
[(x + 2)x̂ + 2ẑ] · (−ẑ) dx dy = −2
0
0
The flux through the remaining vertical side is (at x = 1):
Z 1Z 1
[3x̂ + (z + 2)ẑ] · (x̂) dy dz = 3
0
0
Therefore, the left-hand side of the divergence theorem is −2 + 3 = 1.
Calculating the right-hand side, we have a constant divergence of 2, integrated
over the prism shape which is half of the unit cube. Therefore, we get just the
divergence, 2, times the volume of the prism, 1/2. So the right-hand side of the
divergence theorem also is 1, and we have verified the divergence theorem for this
special case.
2