Math 263 Fall 2008, Test 2 Solutions
1. Let F(x, y, z) = (sin x, 2 cos x, 1 − y 2 ).
(a) Calculate curl F.
(b) Calculate div F.
(c) Calculate div(curl F).
Solution:
(a) curl F = (−2y)~i − (2 sin x)~k.
(b) div F = cos x.
(c) div(curl F)=0. This is true for any vector field.
2. Sketch the domain of integration for the integral given below. Then convert the integral
to spherical coordinates and evaluate it.
Z 3 Z √9−x2 Z √9−x2 −y2 p
z x2 + y 2 + z 2 dzdydx
√
−3
− 9−x2
0
Solution:
The integral represents the top half of a sphere of radius 3, centred at the origin. Converting to spherical coordinates, we get:
Z 2π Z π/2 Z 3
(ρ cos φ)ρ ρ2 sin φ dρdφdθ
0
0
0
!
Z
Z
π/2
3
ρ4 dρ
=2π
35
5
cos φ sin φdφ
0
0
Z
π/2
1
sin 2φdφ
2
0
5
π/2
3
1
=2π
− cos 2φ
5
4
5
0
3
1 1
243π
=2π
+
=
5
4 4
5
=2π
3. Is the vector field F(x, y, z) = (2xy + y 2 )i + (x2 + 2xy + z 2 )j + 2zyk conservative? If so,
find a function f so that F = ∇f . If not, explain clearly why.
Solution:
We can check that F is conservative by checking that the curl is zero. Alternatively, you
could just try to create a function f (x, y, z) so that F = ∇f . By partially integrating
(2xy + y 2 ) with respect to x, you get f = x2 y + y 2 x + ... By partially integrating (x2 +
2xy + z 2 ) with respect to y, you get f = x2 y + xy 2 + z 2 y + .... By partially integrating 2zy
with respect to z, you get f = z 2 y + .... Putting these all together, f = x2 y + y 2 x + z 2 y
is a potential function for the vector field, and the vector field is therefore conservative.
4. Find the line integral of F(x, y, z) = (yz)i + (xz)j + (xy + 1)k around the square with
corners at (0, 0, 1), (1, 0, 1), (1, 1, 1) and (0, 1, 1) (taken in that order).
Solution:
This is a vector integral around a closed curve in 3-d. Green’s theorem does not apply
because it is in 3-d. However, if F is conservative, then we know that the integral will be
1
0. We can check if F is conservative using the curl test: curlF = 0 (check it!). Therefore
F is conservative and the integral of F around any closed curve is zero.
2
R
5. (a) State Green’s theorem for C F · dr where C is a simple, positively oriented, closed
curve in the (x, y) plane and F(x, y) = P (x, y)i + Q(x, y)j is a two dimensional vector
field.
(b) Compute the work done by the force field F(x, y) = i + xj on a particle that makes
one counterclockwise revolution around the circle x2 + y 2 = 1.
(c) Compute the work done by the force field F(x, y) = i + xj on a particle that travels
from (1, 0) to (0, 1), counterclockwise along part of the circle x2 + y 2 = 1.
Solution:
(a) Green’s theorem says that
Z
Z Z ∂Q ∂P
−
dA
F · dr =
∂x
∂y
C
D
where D is the region enclosed by C.
(b) Directly apply Green’s theorem.
Z
Z Z
F · dr =
(1 − 0) dA
W =
C
D
2
2
where D is the circular region x + y < 1. The double integral is just the area of D
so the answer is W = π.
(c) The curve is not closed in this case. Therefore Green’s theorem does not apply
and we have to parameterize the curve. We use polar coordinates for simplicity:
~r(θ) = cos θ~i + sin θ~j on 0 ≤ θ ≤ π/2.
Z
Z π
W =
F · dr =
F(~r(θ)) · ~r0 (θ)dθ
0 ZCπ Z π
1
− sin θ
=
·
dθ =
cos2 θdθ
cos θ
cos θ
0
0
Z π
1
π
=
(1 + cos 2θ) dθ = .
2
0 2
3