Stokes’ theorem
Math 21a
Fall 2016
1 Suppose F~ (x, y, z) = h−y, x, zi and S is the part of the sphere x2 + y 2 + z 2 = 25 below
the plane z = 4, oriented with the outward-pointing normal. Compute the flux integral
RR
~
curl F~ · dS.
S
Solution. By Stokes’ theorem, it suffices to compute the line integral of F~ along the boundary
C of the surface S. The curve C is a circle of radius 3 in the plane z = 4 centered at (0, 0, 4),
oriented clockwise, and it has parametrization
~r(t) = h3 cos t, −3 sin t, 4i,
0 ≤ t ≤ 2π.
Computing the line integral we have
Z
Z 2π
Z
~
F · d~r =
h3 sin t, 3 cos t, 4i · h−3 sin t, −3 cos t, 0i dt =
C
0
2π
−9 dt = −18π.
0
So the flux of curl F~ through S is −18π.
2 Suppose S is a “light bulb-shaped surface” as follows. Imagine a light bulb cut off at the base so that its boundary is the
unit circle x2 + y 2 = 1 in the xy-plane, oriented with the outwardpointing normal. (You can use either an old-fashioned light bulb or
a compact fluorescent if you’re feeling green.) Suppose F~ (x, y, z) =
z 2 −2z
z2
he
x,
sin(xyz)
+
y
+
1,
e
sin(z 2 )i. Compute the flux integral
RR
~
curl F~ · dS.
S
Solution. The boundary of S, no matter whether you take an incandescent or a fluorescent
light bulb, is the oriented curve C with parametrization
~r(t) = hcos t, sin t, 0i,
0 ≤ t ≤ 2π.
By Stokes’ theorem, the flux integral is equal to
Z
Z 2π
Z
~
F · d~r =
hcos t, sin t + 1, 0, 0i · h− sin t, cos t, 0i dt =
C
0
2π
cos t dt = 0.
0
There is no flux!
3 Let C be the oriented curve parametrized by ~r(t) = hcos t, sin t, 8 − cos2 t − sin ti,
√
R
0 ≤ t ≤ 2π, and let F~ (x, y, z) = hz 2 − y 2 , −2xy 2 , e z cos zi. Evaluate C F~ · d~r. (Hint: use
Stokes’ theorem.)
Solution. In order to use Stokes’ theorem, we need to find a surface S whose boundary is C.
By inspection, one notes that the curve C is the intersection of the cylinder x2 + y 2 = 1 and
the paraboloid z = 8 − x2 − y. We can take as our surface S the paraboloid, i.e., the part
of the graph z = 8 − x2 − y above the unit disk, with the upward-pointing normal (in order
for the orientations to be compatible). The surface S has parametrization
~r(u, v) = hu, v, 8 − u2 − vi
where (u, v) belong to the unit disk. We have
curl F~
~ru
~rv
~ru × ~rv
= h0, 2z, −2y 2 + 2yi = h0, 16 − 2u2 − 2v, −2v 2 + 2vi
= h1, 0, −2ui
= h0, 1, −1i
= h2u, 1, 1i
which has the correct upwards orientation. Therefore, by Stokes’ theorem, the desired line
integral is equal to
ZZ
ZZ
~
F · ~ru × ~rv du dv =
16 − 2u2 − 2v 2 du dv
{u2 +v 2 ≤1}
Z
{u2 +v 2 ≤1}
2π Z 1
=2
0
0
1
(8 − r2 )r dr dθ = 4π(4 − ) = 15π.
4
4
(a) Suppose S1 and S2 RR
are two oriented surfaces
that share C as oriented boundary. What
RR
~
~
~
can you say about S1 curl F · dS and S2 curl F~ · dS?
(b) Suppose SRR
1 and S2 are two oriented surfaces that share C as oriented boundary. Is it
RR
~ · dS
~ =
~ · dS
~ for any vector field G?
~ That is, can you always
true that S1 G
G
S2
choose the easiest surface to work with when computing flux integrals over a surface
with boundary?
RR
~ = 0?
(c) Suppose S is a closed surface (a surface without a boundary). Must S F~ · dS
Solution.
(a) They are the same. By Stokes’ theorem, we have
ZZ
Z
ZZ
~
~
~
~
curl F · dS =
F · d~r =
curl F~ · dS.
S1
C
S2
~ = curl F~ for some F~ , but not all vector fields are curl vector fields. (In
(b) This works if G
~ measures the defect – see
fact, this holds if div F~ = 0, and in fact the divergence of G
the divergence theorem, discussed later.)
~ = h0, 0, zi, S1 to be the unit disk in the xy-plane,
As a concrete counterexample, take G
and S2 to be the upper hemisphere of the unit sphere.
(c) F~ may not be a curl vector field, so no. For a counterexample, take S to be the unit
sphere and F~ = hx, y, zi.