MA 242 - Fall 2010
Worksheet X1
Stokes’ Theorem
Remark 0.1. Figure X corresponds to Problem X.
Remark 0.2. To maximize the effectiveness of this worksheet, you should use Stokes’ Theorem for all problems.
Remark 0.3. Repeatedly throughout this worksheet you will want to use the fact that
cos2 (θ) = 21 (1 + cos(2θ)).
1. Let S be the part of the paraboloid z = 5 − x2 − y 2 that lies above the plane z = 1.
~
Assume that S has the upwards orientation. Let F(x,
y, z) = x2 z, x + z, xyez be a
3
vector field on R . Use Stokes’ Theorem to evaluate the surface
integral of the vector
ZZ
~ · d~S using Stokes’
~
curlF
field curl F(x,
y, z) over the surface S. That is, evaluate
S
Theorem.
Solution Steps 0.1. Parametric equations for the surface S can be taken to be
x = u cos(v),
y = u sin(v),
z = 5 − u2 ,
2
0 ≤ u ≤ 2, 0 ≤ v ≤ 2π.
The boundary curve C is the circle of radius 2 centered at (0,0,1) that lies in the
plane z = 1. Since S has been oriented upwards, then the boundary curve C needs
to be traversed counterclockwise when viewed from above. The appropriate parametric
equations for C can be taken to be
x = 2 cos(t),
y = 2 sin(t),
z = 1,
0 ≤ t ≤ 2π.
Figure 1: The plotZ Z
of the surface S and the boundary curve
Z C. Stokes’ theorem says that in
~ · d~S, one can instead evaluate F
~ · d~r.
order to evaluate
curlF
S
1
2
C
A.K.A. ’The Final Worksheet’
If you use Stokes’ Theorem, then these parametric equations are not needed.
2. Let S be the part of the sphere x2 + y 2 + z 2 = 5 that lies above
2 the plane zz = 1.
~
Assume that S has the upwards orientation. Let F(x, y, z) = x z, x + z, xye be a
vector field on R3 . Use Stokes’ Theorem to evaluate the surface
integral of the vector
ZZ
~ · d~S using Stokes’
~
curlF
field curl F(x,
y, z) over the surface S. That is, evaluate
S
Theorem.
Solution Steps 0.2. Parametric equations for the surface S can be taken to be
x=
√
5 sin(v) cos(u),
y=
√
5 sin(v) sin(u),
z=
!
√
5
0 ≤ u ≤ 2π, 0 ≤ v ≤ arccos
.
5
3
√
5 cos(v),
The boundary curve C is the circle of radius 2 centered at (0,0,1) that lies in the
plane z = 1. Since S has been oriented upwards, then the boundary curve C needs
to be traversed counterclockwise when viewed from above. The appropriate parametric
equations for C can be taken to be
x = 2 cos(t),
y = 2 sin(t),
z = 1,
0 ≤ t ≤ 2π.
Figure 2: The plotZ Z
of the surface S and the boundary curve
Z C. Stokes’ theorem says that in
~ · d~S, one can instead evaluate F
~ · d~r.
order to evaluate
curlF
S
3
C
If you use Stokes’ Theorem, then these parametric equations are not needed.
3. Use you answers to questions one and two to answer each of the following.
(a) Compare your answers from problem 1 and problem 2. What do you notice?
(b) Observe that the vector fields in problems 1 and 2 are identical. Furthermore,
observe that the surfaces in problems 1 and 2 share a common boundary and that
they are also oriented in the same manner.
(c) Explain how all of your observations make sense and explain why Stokes’ Theorem
guarantees that all of your observations will hold.
Figure 3: The surfaces in problems 1 and
Z Z 2 share a common boundary curve C.
Z Stokes’
~ · d~S, one can instead evaluate
~ · d~r.
theorem says that in order to evaluate
curlF
F
C
S
ZZ
This means that
S1
~ · dS~1 =
curlF
Z
C
~ · d~r =
F
ZZ
S2
~ · d~S2 . (Note: orientations agree)
curlF
~
4. (Turning the tables) Let F(x,
y, z) = x2 z, x + z, xyez be a vector field on R3 . Let C
be the curve given by the parametric equations
x = 2 cos(t),
y = 2 sin(t),
0 ≤ t ≤ 2π.
z = 1,
~ along the curve C.
Use Stokes’ Theorem to evaluate the line integral of the vector field F
ZZ
Solution Steps 0.3. In this case, we want to evaluate
~ · d~S, where S is some
curlF
S
surface oriented upwards with boundary curve C. We should pick S to be a very simple
surface. Let S to be the disk of radius 2 that lies in the plane z = 1 with boundary
curve C.
Parametric equations for S are
x = v cos(u),
y = v sin(u),
z = 1,
0 ≤ u ≤ 2π, 0 ≤ v ≤ 2,
and the unit normal vector for S is ~n = h0, 0, 1i .
Z
ZZ
~
~ · d~S.
Now use Stokes’ theorem to evaluate
F · d~r by evaluating
curlF
C
S
Figure 4: When you want to use Stokes Theorem to evaluate the line integral of a vector
field along a curve C, you should pick the surface that best fits the problem. The surface
~ will be the same along any surface that shares the common boundary
integral of the curl F
curve (and is oriented appropriately).
Remark 0.4. The following two problems have been taken (roughly) straight from the
text. They are intended to offer additional practice in computation. The integration
should be tolerable, but at some point in problem 5 you will need to use trig identities
in order to complete the integral by hand. My recommendation is to get everything set
up in terms of t and then use integration tables, maple, or your graphing calculator to
complete the integral
ZZ
5. Use Stokes theorem to evaluate
~ · d~S, where F
~ is the vector field F(x,
~
curlF
y, z) =
S
x2 y 3 z, sin(xyz), xyz and S is the part of the cone y 2 = x2 + z 2 that lies between
the the planes y = 0 and y = 2 and is oriented in the direction the positive y axis.
Solution Steps 0.4. Note that the boundary curve is the circle of radius 1 that is
centered at (0, 1, 0) and lies in the plane y = 1. This curve can be given a positive
orientation with the parametric equations
x = sin(t),
y = 1,
z = cos(t),
0 ≤ t ≤ 2π.
ZZ
6. Use Stokes theorem to evaluate
~ · d~S, where F
~ is the vector field F(x,
~
curlF
y, z) =
S
h−y, x, −2i and S is the part of the cone z 2 = x2 + y 2 that lies between the the planes
z = 0 and z = 4 and is oriented downwards.
Solution Steps 0.5. Note that the boundary curve is the circle of radius 4 that is
centered at (0, 0, 4) and lies in the plane z = 4. This curve can be given a positive
orientation with the parametric equations
x = 4 sin(t),
y = 4 cos(t),
z = 4,
0 ≤ t ≤ 2π.