Math 2263 Multivariable Calculus
Homework 32: 16.8 #8, 14
July 29, 2011
16.8#8
RR
Use Stokes’ Theorem to evaluate C F · dr, where
F(x, y, z) = e−x i + ex j + ez k
and C is the boundary of the part of the plane 2x + y + 2z = 2 in the first octant, oriented
counterclockwise as viewed from above.
The curve C is the boundary of the surface S parmetrized by
r(x, z) = hx, 2 − 2x − 2z, zi
with
rx × rz = h−2, 0, −2i.
To orient the surface correctly, we use the Right Hand Rule to determine that the
normal vector must point in the positive z-direction, so we will use
rz × rx = h2, 0, 2i.
When we apply Stokes’ Theorem, we will need to know the curl of F:
curl(F) = ∇ × F = h0, 0, ex i.
ZZ
Z
∇ × F · dS
F · dr =
S
C
1Z
Z
1−z
h0, 0, ex i · h2, 0, 2i dx dz
=
0
0
1
Z
Z
=
0
Z
=
1−z
2ex dx dz
0
1
(2e1−z − 2) dz
0
= 2e − 4
16.8#14
Verify that Stokes’ Theorem is true for the vector field
F(x, y, z) = xi + yj + xyzk,
where S is the part of the plane 2x + y + z = 2 that lies in the first octant, oriented upward.
To verify Stokes’ Theorem, we will need to compute both a surface integral and a line
integral. We will begin with the surface integral.
The surface is parametrized by
r(x, y) = hx, y, 2 − 2x − yi with 0 ≤ y ≤ 2 − 2x and 0 ≤ x ≤ 1.
We have rx × ry = h2, 1, 1i, which gives the correct (upward) orientation.
We find ∇ × F = hxz, −yz, 0i.
Z 1 Z 2−2x
ZZ
hx(2 − 2x − y), −y(2, 2x − y), 0i · h2, 1, 1i dy dx
curl(F) · dS =
0
S
0
Z
1
Z
2−2x
(4x − 2y + y 2 − 4x2 ) dy dx
=
0
0
=0
Now we turn to the line integral. The boundary of the surface S consists of three line
segments. Let L1 be the line from (1, 0, 0) to (0, 2, 0); let L2 be the line from (0, 2, 0) to
(0, 0, 2); and let L3 be the line from (0, 0, 2) to (1, 0, 0).
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
∂S
L1
L2
L3
I will parametrize each line with r(t) = (1 − t)a + tb.
L1 : r1 (t) = h1 − t, 2t, 0i, 0 ≤ t ≤ 1
L2 : r2 (t) = h0, 2 − 2t, 2ti, 0 ≤ t ≤ 1
L3 : r3 (t) = ht, 0, 2 − 2ti, 0 ≤ t ≤ 1
Along L1 we have
Z
Z
1
h1 − t, 2t, 0i · h−1, 2, 0i dt
F · dr =
L1
0
1
Z
(5t − 1) dt
=
0
3
= .
2
Along L2 we have
Z
Z
F · dr =
L2
1
h0, 2 − 2t, 0i · h0, −2, 2i dt
0
1
Z
(4t − 4) dt
=
0
= −2.
Along L3 we have
Z
Z
F · dr =
L3
1
ht, 0, 0i · h1, 0, −2i dt
0
Z
1
t dt
=
0
1
= .
2
R
The sum of the values of the three line integrals which make up ∂S F · dr is 0, so both
sides of Stokes’ Theorem agree.