Math 2263 Multivariable Calculus
Homework 23: 16.2 #8,20
July 18, 2011
16.2#8
R
Evaluate the line integral C sin x dx + cos y dy, where C consists of the top half of the circle
x2 + y 2 = 1 from (1, 0) to (−1, 0), and the line segment from (−1, 0) to (−2, 3).
We can parametrize C as two curves C1 and C2 , where C1 is parametrized by
r1 (t) = hcos t, sin ti, 0 ≤ t ≤ π
and C2 is parametrized by
r2 (t) = h−1 − t, 3ti, 0 ≤ t ≤ 1.
Z
Z
sin x dx + cos y dy =
=
π
Z
sin x dx + cos y dy +
C
Z
C1
sin x dx + cos y dy
C2
Z
sin(cos t)(− sin t) + cos(sin t)(cos t) dt +
0
1
sin(−1 − t)(−1) + cos(3t)(3) dt
0
=
π
π 1
1 − cos(cos(t)) + sin(sin(t)) + − cos(−1 − t) + sin(3t)
0
0
0
0
= (− cos(−1)) − (− cos(1)) + sin(0) − sin(0) + (− cos(−2)) − (− cos(−1)) + sin(3) − sin(0)
= cos(1) + sin(3) − cos(2)
16.2#20
R
Evaluate the line integral C F · dr, where C is given by the vector function
r(t) = t2 i + t3 j + t2 k, 0 ≤ t ≤ 1, and where F (x, y) = (x + y)i + (y − z)j + z 2 k.
Z
Z 1
F · dr =
F (r(t)) · r0 (t) dt
Z
C
1
=
Z
0
1
=
0
ht2 + t3 , t3 − t2 , t4 i · h2t, 3t2 , 2ti dt
3
4
5
4
5
(2t + 2t ) + (3t − 3t ) + 2t dt
0
1
Z
(2t3 − t4 + 5t5 ) dt
=
0
=
1 4 1 5
t − t −
2
5
1 1
= − +
2 5
17
=
15
1
5 6 t 6
0
5
6