Answers and Solutions to Section 13.2 Homework Problems 1"27 ...

Answers and Solutions to Section 13.2 Homework
Problems 1-27 (odd), 35, 37, and 39.
S. F. Ellermeyer
April 22, 2008
1. For C : x = t2 ; y = t; 0
Z
t 2, we have
Z 2
y ds = t jr 0 (t)j dt
where
Since
and
we see that Z
C
y ds =
Z 2
0
C
0
r (t) = t 2 i + tj.
r 0 (t) = 2ti + j
jr 0 (t)j = p 4t 1 + 1,
t p 4t 2 + 1 dt = 1
12 17p 17 1 .
3. C is the curve with parametric representation
Note that
so
Z
C
r (t) = 4 cos (t) i + 4 sin (t) j
2
t
2 .
r 0 (t) = 4 sin (t) i + 4 cos (t) j
jr 0 (t)j = 4
xy 4 Z
2
ds = (4 cos (t)) (4 sin (t)) 4 4 dt
= 4 6
2
Z
2
2
sin 4 (t) cos (t) dt
= 46
5 sin5 t= =2
(t) t= =2
= 2 (46 )
.
5
1

5. In this problem, the curve of integration (C) consists of the line segments
C1 : x = t; y = 0, 0 t 2
and
C2 : x = 2 + t; y = 2t; 0 t 1.
Note that for C1, we have dx = dt; dy = 0, and for C2 we have dx = dt,
dy = 2 dt. Thus
Z
Z
Z
xy dx + (x y) dy = xy dx + (x y) dy + xy dx + (x y) dy
C
7. For the curve
we have
so Z
C
9. For the curve
= 0 +
=
C1
Z 1
Z 1
0
= 17
3 .
0
C2
(2 + t) (2t) dt + (2 + t 2t) 2 dt
2t 2 + 2t + 4 dt
r (t) = 4 sin (t) i + 4 cos (t) j+3tk
0 t
2 ,
r 0 (t) = 4 cos (t) i 4 sin (t) j+3k
jr 0 (t)j = 5
xy 3 Z =2
ds = (4 sin (t)) (4 cos (t)) 3 5 dt = 320.
0
r (t) = ti + 2tj + 3tk
0 t 1
2

we have
so Z
C
xe yz ds = p 14
r 0 (t) = i + 2j + 3k
jr 0 (t)j = p 14
Z 1
0
te 6t2
dt =
p 14
11. (This is di¤erent from the problem 11 in the book.) The curve of
integration (C) consists of the line segments
12
C1 : x = 0; y = t; z = t; 0 t 1
e 6
1 .
C2 : x = t; y = 1 + t; z = 1 + 2t; 0 t 1
C3 : x = 1; y = 2; z = 3 + t; 0 t 1.
The di¤erentials on these curves are
For the curve C1 we have
Z
z 2 dx z dy + 2y dz =
C1
For the curve C2 we have
Z
z 2 dx z dy + 2y dz =
C2
C1 : dx = 0; dy = dt; dz = dt
C2 : dx = dt; dy = dt; dz = 2 dt
C3 : dx = 0; dy = 0; dz = dt.
=
=
Z 1
0
Z 1
0
Z 1
0
= 25
3 .
Z 1
0
t dt + 2t dt =
Z 1
0
t dt = 1
2 .
(1 + 2t) 2 dt (1 + 2t) dt + 2 (1 + t) 2 dt
(1 + 2t) 2 dt (1 + 2t) dt + 2 (1 + t) 2 dt
4t 2 + 6t + 4 dt
3

For the curve C3 we have
Z
z 2 dx z dy + 2y dz =
Thus Z
C
C3
Z 1
0
4 dt = 4.
z 2 dx z dy + 2y dz = 1 25 77
+ + 4 =
2 3 6 .
13. This is di¤erent from the number 13 in the book.
(a) The curve C1 is
Thus
Therefore Z
C1
r (t) = 3i + (6t 3) j
0 t 1.
F dr =
r 0 (t) = 6j.
Z 1
0
F (r (t)) 6j dt.
Since the vector 6j makes an acute angle with any of the vectors
F (r (t)), 0 t 1, the dot product F (r (t)) 6j is positive for all
t such that 0 t 1, meaning that the integrand is a positive–
valued function of t. Therefore
Z
F dr > 0.
(b) The curve C2 is
Thus
Therefore
Z
C2
F dr =
C1
r (t) = 3 cos (t) i + 3 sin (t) j
0 t 2 .
r 0 (t) = 3 sin (t) i + 3 cos (t) j.
Z 1
0
F (r (t)) ( 3 sin (t) i + 3 cos (t)) dt.
4

Recall that, since r (t) is a circle, the tangent vectors r 0 (t) are
orthogonal to this circle. However, the circle is oriented counterclockwise
and the vectors of the vector …eld F are all pointing the
the clockwise direction. In fact, it appears that the angle between
F (r (t)) and r 0 (t) is equal to 180 for each t: This means that the
integrand F (r (t)) r 0 (t) is a negative–valued function for each t
such that 0 t 2 . Therefore
Z
C2
F dr < 0.
15. (This is di¤erent from the number 15 in the book.) For
19. For
we have
Thus, for the vector …eld
we have
Therefore
we have
r (t) = t 2 i t 3 j, 0 t 1,
r 0 (t) = 2ti 3t 2 j.
F (x; y) = x 2 y 3 i y p xj
F (r (t)) r 0 (t) = t 13 i + t 4 j 2ti 3t 2 j = 2t 14
Z
C
F dr=
=
Z 1
0
Z 1
0
= 59
105 .
F (r (t)) r 0 (t) dt
2t 14
3t 6 dt
r (t) = t 3 i t 2 j + tk, 0 t 1,
r 0 (t) = 3t 2 i 2tj + k.
5
3t 6 .

Thus, for the vector …eld
we have
F (x; y; z) = sin (x) i + cos (y) j+xzk
F (r (t)) r 0 (t) = sin t 3 i + cos t 2 j+t 4 k 3t 2 i 2tj + k
Therefore
Z
C
F dr=
= 3t 2 sin t 3
=
Z 1
0
Z 1
0
= 6
5
F (r (t)) r 0 (t) dt
3t 2 sin t 3
2t cos t 2 + t 4 .
cos (1) sin (1) .
2t cos t 2 + t 4 dt
21. The vector …eld F (x; y) = (x y) i + xyj and curve are shown below.
32123
32123
6

Since the curve is traversed counterclockwise, it appears that the tangent
vectors to the curve are mostly pointing in the same general direction
as the vectors of the vector …eld. We thus suspect that the line
integral will be positive. We now evaluate the integral. Since the curve
C is given by
we have
and
r (t) = 2 cos (t) i + 2 sin (t) j
0 t
3
2
r 0 (t) = 2 sin (t) i + 2 cos (t) j
F (r (t)) r 0 (t) = ((2 cos (t) 2 sin (t)) i + 4 sin (t) cos (t) j) ( 2 sin (t) i + 2 cos (t) j)
Thus
Z
F dr =
C
=
= 4 sin (t) cos (t) + 4 sin 2 (t) + 8 cos 2 (t) sin (t) .
Z 3 =2
0
Z 3 =2
0
= 2
+ 3 .
3
F (r (t)) r 0 (t) dt
4 sin (t) cos (t) + 4 sin 2 (t) + 8 cos 2 (t) sin (t) dt
25. The portion of the astroid in the …rst quadrant is shown here. It is
traced out as t ranges from 0 to =2.
7

1.5 1.0 0.5 0.5 1.0 1.5 1.5 1.0 0.5 0.5 1.0 1.5
Since
we see that
and thus
Z
r 0 (t) = 3 cos 2 (t) sin (t) i + 3 sin 2 (t) cos (t) j,
jr 0 (t)j =
x
C
3 y 5 ds =
q
9 cos 4 (t) sin 2 (t) + 9 sin 4 (t) cos 2 (t)
= 3 sin (t) cos (t)
Z =2
0
Z =2
= 3
=
0
945
16; 777; 216
cos 3 (t) 3 sin 3 (t) 5 (3 sin (t) cos (t)) dt
cos 10 (t) sin 16 (t) dt
(This integral was computed by Maple.)
8

29. (a)
x = 1
Z
m
y = 1
m
z = 1
m
Z
Z
C
C
C
x (x; y; z) ds
y (x; y; z) ds
z (x; y; z) ds
(b) For the helical wire C : x = 2 sin (t) ; y = 2 cos (t) ; z = 3t;
0 t 2 and the density function (x; y; z) = k (constant),
the total mass of the wire is
Z
m =
ZC
(x; y; z) ds
= k ds
= k
= k
C
Z 2
0
Z 2
0
= 2 p 13 k
jr 0 (t)j dt
p 13 dt
and the center of mass is at (x; y; z) where
Z 2
1
x =
2 p 13 k
1
y =
2 p 13 k 0
1
z =
2 p 13 k 0
= 3
Z 2
t dt
2 0
= 3
(2 )2
4
= 3 .
0
Z 2
Z 2
9
k (2 sin (t)) p 13 dt = 0
k (2 cos (t)) p 13 dt = 0
k (3t) p 13 dt

31. The work done by the force …eld
F (x; y) = xi+ (y + 2) j
in moving an object along an arch of the cycloid
r (t) = (t sin (t)) i+ (1 cos (t)) j
(corresponding to the positive t direction for 0 t 2 ) is
Z Z 2
F dr = F (r (t)) r 0 (t) dt.
Since
and
we have
C
0
F (r (t)) = (t sin (t)) i+ ((1 cos (t)) + 2) j
= (t sin (t)) i+ (3 cos (t)) j
r 0 (t) = (1 cos (t)) i+ sin (t) j,
F (r (t)) r 0 (t) = ((t sin (t)) i+ (3 cos (t)) j) ((1 cos (t)) i+ sin (t) j)
= (t sin (t)) (1 cos (t)) + (3 cos (t)) sin (t)
= t t cos (t) + 2 sin (t) .
Therefore, the work done is
Z Z 2
F dr = (t t cos (t) + 2 sin (t)) dt = 2 2 .
C
33. The work done by the force …eld
0
F (x; y; z) = xzi+yxj+zyk
on a particle that moves along the curve
is Z
r (t) = t 2 i t 3 j+t 4 k, 0 t 1
C
F dr =
Z 1
0
10
F (r (t)) r 0 (t) dt.

Since
and
we have
F (r (t)) = t 6 i t 5 j t 7 k
r 0 (t) = 2ti 3t 2 j + 4t 3 k,
F (r (t)) r 0 (t) = t 6 i t 5 j t 7 k 2ti 3t 2 j + 4t 3 k
= 2t 7 + 3t 7
= 5t 7
Therefore, the work done is
Z
C
Z 1
F dr =
0
4t 10
5t 7
4t 10
4t 10 dt = 23
88 .
37. The magnitude of the force exerted by gravity on the man and paint
can is the combined weight of the man and the paint can. This is
185 lb. This force acts in the downward (negative z) direction. Thus,
the gravitational force …eld acting on the man and paint can is
F (x; y; z) = 185k.
The helical staircase has parametric representation
r (t) = 20 cos (t) i+20 sin (t) j+ 15 tk
0 t 6 .
We want to …nd the work done by the gravitational force …eld on the
man and the paint can (considered as one object moving from the
bottom to the top of the staircase). The work done is
Since
Z
C
Z 6
F dr =
0
F (r (t)) r 0 (t) dt.
r 0 (t) = 20 sin (t) i+20 cos (t) j+ 15 k
11

we have
F (r (t)) r 0 (t) = ( 185k) 20 sin (t) i+20 cos (t) j+ 15 k = 2; 775 lb .
Thus, the work done by gravity here is
Z
C
Z 6
F dr =
0
2; 775
dt = 16; 650 ft lb .
The work done by the man against gravity is 16; 650 ft lb.
12