Thomas Calculus 13th [Solutions]

502 Chapter 6 Applications of Definite Integrals
4. (a) The graph of f ( x) sin x traces out a path from (0, 0) to ( , sin ) whose length is
0
2
L 1 cos d . The line segment from (0, 0) to ( , sin ) has length
2 2 2 2
( 0) (sin 0) sin . Since the shortest distance between two points is the length of
the straight line segment joining them, we have immediately that
0
2 2 2
d
1 cos sin if 0 .
(b) In general, if y f ( x ) is continuously differentiable and f (0) 0, then
0
2 2 2
1 f ( t) dt f ( ) for 0.
2
5. We can find the centroid and then use Pappus Theorem to calculate the volume.
2 2
f ( x) g( x) x x x x 0 x 0, x 1; 1;
1 1
0
1
;
2 3 6
2
f ( x) x, g( x) x ,
M 1 2 2 3
1
1 1
0 x x dx 2 x 3
x 0
1 2 1 2 3 3 4
1
x 1 x x x dx x x dx 1 x 1 x 1 1 1
1/6 0 0 3 4 0 3 4 2
6 6 6 0 ;
1 2 2
2 1 2 4 3 5
1
1 1
3 3
1 1
3
1 1
0
2
1/6 0 2 0 3 5 0 3 5 3
y x x dx x x dx x x The centroid is
1
,
2
.
is the distance from
1
,
2
to the axis of rotation, y x . To calculate this distance we must find the point
2 5
on y x that also lies on the line perpendicular to y x that passes through
1
,
2
. The equation of this line
2 5
is y 2
1 x 1 x y 9
. The point of intersection of the lines x y 9
and y x is
9
,
9
.
Thus,
5 2 10
9
2
9
2
1 2 1
10 2 20 5 . Thus V 2 1 1
10 2 6
.
10 2 30 2
10
20 20
6. Since the slice is made at an angle of 45 , the volume of the wedge is half the volume of the cylinder of radius
1
2 and height 1. Thus, 1 1
2
V
(1) .
2 2 8
2 5
7.
y 2 x ds 3 3/2
3
1 1 4
28
x
1 dx A 0 2 x x
1 dx 3 (1 x)
0 3
8. This surface is a triangle having a base of 2 a and a height of 2 ak . Therefore the surface area is
1
2 2
(2 )(2 ) 2 .
2 a ak a k
9.
2 2 3
F ma t 2 d a t dx t
;
2
dt m v dt 3m
C 0
x 0 when
t
1 12m
t 0 C 0 x . Then
4
2 4
dx t t
dt 3m 12m
v when t 0 C 0 x C ;
1/4
x h t (12 mh ) . The work done is
1/4 1/4 mh 1/4
3/2
(12 mh) (12 mh) 3 6 (12 )
2 1 1 6/4 (12 mh)
W F dx F( t) dx dt t t dt t
(12 mh)
0 dt 0 3m 3m 6
0
18m 18m
1
12 mh 12 mh 2 h mh 4 h
18m
3 3
2 3 3mh
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