Page 246 - Calculus Demystified

CHAPTER 8
Applications of the Integral
2
generated has radius y −2, height 8−2(y −2) , and thickness y. It therefore 233
generates the element of volume given by
2
2π · (y − 2) ·[8 − 2(y − 2) ]· y.
The aggregate volume that we seek is therefore
4

2
V = 2π · (y − 2) ·[8 − 2(y − 2) ] dy
0
4
3
= 16π(y − 2) − 4π(y − 2) dy
0
2
= 8π(y − 2) − π(y − 4) 4 4
0
= 256π.
You Try It: Calculate the volume enclosed when the curve y = cos x is rotated
about the line y = 4, π ≤ x ≤ 3π.

8.3 Work

One of the basic principles of physics is that work performed is force times distance:
If you apply force F pounds in moving an object d feet, then the work is
W = F · d foot-pounds.
The problem becomes more interesting (than simple arithmetic) if the force is
varying from point to point. We now consider some problems of that type.
EXAMPLE 8.13
A weight ispushed in the plane from x = 0to x = 10. Because of a
2
prevailing wind, the force that must be applied at point x is F(x) = 3x −
x + 10 foot-pounds. What is the total work performed?

0 10
Fig. 8.23

SOLUTION
Followingthewaythatweusuallydothingsincalculus,webreaktheproblem
up into pieces. In moving the object from position x to position x + x, the