The region in the first quadrant bounded by the x-axis, the line x = π, and the curve y = cos(cos(x)) is rotated about the x-axis. What is the volume of the generated solid?

Question

a. 1.921
b. 3.782
c. 6.040
d. 8.130

Steve · Accepted Answer

think of the volume as a stack of discs of thickness dx, and you have

v = ?[0,?] ?r^2 dx
where r=y=cos(cos(x))
v = ?[0,?] ?cos^2(cosx) dx

This is not an elementary integral, so some numeric method is needed. However, using some symmetry,

v = 2?[0,?/2] ?cos^2(cosx) dx
and you can see that the volume is close to that of a truncated cone.

http://www.wolframalpha.com/input/?i=cos(cos(x))

The curve is close to the line

y = cos(1) + (1-cos(1))/(?/2) x
= 0.54 + 0.29x

So, the peak of the cone, which has been removed has a base radius of 0.54 and a height of 1.862

That makes the semi-volume nearly

?/3 (1^2)(1.862+?/2) - ?/3 (0.54^2)(1.862) = 3.02

So the whole volume is 6.04

This can be checked here:

http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,%CF%80%5D+%CF%80cos%5E2(cosx)+dx