# Another Volume Problem

• Oct 4th 2008, 01:00 PM
veronicak5678
Another Volume Problem
I would like to know if I have set this up correctly:

Find the volume of a solid obtained by rotating the region bound by y = cos^2 (x) , -pi/2 < x < pi/2, y = 1/4 about the line x = pi/2.

I came up with

2 pi Integral from -pi/2 to 0 of (1+x)(cos^2x - pi/2)
+
2 pi Integral from 0 to pi/2 of (1-x)(cos^2x - pi/2)

Does this seem OK?
• Oct 4th 2008, 02:24 PM
skeeter
ever use the method of cylindrical shells?

$\displaystyle V = 2\pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left(\frac{\pi}{2} - x\right)\left(\cos^2{x} - \frac{1}{4}\right) \, dx$
• Oct 4th 2008, 02:35 PM
veronicak5678
I thought I was using shells!

Dont I need two different integral since the radius changes from positive to negative?
• Oct 4th 2008, 02:45 PM
skeeter
no ... the radius is $\displaystyle r = \left(\frac{\pi}{2} - x\right)$ no matter what x is.

think about the value of $\displaystyle \left(\frac{\pi}{2} - x\right)$ for $\displaystyle x > 0$ and $\displaystyle x < 0$.
• Oct 4th 2008, 02:53 PM
veronicak5678
OK, I think I understand. I don't really know what I was trying to do! Thanks a lot for helping.