
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 (1x)(cos^2x  pi/2)
Does this seem OK?

ever use the method of cylindrical shells?
$\displaystyle V = 2\pi \int_{\frac{\pi}{3}}^{\frac{\pi}{3}} \left(\frac{\pi}{2}  x\right)\left(\cos^2{x}  \frac{1}{4}\right) \, dx$

I thought I was using shells!
Dont I need two different integral since the radius changes from positive to negative?

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 $\displaystyle x < 0$.

OK, I think I understand. I don't really know what I was trying to do! Thanks a lot for helping.