Another Volume Problem

**veronicak5678** · October 4th 2008, 01:00 PM

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?

**skeeter** · October 4th 2008, 02:24 PM

ever use the method of cylindrical shells?

$V = 2\pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left(\frac{\pi}{2} - x\right)\left(\cos^2{x} - \frac{1}{4}\right) \, dx$

**veronicak5678** · October 4th 2008, 02:35 PM

I thought I was using shells!

Dont I need two different integral since the radius changes from positive to negative?

**skeeter** · October 4th 2008, 02:45 PM

no ... the radius is $r = \left(\frac{\pi}{2} - x\right)$ no matter what x is.

think about the value of $\left(\frac{\pi}{2} - x\right)$ for $x > 0$ and $x < 0$ .

**veronicak5678** · October 4th 2008, 02:53 PM

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

Thread: Another Volume Problem

LinkBack

Thread Tools

Display

Another Volume Problem

Similar Math Help Forum Discussions

How to set this volume problem?

Volume Word Problem (geometry-type problem for physics)

Volume problem...

Volume problem

A volume/SA problem

Search Tags