Originally Posted by

**mathgeek777** Good afternoon.

I am working on a problem very similar to the following.

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Find the volume of the solid generated by revolving $\displaystyle x=3-y^2$, bounded by the x-axis, the y-axis, x=3 , and $\displaystyle y=\sqrt{2}$, about the x-axis.

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Based on the information given in this problem, I have been able to set up the integral to show what is below.

$\displaystyle \int_0^3 2\pi r (3-y^2) dy$

My one problem is the r in the integral. I know how to obtain it when the region is being revolved around the y-axis, but I am not sure how to obtain it when the region is being revolved around the x-axis. Is it the distance from the x-axis to the top of the bounded region, or do you have to obtain it via another method?

Thank you in advance for any assistance given.