# Volume of Rotation of a Quarter Circle

• Mar 2nd 2010, 06:47 PM
Volume of Rotation of a Quarter Circle
I am trying to find the volume of a quarter circle that is rotated about $x=1$. So the equation should look like this:
$
\int_0^{\frac {\pi} {2}} \pi(\sqrt {1^2-y^2})^2dy$

This should simplify to:
$\int_0^{\frac {\pi} {2}} \pi(1-y^2)dy$

If I am doing this right thus far, then the integral is:
$\pi(y-\frac {y^3} {3})$ from 0 to $\frac {\pi} {2}$

And when I plug these values in I some how end up with $\frac {\pi^2} {2} - \frac {\pi^4} {24}$

I feel like I am missing something in between here, or maybe just made an error plugging in the values, although I have tried it a few times and end up with the same answer. Is this right? Thanks for the help.
• Mar 3rd 2010, 03:58 AM
Quote:

I am trying to find the volume of a quarter circle that is rotated about $x=1$. So the equation should look like this:
$
\int_0^{\frac {\pi} {2}} \pi(\sqrt {1^2-y^2})^2dy$

This should simplify to:
$\int_0^{\frac {\pi} {2}} \pi(1-y^2)dy$

If I am doing this right thus far, then the integral is:
$\pi(y-\frac {y^3} {3})$ from 0 to $\frac {\pi} {2}$

And when I plug these values in I some how end up with $\frac {\pi^2} {2} - \frac {\pi^4} {24}$

I feel like I am missing something in between here, or maybe just made an error plugging in the values, although I have tried it a few times and end up with the same answer. Is this right? Thanks for the help.

The limits of the integral are $0$ to $1$, not $0$ to $\pi/2$.