Volume of Rotation of a Quarter Circle

**Spudwad** · March 2nd 2010, 06:47 PM

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

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.

**Grandad** · March 3rd 2010, 03:58 AM

Hello Spudwad

Originally Posted by Spudwad

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

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.

Get ready to kick yourself:

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

Grandad

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