# Integration problem - Volume by Revolution

• Apr 24th 2010, 07:51 AM
bobbert116
Integration problem - Volume by Revolution
Hello everyone. I'm having trouble solving this problem:

Finding the volume by revolution of y=3-2x, rotated around the y-axis. I have the integration set up as $\displaystyle pi* ((y-3)/-2)^2 dy$ from $\displaystyle -3/2...3/2$ . Did I set this up wrong since I'm dealing with a straight line and not a curve? I can't seem to get the answer that my book gives.

Thanks in advance for any help!
• Apr 24th 2010, 09:09 AM
Quote:

Originally Posted by bobbert116
Hello everyone. I'm having trouble solving this problem:

Finding the volume by revolution of y=3-2x, rotated around the y-axis. I have the integration set up as $\displaystyle pi* ((y-3)/-2)^2 dy$ from $\displaystyle -3/2...3/2$ . Did I set this up wrong since I'm dealing with a straight line and not a curve? I can't seem to get the answer that my book gives.

Thanks in advance for any help!

Hi bobbert116,

Your limits are on the x-axis while they should be on the y-axis.

You are integrating horizontal discs parallel to the x-axis,
$\displaystyle \int_{0}^3{{\pi}\left(\frac{3-y}{2}\right)^2}dy$
$\displaystyle \frac{{\pi}r^2h}{3}=\frac{{\pi}\left(\frac{3}{2}\r ight)^2(3)}{3}$