# Integration via shell method

• July 27th 2010, 08:58 AM
Poptimus
Integration via shell method
Reviewing for my final and got one (of many) problems I keep getting a weird answer for. The problem reads:

Use the shell method to integrate. Find the volume generated by revolving the region bounded by the graphs of $y = \sqrt{x-1}$ and $y = x - 1$ about the line $x = 3$.

Using the shell method I set it up as $\int{(3 - x)(\sqrt{x - 1} - x - 1)$ from [1,2]. However, I get a negative volume. Any help would be greatly appreciated.

Thanks
• July 27th 2010, 09:09 AM
Failure
Quote:

Originally Posted by Poptimus
Reviewing for my final and got one (of many) problems I keep getting a weird answer for. The problem reads:

Use the shell method to integrate. Find the volume generated by revolving the region bounded by the graphs of $y = \sqrt{x-1}$ and $y = x - 1$ about the line $x = 3$.

Using the shell method I set it up as $\int{(3 - x)(\sqrt{x - 1} - \Big(x - 1\Big))$ from [1,2]. However, I get a negative volume. Any help would be greatly appreciated.

Thanks

Well, you have forgotten to put parentheses around (x-1), that's why you get a negative result. Also, you have left out a factor of $\scriptstyle 2\pi$, I believe.
• July 27th 2010, 09:11 AM
Ackbeet
Well, one possible reason you're getting negative numbers is that you should have

$V=2\pi\int_{1}^{2}(3 - x)(\sqrt{x - 1} - x + 1)\,dx.$

Sign error in there will do it every time. I think you also forgot the $2\pi$ factor out front. Don't forget: with the shell method, you could slice up one shell, flatten it out, and you'd have a rectangular prism whose volume is quite easy to compute: $dV=2\pi\,r\,h\,dx.$ The radius $r=3-x$, like you had.

[EDIT]: Failure failed to allow my post to go first. (Wink)
• July 27th 2010, 09:29 AM
Poptimus
Appreciate it guys, such a simple mistake. I can't believe I overlooked it.