volume by shell; find my error

**Tclack** · April 6th 2010, 08:27 PM

Find volume of shape between: $y= \sqrt{x}, y=0 , x=9$
rotated about line x=9
So, since I'm going around vertically, I rearranged $x=y^2$

$\pi \int_0^3 (9-y^2)^2 dy$
$=\pi \int_0^3 (81-18y^2+y^4) dy$
$=\pi(81y-6y^2+\frac{y^5}{5})\mid_0^3$
$=\frac{1188}{5}\pi$

the answer is supposed to be 648pi/5

The similarities between my answer and their answer makes me think my method is correct. See attached picture. I'm trying to get the shape to be like a. But I suspect my equation is giving me b. Especially because my answer is larger, and b looks like it would have a larger volume.
Can anyone clarify?

**HallsofIvy** · April 7th 2010, 03:10 AM

Originally Posted by Tclack

Find volume of shape between: $y= \sqrt{x}, y=0 , x=9$
rotated about line x=9
So, since I'm going around vertically, I rearranged $x=y^2$

So you are using disks not "shells".

$\pi \int_0^3 (9-y^2)^2 dy$
$=\pi \int_0^3 (81-18y^2+y^4) dy$
$=\pi(81y-6y^2+\frac{y^5}{5})\mid_0^3$

You have a wrong power:
$=\pi(81y-6y^3+\frac{y^5}{5})\mid_0^3$

$=\frac{1188}{5}\pi$

the answer is supposed to be 648pi/5

The similarities between my answer and their answer makes me think my method is correct. See attached picture. I'm trying to get the shape to be like a. But I suspect my equation is giving me b. Especially because my answer is larger, and b looks like it would have a larger volume.
Can anyone clarify?

**Tclack** · April 7th 2010, 04:02 PM

Damn, thanks, I went over it 4 times and I didn't catch it. So I thought the error was in the setup of the integral.

Do the volumes of A and B equal each other?

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