Okay so I'm having trouble formulating this integral. The region is bounded by:

$\displaystyle {x}^{2}-{y}^{2}=7$ and $\displaystyle x=4$

rotated about: $\displaystyle y=5$

I tried using shells.. It didn't really work for me

Thanks,

CC

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- May 1st 2010, 06:56 AMCalculusCrazedRotated about the...
Okay so I'm having trouble formulating this integral. The region is bounded by:

$\displaystyle {x}^{2}-{y}^{2}=7$ and $\displaystyle x=4$

rotated about: $\displaystyle y=5$

I tried using shells.. It didn't really work for me

Thanks,

CC - May 1st 2010, 11:06 AMFailure
- May 1st 2010, 11:11 AMskeeter
- May 2nd 2010, 12:07 PMCalculusCrazed
Okay so you did R meaning the radius of the big circle and r for the little. How did you find the bounds?

- May 2nd 2010, 12:17 PMskeeter
- May 2nd 2010, 12:26 PMCalculusCrazed
Okay, so why did you do R^2 - r^2?

- May 2nd 2010, 12:35 PMskeeter
volume by the method of washers ... a washer is a disk with a hole in it.

$\displaystyle V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx$

The Washer Method for Solids of Revolution - May 2nd 2010, 02:57 PMCalculusCrazed
The little circle minus the big circle. Okay I guess I just don't get how you got root 7 to 4

- May 2nd 2010, 05:35 PMskeeter
$\displaystyle x = \sqrt{7}$ is the vertex of the hyperbola at the left end of the defined region to be rotated.

$\displaystyle x = 4$ is the right boundary of the region to be rotated.

I recommend you meet with your instructor ... this is all very basic material which should have been covered in class. - May 2nd 2010, 05:47 PMCalculusCrazed
...plug in zero for y in other words.

Thank you for your concern.