• May 1st 2010, 06:56 AM
CalculusCrazed
Okay so I'm having trouble formulating this integral. The region is bounded by:

${x}^{2}-{y}^{2}=7$ and $x=4$
rotated about: $y=5$

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

Thanks,
CC
• May 1st 2010, 11:06 AM
Failure
Quote:

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

${x}^{2}-{y}^{2}=7$ and $x=4$
rotated about: $y=5$

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

Thanks,
CC

$V=2\pi \int_{\sqrt{7}}^4 x \cdot {\color{red}2}\sqrt{x^2-7}\, dx=\ldots =2\cdot 18\pi=36\pi$?
(use substitution $u := x^2-7$ to solve this integral)
• May 1st 2010, 11:11 AM
skeeter
Quote:

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

${x}^{2}-{y}^{2}=7$ and $x=4$
rotated about: $y=5$

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

Thanks,
CC

washers w/r to x ...

$R = 5 + \sqrt{x^2-7}
$

$r = 5 - \sqrt{x^2-7}$

$R^2 - r^2 = 25 - (x^2 - 7) = 32 - x^2$

$V = \pi \int_{\sqrt{7}}^4 32 - x^2 \, dx$
• May 2nd 2010, 12:07 PM
CalculusCrazed
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 PM
skeeter
Quote:

Originally Posted by CalculusCrazed
Okay so you did R meaning the radius of the big circle and r for the little. How did you find the bounds?

I sketched a graph ...

upper branch is $y = \sqrt{x^2-7}$

lower branch is $y = -\sqrt{x^2-7}$

R = 5 - lower branch

r = 5 - upper branch
• May 2nd 2010, 12:26 PM
CalculusCrazed
Okay, so why did you do R^2 - r^2?
• May 2nd 2010, 12:35 PM
skeeter
Quote:

Originally Posted by CalculusCrazed
Okay, so why did you do R^2 - r^2?

volume by the method of washers ... a washer is a disk with a hole in it.

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

The Washer Method for Solids of Revolution
• May 2nd 2010, 02:57 PM
CalculusCrazed
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 PM
skeeter
Quote:

Originally Posted by CalculusCrazed
The little circle minus the big circle. Okay I guess I just don't get how you got root 7 to 4

$x = \sqrt{7}$ is the vertex of the hyperbola at the left end of the defined region to be rotated.

$x = 4$ 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 PM
CalculusCrazed
...plug in zero for y in other words.