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Use the Shell Method to compute the volume of the solid obtained by rotating the region bound by the graph $4y=4^2-x^2$ and the $x$-axis about the line $x=6$.

Hello, I've been having a lot of problems with this question. I feel fairly certain about what I am doing, but the answer is not correct. I'm not really looking for the answer, but rather help in determining where I am going wrong. Currently I'm working with the idea that the radius is $(6-x)$ and the height is $16-x^2$. I'm using the shell method so my area before integrating is $2\pi(6-x)(16-x^2)$, and the bounds being $0$ and $4$.

The answer I keep getting is $384\pi$, which is incorrect.

My guess is I'm making a mistake with the radius, but everything I'm reading points to it being correct. Any help would be appreciated.

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  • $\begingroup$ you are doing the question as if the solid were bounded by y=16-x^2 and y=0 and x=0. But I don't see anything about x=0 (aka y-axis). $\endgroup$ – randomgirl Feb 3 '15 at 3:02
  • $\begingroup$ You're absolutely right. Thank you so much, that got me the right answer. I need to stop assuming boundaries. $\endgroup$ – Steve Feb 3 '15 at 3:11
  • $\begingroup$ I just want to add one more thing to my comment. "y=16-x^2 and y=0 and x=0 in the 1st quadrant" But I guess it pointless now since you figured out the whole correct boundary thing. Awesome job @steve. $\endgroup$ – randomgirl Feb 3 '15 at 3:19
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Your radius function is correct: $r = 6-x$. Your height should be $f(x) = y = \frac{1}{4}(16-x^2)$. The interval of integration is $x \in [-4,4]$, because these are the solutions to $0 = y = 16-x^2 = (4-x)(4+x)$. Therefore, the volume integral is $$V = \int_{x=-4}^4 2\pi r f(x) \, dx = \int_{x=-4}^4 \frac{\pi}{2} (6-x)(16-x^2) \, dx.$$

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