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Find the volume of the solid obtained by rotating the region bounded by the curve: $y=\sqrt{x-6}, y=0, x= 15$, spin about the line $y=4$.

The following formula I have used was

$$ 2\pi\int_0^{15}xf(x))\,dx $$

and I came up with $$237.6\pi$$ I am not sure if this is the correct answer. Looking over the question and noting that there was only one $x = 15$, I'm assuming that $a=0$ and $b=15$ and after solving the integration I use both symbols to plug them to the answer and then add them together.

Am I on the right track or way off base? Please let me know. Thank you.

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  • $\begingroup$ Do you mean $y=\sqrt{x-6}$? $\endgroup$ – Juniven Mar 21 '17 at 23:57
  • $\begingroup$ Anyway, an answer is given below. Just ask if you want for clarification. $\endgroup$ – Juniven Mar 22 '17 at 0:47
  • $\begingroup$ Yes, that's what I meant. Thank you! $\endgroup$ – miiworld2 Mar 22 '17 at 2:56
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The required region is skecthed below: enter image description here

The formula $$ V=2\pi\int_0^{15}xf(x)\,dx $$ that you mentioned is wrong.


Correct Solution: Using the Shell Method, $$V=\int_0^32\pi(4-y)\cdot(15-x)dy=\int_0^3 2\pi(4-y)\cdot[15-(y^2+6)]dy=\dots$$

Alternately, using the Washer Method we get $$V=\int_6^{15}\pi[4^2-(4-y)^2]dx=\pi\int_6^{15}[4^2-(4-\sqrt{x-6})^2]dx=\dots$$

Wolfram Alpha gives both $V=\frac{207\pi}{2}$.

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Are you using the disk or washer method? If so you should probably review your formula. However the formula is right should you choose to use the cylinder method. You only need to change the integration limits.

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