# Find the volume of the solid obtained by rotating the region about the the line $y=4$

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.

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

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 $V=\frac{207\pi}{2}$.