# Volume of $y = 6\sqrt{\sin(x)}$ rotated around $y$-axis using triple integrals

The problem is to find the volume of $y = 6\cdot \sqrt{\sin (x)}$ rotated around the $y$-axis when $0 \leq y \leq 6$.

I know this can be done by the sv-calc method of volumes of revolution but I wanted to see if a problem like this can be done by triple integrals. I tried it a few times myself but could not seem to get the limits of the integrals set up correctly.

• Visualize the object. Then decide on an order of integration. I would suggest integrating the $z$ variable first so that we can get the $z$ axis out of the way and return to the $xy$ plane. To set up the bounds for the $z$ variable, ask yourself: "What two surfaces does the volume lie between?" – nukeguy Jan 30 '15 at 1:22
• I'm not sure this can actually be done. I end up with the integral $\int_{0}^{6} \int_{0}^{arcsin(y^2/36)}\int_{0}^{2\pi }rd\Theta drdy$ which I can't find a way to solve. – user204299 Jan 30 '15 at 3:35
• Hmm... what if you integrate the $y$ variable first? – nukeguy Jan 30 '15 at 4:47
• You can use polar coordination. Solution in any other coordination is somehow similar to that just with a different representation. – Arashium Feb 5 '15 at 11:01
• Did you say there was a second method that worked ? Would you mind giving a reference to it ? Thanks. – Sary Feb 5 '15 at 13:31

Maybe I miss the point of your question, but I think that the rotation of a function $y=f(x)$ around the $y$-axis using the formula \begin{align*} V=\pi\int_a^bx^2(y) dy\tag{1} \end{align*} and your Ansatz in the comment using a triple integral is essentially the same.

Since for