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.

  • $\begingroup$ 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?" $\endgroup$ – nukeguy Jan 30 '15 at 1:22
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    $\begingroup$ 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. $\endgroup$ – user204299 Jan 30 '15 at 3:35
  • $\begingroup$ Hmm... what if you integrate the $y$ variable first? $\endgroup$ – nukeguy Jan 30 '15 at 4:47
  • $\begingroup$ You can use polar coordination. Solution in any other coordination is somehow similar to that just with a different representation. $\endgroup$ – Arashium Feb 5 '15 at 11:01
  • $\begingroup$ Did you say there was a second method that worked ? Would you mind giving a reference to it ? Thanks. $\endgroup$ – 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

\begin{align*} y=6\sqrt{\sin (x)}\qquad\longleftrightarrow\qquad x=\arcsin\left(\frac{y^2}{36}\right) \end{align*} We observe according to your comment \begin{align*} &\int_0^6\int_0^{\arcsin(\frac{y^2}{36})}\int_0^{2\pi}r d\Theta dr dy\\ &\qquad=\int_0^6\int_0^{\arcsin(\frac{y^2}{36})}\left. r\cdot\Theta\right|_0^{2\pi} dr dy\\ &\qquad=2\pi\int_0^6\int_0^{\arcsin(\frac{y^2}{36})}r dr dy\\ &\qquad=2\pi\int_0^6\left.\left(\frac{1}{2}r^2\right)\right|_0^{\arcsin(\frac{y^2}{36})}dy\\ &\qquad=\pi\int_0^6\arcsin^2\left(\frac{y^2}{36}\right)dy\tag{2}\\ \end{align*}

and (2) corresponds to the volume formula (1).

I think the difficulty lies in solving the integral (2) which seems to allow no simple closed representation. Wolfram alpha provides following solution:

\begin{align*} &\pi\int\arcsin^2\left(\frac{y^2}{36}\right)dy\\ &\qquad=\frac{\pi^2y^5}{5184\sqrt{2}\Gamma\left(\frac{7}{4}\right)\Gamma\left(\frac{9}{4}\right)} _{3}F_{2}\left(1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};\left(\frac{y}{6}\right)^4\right)\\ &\qquad\qquad-\frac{2\pi y}{3} _2F_1\left(1,\frac{5}{4};\frac{7}{4};\left(\frac{y}{6}\right)^4\right) \arcsin\left(\frac{y^2}{36}\right)\sin\left(2\arcsin\left(\frac{y^2}{36}\right)\right)\\ &\qquad\qquad+\pi y\arcsin^2\left(\frac{y^2}{36}\right)+C \end{align*}

Note: You could perform a plausibility check, take a simpler integrand and you'll be able to calculate the volume in both ways.

  • $\begingroup$ @JakeLebovic: That's high speed, Jake! Thanks for granting the bounty! :-) $\endgroup$ – Markus Scheuer Feb 7 '15 at 21:24

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