Find the volume of the following solid.

Question

I need to find the volume of that solid: $$(x^2+y^2+z^2)^3=4y^2z^2$$ I obviously tried spherical coordinates, but that simply led me to nowhere. I used: $$\begin{cases}x=r\cos\alpha \cos\beta\\[2ex]y=r\sin\alpha \cos\beta\\[2ex]z=r\sin\beta\end{cases}$$ and it got me to the point where I have: $$r^2=4\cos\alpha\sin\alpha\cos^2\beta.$$ and honestly I don't know what to do next.

Answer 1

In spherical coordinates $(R,\theta,\phi)$ the Jacobian of the spherical coordinates is $R^2\sin\theta$. Also the equation of the surface enveloping the solid is:$$R^2=4\sin^2\theta\sin^2\phi\cos^2\phi=\sin^2\theta\sin^2(2\phi)$$ For calculating the volume we have $$\iiint_VR^2\sin\theta dR d\theta d\phi= \int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{\sqrt{\sin^2\theta\sin^2(2\phi)}}R^2\sin\theta dR d\phi d\theta= \int_{0}^{\pi}\int_{0}^{2\pi}\dfrac{R^3}{3}|_{0}^{\sqrt{\sin^2\theta\sin^2(2\phi)}}\sin\theta d\phi d\theta= \int_{0}^{\pi}\int_{0}^{2\pi}\dfrac{|\sin\theta|^3|\sin(2\phi)|^3}{3}\sin\theta d\phi d\theta$$ since $\sin\theta\ge0$ when $0\le\theta\le\pi$, the period of $|\sin(2\phi)|^3$ is $\dfrac{\pi}{4}$ and $\sin(2\phi)\ge0$ when $0\le\phi\le\dfrac{\pi}{2}$ we can write the integral formula as following: $$I=4\int_{0}^{\pi}\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^4\theta}{3}\sin^3(2\phi) d\phi d\theta$$ also $$\int_{0}^{\frac{\pi}{2}}4\sin^3(2\phi) d\phi=\int_{0}^{\frac{\pi}{2}}3\sin(2\phi)-\sin(6\phi) d\phi=\dfrac{8}{3}$$ therefore the integral can be simplified as follows: $$I=\dfrac{8}{9}\int_{0}^{\pi}\sin^4\theta d\theta$$ which by substituting $\sin^2p=\dfrac{1-\cos(2p)}{2}$ gives us the final result: $$I=\dfrac{\pi}{3}$$