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.

  • $\begingroup$ what kind of solid is it? $\endgroup$ – user Jan 31 '18 at 15:59
  • $\begingroup$ I don't know, and that's the point. I only got the formula. $\endgroup$ – R.K. Jan 31 '18 at 16:03
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    $\begingroup$ You are interested in the volume bounded by the surface given by your equation. Something has gone slightly wrong in your calculation. You should get something more like $r=2\sin\alpha\sin\beta\cos\beta$. The surface is four squashed spheres meeting at the origin. All have the same volume, so just find half the volume of one by integrating over $[0,\pi/2]$ for each of $\alpha,\beta$. $\endgroup$ – almagest Jan 31 '18 at 17:24
  • $\begingroup$ The given equation defines a set $S\subset{\mathbb R}^3$. This set is not a "solid", but a surface (maybe with singularities). $\endgroup$ – Christian Blatter Jan 31 '18 at 17:31

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: $$I=\dfrac{\pi}{3}$$

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  • $\begingroup$ clever (+1), but you shall adjust better the integral bounds for $\theta$ and $\phi$. Also put in brackets $sin(2\theta)$ etc. $\endgroup$ – G Cab Jan 31 '18 at 21:35
  • $\begingroup$ You're right...I'm not very good at mathjax :) $\endgroup$ – Mostafa Ayaz Jan 31 '18 at 21:51
  • $\begingroup$ I do not master mathjax as well, but in this case it is enough that you just put normal brackets : "\sin(2\theta)" $\endgroup$ – G Cab Jan 31 '18 at 22:08
  • $\begingroup$ Would you plz do that yourself? I'm currently on my phone and can't do that $\endgroup$ – Mostafa Ayaz Jan 31 '18 at 22:14

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