<s>[INST] Find the volume of the solid bounded by $(x^2+y^2+z^2)^2=x^2+y^2-z^2$ As Title.
Find the volume of the solid bounded by $(x^2+y^2+z^2)^2=x^2+y^2-z^2$
--
By Spherical Coordinates, I got $\rho^4=\rho^2(\sin^2\phi-\cos^2\phi)$
Then $\rho^2=-\cos2\phi$
How to find the integral bounded?
 [/INST] As $\rho^2>0$
It must be $-1 \leq \cos{2\phi} \leq 0$, then $\frac{\pi}{4} \leq \phi \leq \frac{3\pi}{4}$
--
$\displaystyle V=\int_{0}^{2\pi} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_{0}^{\sqrt{-\cos2\phi}} \rho^2 \sin\phi \, d\rho d\phi d\theta = \frac{-2\pi}{3} \int_{\frac{\pi}{3}}^{\frac{4\pi}{3}} (-\cos{2\phi})^{3/2} \sin{\phi} \, d\phi \\ \displaystyle = \frac{2\pi}{3} \int_{\frac{\pi}{3}}^{\frac{4\pi}{3}} (1-2\cos^2{\phi})^{3/2} \, d(\cos{\phi}) = \frac{2\pi}{3} \times \frac{3\pi}{8\sqrt{2}}=\frac{\pi^2}{4\sqrt{2}}$
</s>