Calculate the volume of the region V bounded by $$(x^2 + y^2 + z^2)^2 = a^2(x^2 + y^2 -z^2)$$
I tried using spherical coordinates $x = r\cos(a)\sin(b)$, $y = r\sin(a)\sin(b)$, $z = r\cos(b)$, the region V then becomes $0 <= a <= 2\pi$, $\pi/4 <= b <= 3\pi/4$, $0 <= r <= \sqrt{a^2.(\sin(b)^2 - \cos(b)^2)}$.
Then I use triple integral to calculate, but it's difficult. Now I need help.
In spherical coordinates, $(x^2 + y^2 + z^2)^2 = a^2(x^2 + y^2 -z^2)$ is converted to $ r^2= a^2(1-2\cos^2\theta )$ with $\theta\in (\frac{\pi}4, \frac{3\pi}4)$, and the triple integral for volume is \begin{align} &\int_0^{2\pi}\int_{\pi/4}^{3\pi/4}\int_0^{a\sqrt{1-2\cos^2\theta}}r^2 \sin\theta \ dr d\theta d\phi\\ =& \ \frac{2\pi a^3}3 \int_{\pi/4}^{3\pi/4}(1-2\cos^2\theta)^{3/2} \sin\theta \ d\theta= \frac{\pi^2 a^3}{4\sqrt2} \end{align}