Volume enclosed by $(x^2+y^2+z^2)^2 = a^2(x^2+y^2-z^2)$

Question

My math problem is the find the volume enclosed by the surface $$(x^2+y^2+z^2)^2 = a^2(x^2+y^2-z^2)$$ I used spherical coordinate substitution, $$\left\{\begin{matrix}x=\rho\cos{\phi}\cos{\theta}\\ y=\rho\cos{\phi}\sin{\theta}\\ z=\rho\sin{\phi} \end{matrix}\right.$$ which gives me $\rho = a \sqrt{\cos{2\phi}}$. So then I tried evaluating the integral: $$V=\int_{0}^{2\pi}\int_{-\pi/4}^{\pi/4}\int_{0}^{a\sqrt{cos2\phi}}\rho^2\cos{\phi} \space \space d\rho d\phi d\theta$$ which gives me $\frac{\pi^2a^3}{4\sqrt{2}}$. However, when I plot the surface in a graphing calculator, it seems like the surface likes like a horn torus and the volume should be $V=2\pi^2r^3=\frac{\pi^2a^3}{4}$. Why would I have the extra $\sqrt{2}$ in the denominator? Did I setup my integral incorrectly?

Answer 1

In spherical coordinates, the surface is

$$r^2= a^2(1-2\cos^2\theta)\>\>\>\>\>\theta\in [\frac\pi4, \frac{3\pi}4]$$

The volume integral is then

\begin{align} V& =2\pi \int_{\frac\pi4}^{\frac{3\pi}4}\int_0^{r(\theta)}r^2\sin\theta drd\theta \\ & = \frac{2\pi}3 a^3\int_{\frac\pi4}^{\frac{3\pi}4} (1-2\cos^2\theta)^{3/2}\sin\theta d\theta \\ & \overset{t=\cos\theta }= \frac{2\pi}3 a^3\int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} (1-2t^2)^{3/2}dt \\ & = \frac{2\pi}3 a^3 \cdot \frac{3\pi}{8\sqrt2}= \frac{\sqrt2\pi^2}{8}a^3\\ \end{align}