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

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?

• It is NOT a horn torus. That has cross-sections formed by two circles touching at the origin. Your surface has cross-sections that look like a lemniscate. May 8 '20 at 3:37
• More importantly (IMHO), why did you think it would be a horn torus? There are infinitely many different shapes. The next exercise may always involve a shape you have never seen earlier. It is not the point of these exercise to build you a database of shapes containing every possible shape in use. Learning a method for calculating the volume is more important than the formula you end up with. May 8 '20 at 3:48
• I didn't mean to "build a database of shapes" in my mind. It's just that when I put the surface into a graphing calculator, it looks like a torus to me at first, so I had doubts about my answer. But thank you for your comment anyway. :) May 8 '20 at 4:07
• Using symmetry, the volume is equal to eight times the volume of the first octant,let $x=\rho \cos \theta \sin \phi,y=\rho \sin \theta \sin \phi,z=\rho \cos \phi$,In the first octant, $\frac{\pi}{4}\leq \phi \leq \frac{\pi}{2}$ May 8 '20 at 4:11

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