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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?

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  • $\begingroup$ 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. $\endgroup$ May 8 '20 at 3:37
  • $\begingroup$ 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. $\endgroup$ May 8 '20 at 3:48
  • $\begingroup$ 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. :) $\endgroup$
    – Potato HY
    May 8 '20 at 4:07
  • $\begingroup$ 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} $ $\endgroup$
    – Eeyore Ho
    May 8 '20 at 4:11
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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}

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