<s>[INST] Volume enclosed by the surface $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}.$ 
Compute the volume enclosed by the surface $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}.$$

My attempt:
I used the following change of coordinates:
$$\begin{aligned}x&=ar\sin\theta\cos\varphi\\y&=br\sin\theta\sin\varphi\\z&=cr\cos\theta,\\\varphi&\in[0,2\pi]\\r&\in[0,1]\\\text{ Jacobian: }J_\psi&=abcr^2\sin\theta.\end{aligned}$$
The equation of the curve becomes: $$\begin{aligned}r^4=r^2-2\frac{c^2r^2\cos^2\theta}{c^2}\\\iff r^2=1-2\cos^2\theta\\r^2=-\cos(2\theta)\end{aligned}$$ so $$0\le r^2\le -\cos(2\theta)\\\implies 0\ge -r^2\ge\cos(2\theta)\ge -1\\\implies \theta\in\left[-\pi,-\frac{\arccos(-r^2)}2\right]\cup\left[\frac{\arccos(-r^2)}2,\pi\right],$$ but $\theta$ must be in the interval $[0,\pi],$ so I eliminated the first possibility. I get the integral $$\begin{aligned}\int_0^{2\pi}\int_0^1\int_{\arccos(-r^2)/2}^\pi abcr^2\sin\theta d\theta drd\varphi&=2abc\pi\int_0^1 r^2\left(-\cos\theta\Big|_{\arccos(-r^2)/2}^\pi\right)dr\\&=-2abc\pi\int_0^1r^2\left(-1-\cos\left(\frac{\arccos(-r^2)}2\right)\right) dr\\&=2abc\pi\int_0^1\left(r^2+r^2\sqrt{\frac{1-\cos(\arccos(-r^2))}2}\right)dr\\&=2abc\pi\left(\int_0^1r^2dr+\frac1{\sqrt 2}\int_0^1r^2\sqrt{1\color{red}+r^2}dr\right)\\&=2abc\pi\left(\frac13+\frac1{\sqrt 2}\frac13 \int_0^1r\sqrt{1+r^2}3rdr\right)\\&=2abc\pi\left(\frac13+\frac1{3\sqrt 2} r\left(1+r^2\right)^{3/2}\Big|_0^1-\int_0^1(1+r^2)^{3/2}dr\right)\\&=2abc\pi\left(1-\frac1{3\sqrt 2}\int_0^1(1+r^2)^{3/2}dr\right)\end{aligned}$$

EDIT:
I found one mistake (marked now with red plus, I had put minus instead). Now, if I proceed with substitution $r=\tan\alpha,$ the procedure continues. Is there a way to avoid all of this?

I compared my procedure with this and mine looks suspicious. I thought this integral would be easier to compute if I changed the order of integration in $\theta$ and $r$. Can anybody tell me what I did wrong?
 [/INST] $$\cos\left(\pi-\frac12\cos^{-1}(-\rho^2)\right) = $$
If $0 \le \rho^2 \le -\cos(2\theta)$, then $\frac{(4n-3)\pi}4 \le \theta \le \frac{(4n-1)\pi}4 \text{ for } n\in\mathbb Z$.
The polar angle must generally be between $0$ and $\pi$, so we take $\theta\in\left[\frac\pi4,\frac{3\pi}4\right]$. Then the volume is
$$\int_0^{2\pi} \int_{\frac\pi4}^{\frac{3\pi}4} \int_0^{\sqrt{-\cos(2\theta)}} abc\rho^2 \sin(\theta) \, d\rho \, d\theta \, d\varphi = \frac{abc\pi^2}{4\sqrt2}$$

Or, preserving the order you chose, we have
$$\rho^2 = -\cos(2\theta) \implies \theta = \frac12 \cos^{-1}(-\rho^2)$$
but this is only true if $\theta\in\left[0,\frac\pi2\right]$. For $\theta\in\left[\frac\pi2,\pi\right]$, we instead would have
$$\rho^2 = -\cos(2\theta) \implies \theta = \pi - \frac12 \cos^{-1}(-\rho^2)$$
Now, if $\theta\le\frac\pi2$, then
$$\rho=0 \implies \theta = \frac\pi4 \text{ and } \rho=1\implies\theta=\frac\pi2$$
Otherwise, if $\frac\pi2<\theta$, then
$$\rho=0\implies\theta=\frac{3\pi}4 \text{ and } \rho=1\implies\theta=\frac\pi2$$
All this tells us that if $\rho\in[0,1]$, then $\theta\in\left[\frac12\cos^{-1}(-\rho^2),\pi-\frac12\cos^{-1}(-\rho^2)\right]$.
The subsequent integral agrees with the previous result:
$$\begin{align}
&\int_0^{2\pi} \int_0^1 \int_{\frac12\cos^{-1}(-\rho^2)}^{\pi-\frac12\cos^{-1}(-\rho^2)} abc\rho^2\sin(\theta) \, d\theta \, d\rho \, d\varphi \\[1ex]
&= 2abc\pi \int_0^1 \rho^2\left(\cos\left(\frac12\cos^{-1}(-\rho^2)\right) - \cos\left(\pi-\frac12\cos^{-1}(-\rho^2)\right)\right) \, d\rho \\[1ex]
&= 4abc\pi \int_0^1 \rho^2\cos\left(\frac12\cos^{-1}(-\rho^2)\right) \, d\rho \\[1ex]
&= 2\sqrt2\,abc\pi \int_0^1 \rho^2 \sqrt{1-\rho^2} \, d\rho \\[1ex]
&= 2\sqrt2\,abc\pi \times \frac\pi{16} = \frac{abc\pi^2}{4\sqrt2}
\end{align}$$
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