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I have to calculate the area of such surface: $$(x^2+y^2+z^2)^2=x^2-y^2$$

I use the spherical coordinates transform, and get $r^2=\sin^2 \varphi \cos 2\theta$, but I don't know how to use this to calculate the area.

I'll be grateful if there's any help. :)

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  • $\begingroup$ It should really be $r^2 = \sin^2\theta\cos2\phi$ (I generally prefer to either use $(\rho,\phi,\theta)$ or $(r,\theta,\phi)$ and not mix the two, but that's just my personal choice). $\endgroup$ May 10 '20 at 3:01
  • $\begingroup$ You're right. I miswrite it. $\endgroup$ May 10 '20 at 3:03
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First notice that by even symmetry in each of the variables we can reduce the integral to just one in the first octant

$$\iint_S \:dS = 8 \iint_{S\:\cap\:\text{First octant}} \:dS$$

Next we will parametrize the surface like so:

$$x = \sin^2\theta \cos\phi \sqrt{\cos 2\phi}$$

$$y = \sin^2\theta \sin\phi \sqrt{\cos 2\phi}$$

$$z = \sin\theta\cos\theta \sqrt{\cos 2\phi}$$

with $\theta \in \left[0,\frac{\pi}{2}\right]$ and $\phi \in \left[0,\frac{\pi}{4}\right]$

Then we can use the following convenient fact about parametrizations in spherical coordinates:

$$\vec{r}(\theta,\phi) = f(\theta,\phi)\hat{r} \implies |\vec{r}_\theta\times\vec{r}_\phi| = f\sqrt{f_\phi^2 + f_\theta^2\sin^2\theta + f^2\sin^2\theta}$$

$$ = f\sin\theta\sqrt{\frac{f_\phi^2}{\sin^2\theta}+f_\theta^2+f^2}$$

Notice the similarity to the spherical coordinates Jacobian, $r^2\sin\theta$, when $f = \text{const}$

Since in this case $f = \sin\theta\sqrt{\cos2\phi}$ we get that

$$|\vec{r}_\theta\times\vec{r}_\phi| = \sin^2\theta\sqrt{\cos2\phi}\sqrt{\frac{\sin^22\phi}{\cos2\phi} + \cos^2\theta\cos2\phi+\sin^2\theta\cos2\phi}$$

$$ = \sin^2\theta\sqrt{\sin^22\phi + \cos^2\theta\cos^22\phi+\sin^2\theta\cos^22\phi} = \sin^2\theta$$

giving us the integral

$$S = 8\int_0^{\frac{\pi}{4}} \int_0^{\frac{\pi}{2}} \sin^2\theta\:d\theta\:d\phi = \frac{\pi^2}{2}$$

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