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