Q:

Calculate: $\int_{(x^2+y^2+z^2)^2\le2xyz} 1 d\lambda_3(x,y,z)$

Calculate: $\int_{(x^2+y^2+z^2)^2\le2xyz} 1 d\lambda_3(x,y,z)$
I have a problem finding a substitution that will give me nice integration limits. I tried as well as $x=r\sin \alpha, y=\cos \alpha, z=z$ as $x=r\cos\alpha \cos \beta, y=r\cos \beta \sin \alpha, z=r \sin \beta$ as for example $s=(x+y)^2=t+2xy, t=x+y, z=z$ but nothing brought the desired result.
Does anyone have any hint?

A:

Notice the region of integration has a nice symmetry that allows us to only compute the integral in the first octant:
$$I = \iiint_{(x^2+y^2+z^2)^2\leq 2xyz} d\lambda_3 = 4 \iiint_{(x^2+y^2+z^2)^2\leq 2xyz \: \cap \: x\geq 0 \: \cap \: y\geq 0 \: \cap \: z\geq 0} d\lambda_3$$
Next, spherical coordinates is definitely the way to go. Plugging in we get that the boundary is given by
$$r^4 = 2r^3\sin\theta\cos\theta\sin\phi\cos\phi \implies r = \sin\theta\cos\theta\sin(2\phi)$$
so the integral is given by
$$I = 4\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^{\sin\theta\cos\theta\sin(2\phi)} r^2 \sin\theta \:dr \:d\theta \:d\phi = \frac{8}{315}$$

