Triple Integral and symmetry The problem is as follows: Compute the intergal
$$I=\iiint_B \frac{x^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz,$$ where $B$ is the unit ball defined by $B=\{(x,y,z) \mid x^2+y^2+z^2 \leq 1\}$.
The official solution is tricky: The change of variable $(x,y,z) \mapsto (z,y,x)$ transforms the integral into $$\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz,\text{ hence }2I= \iiint_B \:dx\:dy\:dz= 4\pi /3,$$ which implies $I=2\pi/3$.
My question is: what is meant by $(x,y,z) \mapsto (z,y,x)$, isn't it ambiguous? and the jacobian $J= -1$, why the integral $$I =\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \,dx\,dy\,dz$$ instead of $$-\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \,dx\,dy\,dz\text{ ??}$$
thank you very much!!

Forget about the Jacobian. From symmetry considerations it is obvious that
$$I:=\iiint_B \frac{x^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz=\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz\ .$$
Therefore $2 I=\int_B 1 \:dx\:dy\:dz= {\rm vol}(B)$ and $I={2\pi\over3}$.
