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!!

One uses the absolute value of the Jacobian.
The point is that if the roles of $x$ and $z$ are interchanged, the value of the integral is still the same because of the symmetry of the situation.  One is in effect merely renaming the variables, so the value of the integral, which doesn't depend on any values of those variables, is the same.  Here it is relevant that the region over which you're integrating is also symmetric: it stays the same if you interchange the $x$- and $z$-axes.
Therefore the sum of the two integrals is just twice the original integral.  But the sum of the two functions is one that is easy to integrate.
