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!!
 A: 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.
A: 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 $I={2\pi\over3}$.
A: Attention: Note that the change of variable theorem has the absolute value of the Jacobian, which is $1$ in this case.
Since the boundary of the closed unit ball has measure $0$, you can restrict to the open unit ball $\mathring{B}$. Now the function
$$
\phi:(x,y,z)\longmapsto (z,y,x)
$$
is a $C^1$ diffeomorphism of $\mathring{B}$ onto itself. Indeed, it is $C^1$, bijective from $\mathring{B}$ onto $\mathring{B}$, and its inverse is (automatically) $C^1$. It is easily seen that the (determinant of) the Jacobian is $-1$. So the change of variable applies and yields
$$
\iiint_Bf=\iiint_{\mathring{B}}f(x,y,z)dxdydz=\iiint_{\mathring{B}}(f\circ \phi)(x,y,z)\;|\mbox{Jac} \;\phi(x,y,z)|\;dxdydz
$$
$$
=\iiint_{\mathring{B}}f(z,y,x)dxdydz=\iiint_{B}f(z,y,x)dxdydz.
$$
That's what you want to use the trick.
Note: Of course, it is not really necessary to restrict to the interior of the unit ball, since there are stronger forms of the change of variable theorem which apply to non necessarily open sets. But a change of variable theorem which is easy to state and covers most situations in $\mathbb{R}^n$ goes as stated below.
Let $U$ be open in $\mathbb{R}^n$ and $\phi:U\longrightarrow\mathbb{R}^n$. If $\phi$ is a $C^1$ diffeomorphism of $U$ onto $\phi(U)$ (equivalently, $\phi$ is injective and $C^1$) and if $f$ is a Lebesgue integrable function on $\phi(U)$, then $(f\circ\phi)|\mbox{Jac}\;\phi|$ is integrable on $U$ and
$$
\int_{\phi(U)} f(v)dv=\int_{U}f(\phi(u))|\mbox{Jac}\;\phi(u)|du.
$$
