Triple Integration The problem is as follows: Compute the intergal 
$$I= \iiint_B \frac{x^4+2y^4}{x^4 +4y^4 +z^4 }\text{d}x\,\text{d}y\,\text{d}z $$
 where $B$  is the unit ball defined by $B=$  {$(x,y,z)∣x^ 2 +y^ 2 +z^ 2 \leq 1$} .
The official solution is tricky: The change of variable $(x,y,z) \mapsto (z,y,x)$  transforms the integral into 
$$ I= \iiint_B \frac{z^4+2y^4}{x^4 +4y^4 +z^4 }\text{d}x\,\text{d}y\,\text{d}z \qquad(1)$$
hence $2I= \iiint_B 1\, \text{d}x\,\text{d}y\,\text{d}z = 4\pi/3$. Therefore, $I=2\pi/3$.
My question is: what is meant by $(x,y,z)\mapsto (z,y,x)$ , isn't it ambiguous? Somebody have another solution saying that by symmetry, we have $I$ equals $(1)$ and adding gives the answer. I wanna ask if symmetric here means that the permutation of the variables preserves the domain $D$? Why if it is symmetric, then the two integrals are the same?
The next question is to compute the integral
$$J=\int_0^1 \int_0^1 \int_0^1 \cos^2(\frac{\pi}{6}(x+y+z)) \text{d}x\,\text{d}y\,\text{d}z$$
The solution uses similar technique in above: substitutes $x=1-u. y=1-v, z=1-w$, then we will get $$J=\int_1^0 \int_1^0\int_1^0  \cos^2(\frac{\pi}{2}-\frac{\pi}{6}(u+v+w)) \text{d}u\,\text{d}v\,\text{d}w$$ but why it also equals to
$$\int_0^1 \int_0^1\int_0^1  \cos^2(\frac{\pi}{2}-\frac{\pi}{6}(u+v+w)) \text{d}u\,\text{d}v\,\text{d}w$$
did I do something wrong??
Thank you for answering such a dumb question! 
 A: For your first question, the solution is not so mysterious.  You are integrating over a region symmetric in $x$ and $z$.  Therefore you can switch $x$ for $z$ and get the same answer.  Specifically:
$$x^2+y^2+z^2=1 \implies z^2+y^2+x^2=1$$
$$\begin{align}I=\iiint_B dx\,dy\,dz \frac{x^4+2 y^4}{x^4+4 y^4+z^4} &= \iiint_B dz\,dy\,dx \frac{z^4+2 y^4}{z^4+4 y^4+x^4}\\ &= \iiint_B dx\,dy\,dz \frac{z^4+2 y^4}{z^4+4 y^4+x^4}\end{align}$$
Note that the volume element is the same after substitution.  You may then add the two equal quantities as you mentioned above:
$$\begin{align}2I &= \iiint_B dx\,dy\,dz \frac{x^4+2 y^4+z^4+2 y^4}{x^4+4 y^4+z^4} \\ &= \iiint_B dx\,dy\,dz\\ &= \frac{4 \pi}{3}\\\implies I&=\frac{2 \pi}{3} \end{align}$$
A: Start with the innermost integral:
$$\int_0^1{cos^2\frac{\pi}{6}(x+y+z)dx}$$
Treat all variables besides x as constants, and using the trigonometric identity, the indefinite integral is
$$\frac{x}{2}+\frac{3}{2\pi}sin(\frac{\pi}{3}(x+y+z))$$
Now plug into x the limits 0 to 1, and you get this function:
$$\frac{1}{2}-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))$$
Now you have a function that is only dependent on y and z, and for your total integral you're left with only two integrals:
$$\int_0^1\int_0^1{(\frac{1}{2}-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z)))}dy dz$$
Take the integral with respect to y of this function(y is the next outermost integral), taking all variables by y to be constant:
$$\int_0^1{(\frac{1}{2}-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z)))}dy$$
Rinse, repeat, until there's no more integrals.
