Triple integrals over a specific region Evaluate:
$$\iiint_S x^2 dx\, dy\,dz$$,
where S is the region bound by
$$4x^2+y^2=4,\,z+x=2, \,z=0$$
Can anyone show me how do to this, I've been trying for ages and don't know what to do
 A: From the first condition $4x^2+y^2=4$ (a right cylinder with an elliptical section in the $xy$ plane) we have:
$$
-2\sqrt{1-x^2}\le y \le 2\sqrt{1-x^2} 
$$
so $y$ is a real number for:
 $$-1 \le x \le 1$$
and these are the limits for $x$. From the other conditions we find the limits for $z$:
$$
0\le z \le 2-x
$$
so the integral becomes:
$$
\int_{-1}^{1} \int_{-2\sqrt{1-x^2}}^{2\sqrt{1-x^2}}\int_0^{2-x} x^2dz dy dx
$$
can you do from this?
A: The first condition is a squashed cylinder. The second is a plane. The third is a coordinate plane. The $z$ integral will be an inner integral (because it is coupled with $x$, which will be varied by the outer integrals). The first two integrals are over an ellipse, which can be turned into a circle by $w=2x$. Your new integral is
$$\frac14\iint \int_0^{2-w/2} w^2\, dz \,dw\, dy $$
and the conditions are $w^2+y^2=4$ and $z+w/2=2$, $z=0$. This is now trivial, as the first condition is a circle with radius 2, which you can solve by integrating in polar coordinates, $(w,y)=r(\cos\phi,\sin\phi)$:
$$\frac14 \int_0^2 \int_0^{2\pi}\int_0^{2-r\cos\phi/2}r^2 \cos^2\phi\, dz \,d\phi\, (r\,dr) $$
