Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$. Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$.
I need to solve this with triple integrals, however I'm having a lot of problems. Anyone knows how to solve?

Let's first find the boundary of the integration region, that is, where both surfaces intersect:
$z=x^2+3y^2=8-x^2-y^2$ thus $2x^2+4y^2=8$ thus it's the ellipse $\frac{x^2}{2^2}+\frac{y^2}{\sqrt 2^2}=1$ in the $xy$ plane.
Now let $E$ be the plane region bounded by that ellipse. The volume of $W$ is then $\displaystyle\iint_E\left(8-x^2-y^2-(x^2+3y^2)\right)\,dx\,dy=\iint_E\left(8-2 x^2-4 y^2\right)\,dx\,dy$.
In order to define the integration limits one should change coordinates as follows: $x=2\,r\cos\theta$, $y=\sqrt2\,r\sin\theta$, so that the bounding ellipse is simply $r=1$, $0\leq\theta\leq2\pi$ in the new coordinates. Now let's calculate the Jacobian:
$\displaystyle J=\begin{vmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}\end{vmatrix}=\begin{vmatrix}2\cos\theta&-2\,r\sin\theta\\\sqrt2\sin\theta&\sqrt2\,r\cos\theta\end{vmatrix}=2\sqrt2\,r$.
Therefore, the volume is
$\displaystyle\iint_E\left(8-2 x^2-4 y^2\right)\,dx\,dy=\int_0^1\int_0^{2\pi}\left(8-2\cdot4r^2\cos^2\theta-4\cdot2r^2\sin^2\theta\right)2\sqrt2\,r\,d\theta\,dr=8\cdot2\sqrt2\int_0^1(1-r^2)r\,dr\int_0^{2\pi}d\theta=8\cdot2\sqrt2\frac14\cdot2\pi=8\sqrt2\pi$.
