Volume bounded by the intersections of $z+x^2=4$ and $z-x^2=3y^2$ I'm having trouble with solving this question.
Find the volume of the solid bounded by the intersections of  $z+x^2=4$ and $z-x^2=3y^2$.
Thanks in advance. 
 A: Besides to answer above which is base on using Cartesian coordinates, you can also use the extended polar coordinates as well. Here; the intersection of paraboloid and parabola is the ellipse $$\frac{x^2}{2}+\frac{y^2}{4/3}=1$$ So we have $$x=\sqrt{2}r\cos t,~~y=\frac{2}{\sqrt{3}}r\sin t,~~0\le t\le2\pi$$ and the Jacobian here is $J=\frac{2}{\sqrt{3}}\times\sqrt{2}~r$ so we get the triple integrals:
$$\int_{t=0}^{2\pi}\int_{r=0}^1\int_{z=2r^2\cos^2 t+4\sin^2 t}^{4-2r^2\cos^2 t}|J|dz~dr~dt$$

A: First, check out a 3D plot of the two curves. You should know off the bat that one is a parabaloid and the other is a parabola. 

Try to intuitively convince yourself that the intersection of these two shapes projected onto the $xy$ plane is an ellipse. If that doesn't work, we can always solve directly for when their $z$-values are equal.
$$\begin{align*}
x^2+3y^2&=4-x^2\\
2x^2+3y^2&=4
\end{align*}
$$
Now we need to find the volume over the region bounded by the two 3D curves and this ellipse. Can you take it from here? If not I can give more hints.
EDIT: Since someone already mentioned the proper technique for solving this, I'll set up the triple integral for you. See here for a thorough explanation and more examples.
We want to find the volume bound by the region I have just described. The triple integral is
$$\int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{\frac{2}{3}}\sqrt{2-x^2}}^{\sqrt{\frac{2}{3}}\sqrt{2-x^2}} \int_{x^2+3y^2}^{4-x^2} dz dy dx $$
Let's step through each of the bounds, from the inside out. The innermost are the bounds of the parabola and parabaloid. The middle bounds is the elliptical projection of the intersection, given as a function $y$ of $x$. The outermost bounds is the range of values assumed by $x$ within the ellipse. 
