I have to calculate the volume between $z = 1+x^2+4y^2$ and $z^2=4x^2+16y^2$ in the half plane $z \le2$.
I have set the bounds as $ -\sqrt{\frac{1}{4}} \le y \le\sqrt{\frac{1}{4}} $ and $-\sqrt{-4y^2+1} \le x \le \sqrt{-4y^2+1}$.
My problem is that I don't know how to handle the $z\le 2$ part of the problem. Can someone help me understand how to solve this problem?
On
The outer volume is
$$V_{\text{outer}}=\int _0^2\int _{-\frac{z}{2}}^{\frac{z}{2}}\int _{-\frac{1}{4} \sqrt{z^2-4 x^2}}^{\frac{1}{4} \sqrt{z^2-4 x^2}} dydxdz=\frac{\pi }{3}$$
The inner volume is
$$V_{\text{inner}}=\int _1^2\int _{-\sqrt{z-1}}^{\sqrt{z-1}}\int _{-\frac{1}{2} \sqrt{-x^2+z-1}}^{\frac{1}{2} \sqrt{-x^2+z-1}} dydxdz=\frac{\pi }{4}$$
Hence the volume between the 2 surfaces is
$$V_{\text{outer}}-V_{\text{inner}}=\frac{\pi }{12}$$
$z_1=1+x^2+4y^2$ is an elliptic paraboloid, while ${z_2}^2=4x^2+16y^2$ is a cone. We can check that the first two surfaces intersect in an ellipse,
$$z_1=z_2 \implies (1+x^2+4y^2)^2 = 4(x^2 + 4y^2) \implies x^2 + 4y^2 = 1$$
which lies in the plane $z=2$. From here we can choose $x\in[-1,1]$ and $y\in\left[-\frac12\sqrt{1-x^2},\frac12\sqrt{1-x^2}\right]$.
I suppose "in the half-plane $z\le2$" simply means that $z\le2$. It's easy to show which surface lies above the other by picking any point inside the cylinder $x^2+4y^2\le1$ and comparing the corresponding $z$'s.
The solid can be parameterized by
$$(x,y,z) \in [-1,1] \times \left[-\frac12\sqrt{1-x^2}, \frac12\sqrt{1-x^2}\right]\times\left[\sqrt{4x^2+16y^2},1+x^2+4y^2\right]$$
so its volume is
$$\int_{-1}^1 \int_{-\frac12\sqrt{1-x^2}}^{\frac12\sqrt{1-x^2}} \int_{\sqrt{4x^2+16y^2}}^{1+x^2+4y^2} dz \, dy \, dx$$
Or, using your order of integration,
$$\int_{-\frac12}^{\frac12} \int_{-\sqrt{1-4y^2}}^{\sqrt{1-4y^2}} \int_{\sqrt{4x^2+16y^2}}^{1+x^2+4y^2} dz\,dx\,dy$$
Consider converting to cylindrical coordinates to make the actual computation a bit easier.