Find the volume of the solid bounded below by $z=x^2+4y^2$ and above by $z=4x-8y-4$ I'm looking for feedback on if I did this question properly.
The surfaces intersect at $x^2+4y^2=4x-8y-4$ which we can rearrange as $(x^2-4x)+(4y^2+8y)=-4$. We simply complete the square:
$$(x^2-4x)+4(y^2+2y)=-4$$
$$\frac{1}{4}(x^2-4x)+(y^2+2y)=-1$$
$$\frac{1}{4}(x^2-4x+4)+(y^2+2y+1)=-1+1+1$$
$$\frac{1}{4}(x-2)^2+(y+1)^2=1$$
Then we use the change of variables $g(u,v)=(2u+2,v-1)$ and we get the area of intersection as
$$\frac{1}{4}(2u+2-2)^2+(v-1+1)^2$$
$$=\frac{1}{4}(2u)^2+(v)^2$$
$$=u^2+v^2=1$$
Now, since $x^2+4y^2$ is the lower bound and $4x-8y-4$ is the upper bound (I'm not sure how to show this...I am simply using the fact that it was stated in the question) then the area of integration under the change of variables is the area $=u^2+v^2\leq1$ (i.e. the unit ball).
Therefore, the area of this solid is simply $$\int\int_{x^2+y^2\leq1}1\cdot(\det Dg)$$.
Note that $\det Dg= \begin{vmatrix} 2 & 0 \\ 0 & 1 \\ \end{vmatrix}=2$
We use the polar change of variables and we get
$$\int\int_{x^2+y^2\leq1}2=\int_0^{2\pi}\int_0^12r\,dr\,d\theta=\int_0^{2\pi}1=2\pi$$
I don't believe I have made an error but I am unpracticed in this technique.

What you've calculated is the area of the ellipse at which the paraboloid and the plane intersect. To see that, note that the ellipse has half-major axis $2$ units long and half-minor axis $1$ unit long and therefore has area $\pi\cdot2\cdot1$ squnits = $2\pi$ squnits. The following calculates the volume of the solid bounded by those surfaces.

Let's combine your transformation with cylindrical polar to get a new transformation in $\mathbb R^3$
$$
x=2r\cos\phi+2,\;y=r\sin\phi-1,\;z=w
$$
The corresponding Jacobian determinant is 
$$\begin{vmatrix}
2\cos\phi&-2r\sin\phi&0\\
\sin\phi&r\cos\phi&0\\
0&0&1
\end{vmatrix}=2r
$$
The new equation of the paraboloid, the plane and the curve of intersection are respectively
$$
w=4r^2+8r(\cos\phi-\sin\phi)+8,\;w=8r(\cos\phi-\sin\phi)+12,\;r=1
$$
The required volume is 
$$\begin{align}
&\int_0^{2\pi}\int_{0}^1\int_{4r^2+8r(\cos\phi-\sin\phi)+8}^{8r(\cos\phi-\sin\phi)+12}2r\;dw\;dr\;d\phi\\
=&\int_0^{2\pi}\int_{0}^18r(1-r^2)\;dr\;d\phi={4\pi}
\end{align}$$

You can see from the above calculation that height of the plane $-$ height of the paraboloid = $8r(1-r^2)$ which is positive when $1<r<0$ and therefore the plane is the upper surface and the paraboloid is the lower surface in our region of interest. 
