Calculate the volume bounded by the surfaces Calculate the volume of the solid bounded by the surfaces $$\begin{aligned}z&=4x^2+4y^2, \\ z&=x^2+y^2, \\z&=4.\end{aligned}$$
I made an equation of $4x^2+4y^2=4-x^2+y^2$ and solved it to get $x^2+y^2=\dfrac{4}{5}$.
Then I did a double integration 
$$\displaystyle
\iint_{x^2+y^2\leq \frac{4}{5}}  \left(4x^2+4y^2\right)-\left(4-x^2-y^2\right)dA,\ 
$$ 
did the subtraction and the changed to polar
$$\displaystyle
\int_0^{2\pi} \int_0^{\frac{4}{5}} \left(3r^2-4\right) r \,dr \,d\theta,\ 
$$
got a result of 
$$
\displaystyle\int_0^{ 2\pi}\Bigg[ \left.\left(\frac{3}{4}r^4-2r^2\right) \right\vert_{r=0}^{\frac{4}{5}} \Bigg]d\theta
$$
Obviously this is not correct. Can you tell me where I have gone wrong?
Thanks in advance :)

Let
$(x,y,z) \in \mathbb{R}^{3}$ and let $S$ be the region enclosed by the surfaces $z = x^{2}+y^{2}$ and $z=4$. Then $(x,y,z) \in S$ if and only if $|x| \leq 4, |y| \leq \sqrt{4-x^{2}}, 0 \leq z \leq 4$. Thus
$$4\int_{0}^{4}\int_{0}^{\sqrt{4-x^{2}}} x^{2} + y^{2} dy dx = 4\int_{0}^{4} x^{2}\sqrt{4-x^{2}} + \frac{(4-x^{2})^{3/2}}{3} dx$$
is the content of $S$.
Let $T$ be the region enclosed by $z=4$ and $z = 4x^{2} + 4y^{2}$ and do the above for $T$. Then the absolute value of the difference of the resulting two contents is the desired content.  
