<s>[INST] 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 :)
 [/INST] Three volumes.
Say $r^2 = x^2 + y^2$, cut through the volume gives the following image.

Cylinder #1

$$
V_{C_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^4 dz r = 16 \pi
$$

Paraboloid #1

$$
V_{P_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^{r^2} dz r = 8 \pi
$$

Cylinder #2

$$
V_{C_1} = \int_0^{2\pi} d\phi \int_0^1 d r \int_0^4 dz r = 4 \pi
$$

Paraboloid #2

$$
V_{P_1} = \int_0^{2\pi} d\phi \int_0^1 d r \int_0^{4 r^2} dz r = 2 \pi
$$

End volume is

$$
\Big( V_{C_1} - V_{P_1} \Big) - \Big( V_{C_2} - V_{P_2} \Big) = 6 \pi.
$$

</s>