Calculate the volume of the solid bounded laterally. How do i find the volume bounded below by the plane $xy$ and bounded above by $x^2+y^2+4z^2=16$ and laterally by the cylinder $x^2+y^2-4y=0$.
Since when i change to polar coordinates $x^2+y^2-4y=0$. is equal to $4sin(\theta)$. 
And for the limits. $z=\frac{\sqrt{16-x^2-y^2}}{4}$ is equal to $z=\sqrt{4-\frac{r^2}{4}}$
So i think the integral for this is:
$$\int_{0}^{\pi}\int_{0}^{\sqrt{4-\frac{r^2}{4}}} 4rsin(\theta)drd\theta$$

You have some mistake in the limits of integration.   Using the symmetry of the solid around the $y-z$ plane ( see the figure), we can take  for $\theta$  the values between $0$ and $\frac{\pi}{2}$ and duplicate the integral, so the limits becomes:
$$
0<\theta<\frac{\pi}{2} \qquad 0<r<4\sin \theta \qquad 0<z< \frac{1}{2}\sqrt{16-r^2}
$$
so the volume is:
$$
V= 2\int_0^{\frac{\pi}{2}}\int_0^{4\sin \theta}\int_0^{\frac{1}{2}\sqrt{16-r^2}}rdzdrd\theta=2\int_0^{\frac{\pi}{2}}\int_0^{4\sin \theta}\frac{1}{2}\sqrt{16-r^2}dr d\theta=
$$
$$
=\frac{1}{3}\int_0^{\frac{\pi}{2}}64(1-\cos^2\theta)d\theta=\frac{32}{3}\pi-\frac{128}{9}
$$
(if my calculations are correct). 

