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 are almost right but you reversed some of the steps:
For these questions always ask yourself: what is the projection of the solid in the $xy$ plane (or any plane actually, but here $xy$ is relevant). You have found that it is the disc
 $$ D= \{ (r,\theta) \mid 0 \le r \le 4 \sin \theta, 0 \le \theta \le \pi \}
$$
So far so good. Then, you have found that the solid is bounded above by the sphere $z=f(r,\theta)=\frac{\sqrt{16-r^2}}{2}$. Therefore, the volume equals
$$
V = \iint_D f(r,\theta)r\; dr d\theta = \int_0^{\pi}\int_0^{4\sin \theta} \frac{\sqrt{16-r^2}}{2}\; r \; dr d\theta
$$
