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:
$$\int_{0}^{\pi}\int_{0}^{\sqrt{4-\frac{r^2}{4}}} 4rsin(\theta)drd\theta$$
On
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 $$ V = \iint_D f(r,\theta)r\; dr d\theta = \int_0^{\pi}\int_0^{4\sin \theta} \frac{\sqrt{16-r^2}}{2}\; r \; dr d\theta $$
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).