4. (10 pts) Use polar coordinates to set up a double integral to calculate the volume of the solid under the surface $z=x^{2}+y^{2}$, above the $x y$ - plane, and inside the cylinder $x^{2}+$ $y^{2}=2 y$.

Step 1 ：Convert the given equations into polar coordinates. In polar coordinates, \(x = rcos(\theta)\) and \(y = rsin(\theta)\). So, the equation of the surface becomes \(z = r^{2}\) and the equation of the cylinder becomes \(r^{2} = 2rsin(\theta)\) or \(r = 2sin(\theta)\).

Step 2 ：Determine the limits of the integral by the region defined by the cylinder. Since the cylinder equation simplifies to \(r = 2sin(\theta)\), the limits for \(r\) are from 0 to \(2sin(\theta)\). The limits for \(\theta\) are from 0 to \(\pi\) because the cylinder is symmetric about the y-axis and covers the entire upper half plane.

Step 3 ：Calculate the volume of the solid by the double integral of the function \(z = r^{2}\) over the region defined by the cylinder. In polar coordinates, the differential area element is \(rdrd\theta\). So, the double integral becomes \(\int_{0}^{\pi}\int_{0}^{2sin(\theta)}r^{3}drd\theta\).

Step 4 ：Compute the double integral to get the volume of the solid. The volume is \(\frac{3\pi}{2}\).

Step 5 ：Final Answer: The volume of the solid under the surface \(z=x^{2}+y^{2}\), above the \(x y\) - plane, and inside the cylinder \(x^{2}+ y^{2}=2 y\) is \(\boxed{\frac{3\pi}{2}}\).