Find the surface area of the solid of revolution. Find the surface area of the solid of revolution obtained by rotating the curve.
$$x= \sqrt{2y-y^2},\quad y\in\left[\frac{15}{16},\frac{25}{16}\right]$$
I get  $$A=\int 2\pi\sqrt{2y-y^2}\sqrt{\left(\frac{-1}{y(y-2)}\right)}\,dy$$ 
and I can't find a way to integrate it :(

There is a very nice simplification to this integral if we simply multiply the two radicals together. 
\begin{align*}
\int_{15/16}^{25/16}2\pi\sqrt{\frac{-(2y-y^2)}{y^2-2y}}\ \mathrm{d}y&=\int_{15/16}^{25/16} 2\pi\sqrt{\frac{y^2-2y}{y^2-2y}}\ \mathrm{d}y\\ &=\int_{15/16}^{25/16}2\pi \ \mathrm{d}y \\&=\bigg[2\pi y\bigg]_{15/16}^{25/16}\\&=\frac{50}{16}\pi-\frac{30}{16}\pi\\
&=\frac54\pi
\end{align*}
