Find the surface area of the solid of revolution.

Question

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 :(

Answer 1

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*}