#### Answer

See the proof below.

#### Work Step by Step

Applying the surface area formula and using the facts that $y=r$, $y'=0$, we have the integral,
$$2\pi\int_0^{h}r\sqrt{1+(y')^2}dx=2\pi\int_0^{h}rdx=2\pi r x|_0^h=2\pi r h.$$
Which is the surface area of a cylinder of radius r and length h.