#### Answer

$$2 \pi R h $$

#### Work Step by Step

Since
$$x^2+y^2=R^2 $$
Then
\begin{aligned}
1+\left(x^{\prime}\right)^{2} &=1+\left(-\frac{y}{\sqrt{R^{2}-y^{2}}}\right)^{2} \\
&=1+\frac{y^{2}}{R^{2}-y^{2}} \\
&=\frac{R^{2}}{R^{2}-y^{2}}
\end{aligned}
Hence
\begin{aligned}
\text { Surface Area }&=2 \pi \int_{R-h}^{R} \sqrt{R^{2}-y^{2}} \sqrt{\frac{R^{2}}{R^{2}-y^{2}}} d y \\
&=2 \pi \int_{R-h}^{R} \sqrt{R^{2}-y^{2}} \frac{R}{\sqrt{R^{2}-y^{2}}} d y \\
&=2 \pi R \int_{R-h}^{R} d y \\
&=2 \pi R[y]_{R-h}^{R} \\
&=2 \pi R[R-(R-h)] \\
&=2 \pi R h
\end{aligned}