Answer
$$\frac{13\pi }{3}$$
Work Step by Step
Given $$y= \sqrt{5-x}, \ \ \ 3 \leq x \leq 5$$
Since the surface area of the surface of revolution
formed by revolving the graph of $f (x)$ around the $x-$axis is given by
\begin{align*}
S_x&=\int_{a}^{b}2\pi f(x)\sqrt{1+[f'(x)]^2}dx\\
&=\int_{3}^{5}2\pi \sqrt{5-x} \sqrt{1+ \frac{1}{4(5-x)}}dx \\
&=\int_{3}^{5}2\pi \sqrt{5-x} \sqrt{ \frac{6-x }{4(5-x)}}dx \\
&=\int_{3}^{5} \pi \sqrt{21-4x} dx \\
&=\frac{-1}{6}\pi (21-x)^{3/2}\bigg|_{3}^{5}\\
&=\frac{13\pi }{3}
\end{align*}
