#### Answer

$\int^{b}_{a} 2\pi[c-f(x)] \sqrt{1+[f'(x)]^{2}}dx$

#### Work Step by Step

The analogue of $f(x_{i}^{*})$in the derivation of (4) is now $c - f(x_{i}^{*})$, so
$S = \lim_{n-> \infty} \Sigma^{n}_{i=1} 2\pi [c-f(x_{i}^{*})] \sqrt{1+[f'(x_{i}^{*})]^{2}}\Delta x = \int^{b}_{a} 2\pi[c-f(x)] \sqrt{1+[f'(x)]^{2}}dx$