Answer
$S_g = S_f + 2\pi cL$
Work Step by Step
$g(x) = f(x) + c$ and $g'(x) = f'(x)$
$S_g = \int ^{b}_{a} 2\pi g(x) \sqrt{1+[g'(x)]^{2}}dx = \int^{b}_{a} 2\pi [f(x) + c] \sqrt{1+[f'(x)]^{2}}dx$
$ = \int ^{b}_{a} 2\pi f(x) \sqrt{1+[f'(x)]^{2}}dx + 2\pi c\int^{b}_{a} \sqrt{1+[f'(x)]^{2}}dx = S_f + 2\pi cL$
