Answer
a) See the explanation below.
b) See the explanation below.
c) See the explanation below.
Work Step by Step
a) Let the surface area of the curve $C$ be $S$ about the $x$-axis:
$S=\int_a^b2 \pi f(x) \sqrt {1+[f'(x)]^2} dx$
b) Let the formula to calculate the surface area of the curve $C$ be $S$ about the $x$-axis:
$S=\int_c^d2 \pi f(y) \sqrt {1+[g'(y)]^2} dy$
c) From part (a) and (b), we have
$S=\int_a^b2 \pi f(x) \sqrt {1+[f'(x)]^2} dx$; for $y=f(x)$
and
$S=\int_c^d2 \pi f(y) \sqrt {1+[g'(y)]^2} dy$ for $x$ as a function of $y$
