Answer
$$S = \int_c^d 2\pi g(y) \sqrt {1 + [g'(y)]^2} dy $$
Work Step by Step
Let us consider that $x=g(y)$ is a non-negative function which defines a continuous first derivative on the interval $[c,d]$.
The area of the surface obtained when the graph of a function $g(y)$ on the interval $\left[ {c,d} \right]$ is revolved about the $y$- axis can be expressed as:
$$S = \int_c^d 2\pi g(y) \sqrt {1 + [g'(y)]^2} dy $$
