Answer
$S(A)=\int_a^b 2 \pi y \sqrt {1+(\dfrac{dy}{dx})^2} \ dx$
Work Step by Step
Let $S(A)$ be the area of the surface swept out by revolving the graph of a smooth continuous function $y=f(x)$ on the interval $[a,b]$ about the x-axis, which can be mathematically expressed as:
$S(A)=\int_a^b 2 \pi y \sqrt {1+(\dfrac{dy}{dx})^2} \ dx$
