Assume f is a nonnegative function with a continuous first derivative on [a, b]. The curve y = f(x) on [a, b] is revolved about the x-axis. Explain how to find the area of the surface that is generated.
What is the area of the surface generated when the graph of f on the interval [a, b] is revolved about the x-axis?
A. To find the area of the surface that is generated, evaluate the integral $S = \int_a^b 2\pi f(x)\sqrt{1 + (f'(x))^2} dx$.
B. To find the area of the surface that is generated, evaluate the integral $S = \int_a^b \pi f(x)\sqrt{1 + f'(x)} dx$.
C. To find the area of the surface that is generated, evaluate the integral $S = \int_a^b 2\pi \sqrt{1 + (f'(x))^2} dx$.
D. To find the area of the surface that is generated, evaluate the integral $S = \int_a^b \pi f(x)\sqrt{1 + (f'(x))^2} dx$.
