# Thread: Surface Area of a Revolution

1. ## Surface Area of a Revolution

Problem: Find the surface area of revolution about the x-axis of ${y}={5}{\sin{{\left({6}{x}\right)}}}$ over the interval ${0}\le{x}\le\frac{\pi}{{6}}$.

I got that y' = 30cos(6x)

and
S.A. = $10\pi \int_0^{\pi/6} sin(6x)\sqrt{1+900cos(6x)^2}$

But can't figure out how to do that integral. Is the integral wrong?

2. Originally Posted by twoteenine
Problem: Find the surface area of revolution about the x-axis of ${y}={5}{\sin{{\left({6}{x}\right)}}}$ over the interval ${0}\le{x}\le\frac{\pi}{{6}}$.

I got that y' = 30cos(6x)

and
S.A. = $10\pi \int_0^{\pi/6} sin(6x)\sqrt{1+900cos(6x)^2}$

But can't figure out how to do that integral. Is the integral wrong?
It should be $cos^2(6x)$ not $cos(6x)^2$.

use $u=cos(6x)$.