# Find an integral for the area of the surface generated by revolving the curve $y=sin(x)$ between $0 \le x \le \pi$, about the x-axis

So here is my problem:

Find an integral for the area of the surface generated by revolving the curve $y=sin(x)$ between $0 \le x \le \pi$, about the x-axis

Just thinking about the problem I feel like I would just do the following:

$$2\int_{0}^{\pi}\sin(x)\mathrm{d}x$$

But I'm not sure if this is the way to go about doing this. Any hints?

• Write $\sin$ using \sin instead of sin and write $\mathrm dx$ using \mathrm{d}x instead of dx.
– user93957
Dec 7 '13 at 21:38

The surface area is found by the formula $$A = 2\pi\int\rho\, ds,$$ where $\rho$ is the radius of rotation and $ds$ is the element of arc-length. You have $$A = 2\pi \int_0^\pi \sin(x)\sqrt{1 + \cos^2(x)}\, dx$$ Can you integrate this?