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?
Best Answer
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?