Surface area of $y = \sin(\pi x)$, from $x=0$ to $2$, rotated about the $x$-axis.

Question

When I use the surface area formula I get 0, and Wolfram got zero as well when I use the bounds 0 to 2, why is this? However the solution manual uses the integral with bounds 0 to 1.. What is going on?!

Answer 1

Surface area of $y=\sin(\pi x)$ from $x=0$ to $2$, rotated about the $x$-axis, is $$S = 2\pi\int_0^2 |\sin(\pi x)|\sqrt{1+(\sin(\pi x))'^2}\,dx$$ Using WA (see link), we obtain $$S = 4\sqrt{1+\pi^2} + \frac{4}{\pi}\ln{(\pi+\sqrt{1+\pi^2})}$$

Since $|\sin(\pi(1+x))|=|\sin(\pi x)|$, you can integrate from $0$ to $1$ and double result.