The given curve is rotated about the y-axis. Find the area of the resulting surface.

$\sqrt[3]{x^2}+\sqrt[3]{y^2}=1 , 0\leq y \leq1 $

We set up the surface area integral:

$\displaystyle \int_{-1}^{1} 2\pi x ds $

I understand how to find the arc-length(ds) and set up the integral but why wouldn't out boundaries stay from 0 to 1? My reasoning is that $\displaystyle 2\pi x $

is taking in account that our radius from 0 to 1 should be doubled. Shouldn't we only care about the arc-length from 0 to 1?

For instance if we were to find the surface area of $\displaystyle y=1/3 \sqrt[2]{x^3}+$ , $0\leq y \leq12 $rotating about the y-axis, our integral would be,

$\displaystyle \int_{0}^{12} 2\pi y ds $