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:

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

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 rotating about the y-axis, our integral would be,