Find surface area generated by $x^{2/3}+y^{2/3}=1$ about $-1 \le x \le 1$
So I got $\frac{dy}{dx} = - \frac{y^{1/3}}{x^{1/3}}, \qquad \sqrt{1+f'(x)}=\frac{1}{x^{1/3}}$
The rest of the answer looks like $A = \color{blue}2 \int^{1}_{0} 2 \pi \frac{y}{x^{1/3}} dx $
How did the 2 appear?
$= 4\pi \int y^{2/3} \color{blue}{\frac{y^{1/3}}{x^{1/3}}} dx= -4\pi \int y^{2/3} dy$
How did they get this $\frac{y^{1/3}}{x^{1/3}}$ removed?