How to find the surface area of revolution of $x^{2/3} + y^{2/3} = 1, 0 \le \le 1$ around the $y$-axis?

Question

I am trying to solve this problem, but I have a hard time understanding how to solve it. The question is to find the area of the resulting surface...

$x^{2/3} + y^{2/3} = 1, 0 \le \le 1$ around the y-axis

I am thinking to put...

The derivative is $-1/x^{1/3}.$

I am stuck on this problem, please help!

Answer 1

The surface area is $$2\pi\int_0^1 x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy.$$ Here, $$x=(1-y^{2/3})^{3/2}$$ so that $$\frac{dx}{dy}=-y^{-1/3}(1-y^{2/3})^{1/2}$$ and $$\left(\frac{dx}{dy}\right)^2=y^{-2/3}(1-y^{2/3})=y^{-2/3}-1.$$ The integral is $$2\pi\int_0^1(1-y^{2/3})^{3/2}y^{-2/3}\,dy.$$ Good luck!

Answer 2

Starting from Lord Shark the Unknown's answer $$I=\int(1-y^{\frac 23})^{\frac32}y^{-\frac23}\,dy$$ let $$y=\sin^3(t)\implies dy=3 \sin ^2(t) \cos (t)\,dt$$ which makes $$I=3 \int \cos^4(t)\,dt$$ and $$\cos^4(t)=\frac{1}{8} \cos (4 t)+\frac{1}{2} \cos (2 t)+\frac{3}{8}$$