Solve Integral: y=(1-x^{2/3})^{3/2}, 0\leqx\leq1

[apoptosis]

Hello
Here is an integral question I have. It's a surface area problem, but i just have issues simplifying the square root before solving. Any help appreciated!

y=(1-x$$^{2/3}$$)$$^{3/2}$$, 0$$\leq$$x$$\leq$$1

My work:
dy/dx=-x$$^{-1/3}$$(1-x$$^{2/3}$$)$$^{1/2}$$
(dy/dx)$$^{2}$$=-x$$^{-2/3}$$(1-x$$^{2/3}$$)
=-x$$^{-2/3}$$+1



S=(insert integral with limits here) 2$$\pi$$xds
=2$$\pi$$x $$\sqrt{-x^{-2/3} +1+1}$$
=2$$\pi$$x $$\sqrt{-x^{-2/3} +2}$$

I'm not sure where to go after this...
thx

 

[kurt.physics]

!The integral you are looking for is S=2\pi\int_0^1 \sqrt{-x^{-2/3}+2}\, dx. Now you can split the integrand into two parts to simplify the square root by using u-substitution: Let u=-x^{-2/3}+2, then du=-2x^{-5/3}dx. Substitute this into your integral and solve it: S=2\pi\int_2^3 \sqrt{u}\cdot (-2x^{-5/3})du=4\pi\int_2^3 u^{1/2}du=4\pi\left[\frac23u^{3/2}\right]_2^3 =\frac{16\pi}{3}\left(3^{3/2}-2^{3/2}\right)=\frac{16\pi}{3}\left(3\sqrt 3 -2\sqrt 2\right).