Pls slove the math

Question

Find the surface area of the region under the curve (x/a)^2/3 + (y/a)^2/3 =1 by revolving about x-axis.

Answer 1

I[0,a ]2 π f(x) √(1+(dy/dx)^2) dx

If (x/a)^(2/3) + (y/a)^(2/3) = 1,

(y/a)^(2/3) = 1 - (x/a)^(2/3), or

y = a(1 - (x/a)^(2/3))^(3/2)

dy/dx = -(x/a)^-1/3 (1 - (x/a)^(2/3))^(1/2), and

I[0,a f(x) √(1+(dy/dx)^2) dx =

2 π I[0, a] a(1 - (x/a)^(2/3))^(3/2) √1+ (x/a)^-2/3 (1 - (x/a)^(2/3)) =

2 π I[0, a] a(1 - (x/a)^(2/3))^(3/2) (x/a)^-1/3 dx =

-6 π a/5 (1 - (x/a)^(2/3))^(5/2) E[0, a] =

6 π a/5