Y=x² limit [0,1] rotated about y=x find area of surface of revolution?

If the function y= x² is rotated about y = x , at some point (x, y) on the parabola, the distance from y=x is equal to

(x-y)/ sqrt(2) = (x- x²) / sqrt (2) ; this will be the radius of surface of revolution at (x, y) on parabola.

Take a differential surface with this radius and width ds along the surface. Surface area of this cylinder will be equal to:

dA = 2 pi R ds = 2 (pi)  [(x- x²) / sqrt (2)] sqrt( 1+(y’)²) dx 

Hece area of surface of revolution is 

= sqrt(2) (pi) ∫ (x- x²) sqrt( 1+4x²) dx  (int. limit 0 to 1)

Use indefinite integration formula for integration of form x sqrt ( a²+x²) , and x² (a²+x²), and limit from 0 to 1.

1/96 sqrt(4 x^2+1) (-24 x^3+32 x^2-3 x+8)+1/64 sinh^(-1)(2 x) ( limit 0 to 1 )