Graph of equation y^2 = 2x from A(0, 0) and B(2, 2) is resolved around the x-axis. The surface area of resulting solid is?

y = √( 2x )
dy/dx = (1/2) * (2x)^(-1/2) * 2 = 1 / √( 2x )
( dy/dx )^2 = 1/(2x)

A
= ∫ 2πy√[ 1 + (dy/dx)^2 ] dx , from x = 0 to x = 2
= 2π * ∫ √( 2x )√[ 1 + 1/(2x) ] dx
= 2π * ∫ √( 2x + 1 ) dx , from x = 0 to x = 2

Let u = √( 2x + 1 ) , then
u^2 = 2x + 1
2u du = 2 dx
dx = u du

A
= 2π * ∫ u * u du , from u = 1 to u = √5
= 2π * ∫ u^2 du
= 2π * [ u^3 / 3 ] , from u = 1 to u = √5
= (2π/3)( 5√5 - 1 ) ..... Ans