Find the area of the surface?

Question

Find the area of the surface obtained when the graph of
y=x^2, 0 <= x <= 1, is rotated around the y-axis.

Answer 1

Given the function:
y=x^2 ; 0≤x≤1
Calculation of the surface of the revolving surface:
First, we calculate the circumference for any x:
O= 2πR & R=x^2 => O=2πx^2

For any infinitesimal curve length we have
ds = sqrt( dx^2+dy^2) & dy/dx = 2x => dy =2xdx =>
ds = sqrt(dx^2+4x^2dx^2) = sqrt( 1+4x^2) *dx

for any infinitesimal strip of the surface we have da = O*ds = 2π*x^2*sqrt( 1+4x^2)*dx = >
A = ∫da=∫ 2πx^2* sqrt( 1+4x^2) *dx =
= 4π *∫x^2* sqrt( 1/4+x^2) *dx
I have divided with 4 under the sqrt and ofcause multiplicatet with 2 outside the sqrt.

The integral is it possible to find in at Integrations table, then calculate the integral for x=0 to x=1