The surface area of a curve y = f(x) generated by revolving about the x-axis on the interval [a,b]
= 2π*∫y*√(1 + (dy/dx)² dx from a to b
In this problem:
dy/dx = 3x²/4
The surface area generated by revolving the curve y = x^3/4 on the interval [0,1] about the x-axis:
= 2π*∫(x^3/4)*√(1 + 9x^4/16) dx from 0 to 1
= π/8*∫x^3*√(16 +9x^4) dx from 0 to 1
u = 16 + 9x^4
du/36 = x^3 dx
= π/288*∫√u du from 16 to 25
= π/288[2/3*u^(3/2) eval. from 16 to 25]
= π/288[250/3 - 128/3]
= π/288[122/3] = 61π/432
