Question

Find the area of the surface generated by revolving the curve x=\frac{y^3}{2} from 0 \leq y \leq 3 about the y-axis.

Answer 1

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