Area of a surface of Revolution

Question

The given curve is rotated about the y-axis. Find the area of the resulting surface.

x^2⁄3 + y^2⁄3 = 4,    0 ≤ y ≤ 8

Answer 1

x= (4 - y^2/3)^3/2, 0 ≤ y ≤ 8; s = 2π∫₀⁸(4 - y^2/3)^3/2√1 + [3/2(4 - y^2/3)^1/2(-2/3y^-1/3]²dy =

2π∫₀⁸(4 - y^2/3)^3/2√1 + (4 - y^2/3)y^-2/3dy = 2π∫₀⁸(4 - y^2/3)^3/22y^-1/3dy = 4π∫₀⁸(4 - y^2/3)^3/2d(4 - y^2/3)·(-3/2) =

-6π·2/5(4 - y^2/3)^5/2₀⁸ = 12π/5·32 = 384π/5

Answer 2

x^2/3 + y^2/3 = 4, 0 ≤ y ≤ 8

As y varies from 0 to 8, x varies from 8 to 0.

y^2/3 = 4 - x^2/3. So, y = (4 - x^2/3)^3/2 and dy/dx = (3/2)(4 - x^2/3)^1/2(-2/3)x^-1/3

S = 2π∫_{(0 to 8)} x√(1 + (dy/dx)²)dx = 2π∫_{(0 to 8)}x(2x^-1/3)dx = 4π∫_{(0 to 8)} x^2/3dx = 4π[(3/5)x^5/3]_{(0 to 8)} = 384π/5