18. Find the exact area of the surface obtained by rotating the given curve about the x-axis. x = a cos3(θ), y = a sin3(θ), 0 ≤ θ ≤ π/2

Answer 1

I will write t for θ. The equation for surface area obtained by rotating through the x-axis a curve given by parametric equations is

1. ∫2πy√[(dx/dt)²+(dy/dt)²]dt with bounds c<=t<=d
2. dx/dt=-3asin(3t) => (dx/dt)²=9a²sin²(3t)
3. dy/dt=3acos(3t) => (dy/dt)²=9a²cos²(3t)
4. (dx/dt)²+(dy/dt)²=9a²(sin²(3t)+cos²(3t))=9a²

Putting it all together:

SA = ∫2πasin(3t)√(9a²)dt with bounds 0<=t<=π/2

SA = 2aπ∫sin(3t)*3adt = 6a²π∫sin(3t)dt = -6a²πcos(3t)/3 = -2a²π(cos(3t)) from 0 to pi/2 = -2a²π(0-1)

=> SA = 2a²π