Question

How do you find the area of the surface generated by rotating the curve about the y-axis `x^(2/3)+y^(2/3)=1` for the first quadrant?

Answer 1

`(6 pi)/5`

Explanation:

We can param this:

Let `x = cos^3 t` and `y = sin^3 t` so that we do indeed have `x^(2/3)+y^(2/3)=1`

Then arc lenth along the curve in Q1 is:

`ds = dot s \ dt = sqrt(dot x^2 + dot y ^2) \ dt`

`= sqrt((-3 cos^2 t sin t)^2 + (3 sin ^2 t cos t) ^2) \ dt`

`= 3 cos t sin t \ dt`

The surface area of the revolution around the y-axis is:

`dS = 2 pi x ds`

`= 6 pi \ cos^4 t \ sin t \ dt`

And:

`S = 6 pi \ int_(t = 0)^(pi/2) \ dt qquad cos^4 t \ sin t`

`= 6 pi (- 1/5cos^5 t)_(t = 0)^(pi/2) = (6 pi)/5`