Find the length of the loop of the curve: 9y^2 = x(x-3)^2?

Question

I think you have to use the equation: L= integral from a to b of sqrt(1 + (dy/dx)^2) dx, but I am having a tough time using this equation.

Answer 1

The curve is not a function so you have to take only the top half of the curve. So use y = +sqrt(x(x-2)^2/9)
y = [(x-2)x^.5]/3
So L = 2*integral 0 to 3 (1+y'^2)^.5 dx

Hope this helps.