how do you find the area of the surface obtained by rotating the curve about the x-axis? Given hint in the question: write $y$ in terms of $x$.
$3y^2 = x(1-x)^2,\ 0\leq x\leq 1$
I got $\frac{1}{3}\int_{0}^{1}\sqrt{12x(1-x)^2+(1-4x+3x^2)^2} \ dx$ and I'm stuck.
Any hints would be appreciated. Thanks!