Question

Find the total area of the shaded region.

Answer 1

The two curves intersect when $y=0$ and $y=1$. Over the range $0\le y\le1$, the curve $x=12y^2$ lies above $x=12y^3$. Hence the area of the shaded region is

$\displaystyle \int_0^1 (12y^2 - 12y^3) \, dy = 4y^3 - 3y^4\bigg|_0^1 = (4-3)-(0-0) = \boxed{1}$