Answer
$\displaystyle \frac{4}{3}$
Work Step by Step
On the interval $y\in[0,1]$, the graph of $x=12y^{2}-12y^{3}$ is above the graph of $x=2y^{2}-2y$
(from the perspective of the y-axis, "above" means "to the right".)
The area between the two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dt$.
Thus,
$A=\displaystyle \int_{0}^{1}[(12y^{2}-12y^{3})-(2y^{2}-2y)]dy$=$\displaystyle \int_{0}^{1}(10y^{2}-12y^{3}+2y)dy$
$=10[\displaystyle \frac{y^{3}}{3}]_{0}^{1}-12[\frac{y^{4}}{4}]_{0}^{1}+2[\frac{y^{2}}{2}]_{0}^{1}$
$=10(\displaystyle \frac{1}{3}-0)-12(\frac{1}{4}-0)+2(\frac{1}{2}-0)$
$=\displaystyle \frac{10}{3}-3+1$
$=\displaystyle \frac{10}{3}-2$
$=\displaystyle \frac{10-6}{3}$
$=\displaystyle \frac{4}{3}$
