Answer
$\displaystyle \frac{1}{12}$
Work Step by Step
On the interval $y\in[0,1]$, the graph of $x=y^{2}$ is above the graph of $x=y^{3}$
(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,
$A=\displaystyle \int_{0}^{1}(y^{2}-y^{3})dy$=$\displaystyle \int_{0}^{1}y^{2}dy-\int_{0}^{1}y^{3}dy$
$=[\displaystyle \frac{y^{3}}{3}]_{0}^{1}-[\frac{y^{4}}{4}]_{0}^{1}$
$=(\displaystyle \frac{1}{3}-0)-(\frac{1}{4}-0)$
$=\displaystyle \frac{1}{3}-\frac{1}{4}$
$=\displaystyle \frac{4-3}{12}$
$=\displaystyle \frac{1}{12}$
