Answer
$18$
Work Step by Step
Graphing the given equations, we take
$f(y)=2y^{2}$ (the curve on the right side) and
$g(y)=0$ (the curve on the left).
When $y\in[c,d]=[0,3]$, the area between the graphs is
$A=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=\displaystyle \int_{0}^{3}2y^{2}dy=$
$=2\displaystyle \left[\frac{y^{3}}{3}\right]_{0}^{3}$
$=\displaystyle \frac{2}{3}(27-0)$
$=18$
