Answer
$4$
Work Step by Step
Rewrite the equations as functions of y:
$\left[\begin{array}{lll}
x-y^{2}=0 & ... & x+2y^{2}=3\\
x=y^{2} & & x=3-2y^{2}\\
& &
\end{array}\right]$
Graphing the given equations, we find the intersections
or we solve the equation $f(y)=g(y), $ where
$ f(y)=3-2y^{2} \quad$ (the curve on the right side) and
$ g(y)=y^{2}\quad$ (the curve on the left).
When $y\in[c,d]=[-1,1]$, the area between the graphs is
$A=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=\displaystyle \int_{-1}^{1}[3-2y^{2}-(y^{2})]dy$
$=\displaystyle \int_{-1}^{1}[3-3y^{2}]dy$
$=\left[3y-y^{3} \right]_{-1}^{1}$
$=3(1+1)-(1+1)$
$=4$
