Answer
$\displaystyle \frac{8}{3}$
Work Step by Step
Rewrite the equations as functions of y:
$\left[\begin{array}{lll}
x+y^{2}=0 & ... & x+3y^{2}=2\\
x=-y^{2} & & x=2-3y^{2}\\
& &
\end{array}\right]$
Graphing the given equations, we find the intersections
or we solve the equation $f(y)=g(y), $ where
$ f(y)=2-3y^{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
$A=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=\displaystyle \int_{-1}^{1}[2-3y^{2}-(-y^{2})]dy$
$=\displaystyle \int_{-1}^{1}[2-2y^{2}]dy$
$=\displaystyle \left[2y-\frac{2y^{3}}{3} \right]_{-1}^{1}$
$=2(1+1)-\displaystyle \frac{2(1+1)}{3}$
$=\displaystyle \frac{8}{3}$
