The area of the plane region bounded by the curves $x + 2y^2=0$ and $x+3y^2=1$ above the $x-$axis is equal to

We first find the points of intersection of the given curves. Put $x = – 2y^2$ in $x+3y2=1$ to obtain $y^2=1 \Rightarrow y=\pm 1\Rightarrow x=-2$.
Thus the points of intersection are $(– 2, 1)$ and $(– 2, – 1)$. Both the curves are parabola with $x+2y^2=0$ having $(0, 0)$ as vertex and $x+3y^2=1$ has vertex as $(1, 0)$.

Required area $=\left[{\int }_{-2}^1 \sqrt{\frac{1-x}{3}}dx-{\int }_{-2}^0 \sqrt{\frac{-x}{2}}dx\right]$
                    $\begin{array}{l}{\left.{\left.=\frac{1}{\sqrt{3}}(1-x{)}^{3/2}\left(\frac{2}{3}\right)(-1)\right]}_{-2}^1-\frac{1}{\sqrt{2}}(-x{)}^{3/2}\left(\frac{2}{3}\right)(-1)\right]}_{-2}^0\\ =\frac{4}{3}\left[\frac{1}{\sqrt{2}}\left(0-2^{3/2}\right)-\frac{1}{\sqrt{3}}\left(0-3^{3/2}\right)\right]\\ =\frac{4}{3}(3-2)=\frac{4}{3}\end{array}$