The area of the plane region bounded by the curves $x+2 y^2=0$ and $x+3 y^2=1$ is equal to
The area of the plane region bounded by the curves $x+2 y^2=0$ and $x+3 y^2=1$ is equal to
$\frac{5}{3}$
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{4}{3}$
Solution
Solving the equations we get the points of intersection $(-2,1)$ and $(-2,-1)$
The bounded region is shown as shaded region.
$
\begin{aligned}
& \text { The required area }=2 \int_0^1\left(1-3 y^2\right)-\left(-2 y^2\right) \\
& =2 \int_0^1\left(1-y^2\right) d y=2\left[y-\frac{y^3}{3}\right]_0^1=2 \times \frac{2}{3}=\frac{4}{3} .
\end{aligned}
$
