The area (in sq units) bounded by the curves $x=-2 y^2$ and $x=1-3 y^2$ is
The area (in sq units) bounded by the curves $x=-2 y^2$ and $x=1-3 y^2$ is
$\frac{2}{3}$
$1$
$\frac{2}{3}$
$\frac{2}{3}$
Solution
Given curves,
$
\text { and } \quad \begin{aligned}
& x=-2 y^2 \\
& x=1-3 y^2
\end{aligned}
$
On solving both curves, we get
$
\begin{gathered}
-2 y^2=1-3 y^2 \\
\Rightarrow \quad y= \pm 1 \text { and } x=-2
\end{gathered}
$
So, the intersection points are $(-2,1)$ and $(-2,-1)$.
$\therefore$ Required area
$
\begin{aligned}
& =2 \int_0^1\left\{\left(1-3 y^2\right)-\left(-2 y^2\right)\right\} d y \\
& =2 \int_0^1\left(1-y^2\right) d y \\
& =2\left(y-\frac{y^3}{3}\right)_0^1=2\left(1-\frac{1}{3}\right) \\
& =2 \times \frac{2}{3}=\frac{4}{3}
\end{aligned}
$
