The area (in sq units) bounded by the curves $x=-2 y^2$ and $x=1-3 y^2$ is

• A

  $\frac{2}{3}$
• B

  $1$
• Correct answer
  C

  $\frac{2}{3}$
• D

  $\frac{2}{3}$

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 $$ \begin{aligned} & =2 \int_0^1\left\{\left(1-3 y^2\right)-\left(-2 y^2\right)\right\} d y \\ & =2 \int_0^1\left(1-y^2\right) d y \\ & =2\left(y-\frac{y^3}{3}\right)_0^1=2\left(1-\frac{1}{3}\right) \\ & =2 \times \frac{2}{3}=\frac{4}{3} \end{aligned} $$

Hence, the correct answer is option: 'C'.