The area of the figure bounded by the parabolas $x=-2 y^{2}$ and $x=1-3 y^{2}$ is

• Correct answer
  A

  $\frac{4}{3} \mathrm{sq}$ units
• B

  $\frac{2}{3} \mathrm{sq}$ units
• C

  $\frac{3}{7}$ sq units
• D

  $\frac{6}{7}$ sq units

Solution

We have, $x=-2 y^{2}$ $\Rightarrow \quad y^{2}=-\frac{1}{2} x$ ...(i) and $\quad x=1-3 y^{2}$ $\Rightarrow y^{2}=-\frac{1}{3}(x-1)$ ...(ii) From Eqs. (i) and (ii),we get. $\begin{aligned} &-\frac{1}{2} x=-\frac{1}{3}(x-1) \\ \Rightarrow & 3 x=2(x-1) \\ \Rightarrow & 3 x=2 x-2 \\ \Rightarrow & x=-2 \\ \Rightarrow & y^{2}=-\frac{1}{2}(-2)=1 \\ \Rightarrow & y=\pm 1 \end{aligned}$ $\therefore$ Point of intersection of two curves is $(-2 \pm 1)$ $\therefore$ Required Area $=2 \int_{0}^{1} \sin x d x$ $=2\left[y-\frac{y^{3}}{3}\right]_{0}^{1}$ $=2\left[1-\frac{1}{3}\right]$ $=\frac{4}{3} \mathrm{sq}$ units

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