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

• A

  $\frac{3}{2}$
• B

  2
• C

  3
• Correct answer
  D

  4

Solution

Given curves are $x=y^2$ and $x=3-2 y^2 \Rightarrow y^2=-\frac{(x-3)}{2}$ Point of intersections are $(1,1)$ and $(1,-1)$
$\begin{aligned} & \therefore \text { Required area }=2 \int_0^1\left(x_2-x_1\right) d y \\ & =2 \int_0^1\left[\left(3-2 y^2\right)-y^2\right] d y=2\left[\left(3 y-\frac{3 y^3}{3}\right)\right]_0^1 \\ & =2[3-1]=4\end{aligned}$

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