The area included between the curves $y^2=2 x$ and $x^2=2 y$ is

• A

  $\frac{3}{4}$ sq unit
• Correct answer
  B

  $\frac{4}{3}$ sq unit
• C

  $\frac{1}{2}$ sq unit
• D

  $\frac{4}{3}$ sq unit

Solution

The point of intersection of the curves $y^2=2 x$ and $x^2=2 y$ are $O(0,0), A(2,2)$. $\begin{aligned} \therefore \text { Required area } & =\int_0^2\left(\sqrt{2 x}-\frac{x^2}{2}\right) d x \\ & =\left[\sqrt{2} \frac{x^{3 / 2}}{3 / 2}-\frac{x^3}{6}\right]_0^2 \\ & =\left[\frac{2}{3} \cdot 2^2-\frac{2^3}{6}\right]=\left[\frac{8}{3}-\frac{4}{3}\right] \\ & =\frac{4}{3} \text { sq unit }\end{aligned}$

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