The area (in sq units) bounded by the curves $y^2=4 x$ and $x^2=4 y$ is

• A

  $\frac{64}{3}$
• Correct answer
  B

  $\frac{16}{3}$
• C

  $\frac{8}{3}$
• D

  $\frac{2}{3}$

Solution

Given curves are $y^2=4 x$ and $x^2=4 y$. The intersection point is $x^4=16 y^2=16(4 x)$
$\begin{array}{rlrl}\Rightarrow & x\left(x^3-64\right) & =0 \\ \Rightarrow & & x & =0,4 \\ \Rightarrow & & y & =0,4\end{array}$ Hence, intersection point is $O(0,0)$ and $A(4,4)$ $\therefore$ Required area $=$ shaded region of the curve $\begin{aligned} & =\int\left(y_2-y_1\right) d x \\ & =\int_0^4\left(\sqrt{4 x}-\frac{x^2}{4}\right) d x \\ & =\left[\frac{2 x^{3 / 2}}{3 / 2}-\frac{x^3}{12}\right]_0^4 \\ & =\left[\frac{4}{3}(4)^{3 / 2}-\frac{(4)^3}{12}-(0-0)\right] \\ & =\left(\frac{32}{3}-\frac{64}{12}\right)=\frac{128-64}{12} \\ & =\frac{64}{12}=\frac{16}{3} \text { sq units }\end{aligned}$

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