Q. The area (in square unit)bounded by the curves $y^{2} = 4x$ and $x^{2} = 4y$ in the plane is
A) $\frac{8}{3}$
B) $\frac{16}{3}$
C) $\frac{32}{3}$
D) $\frac{64}{3}$
Explanation
(b)
Required area of shaded portion $OABCO$
[image:8_image99.jpeg]
$= \int_{0}^{4}{\left( \sqrt{4x} - \frac{x^{2}}{4} \right)dx}$
${= \left\lbrack \frac{{2x}^{3/2}}{3/2} - \frac{x^{3}}{12} \right\rbrack}_{0}^{4}$
$= \left\lbrack \frac{32}{3} - \frac{16}{3} \right\rbrack$
$= \frac{16}{3}sq\ units$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.