Q. The area bounded by the curves $y^2 = 4x$ and $x^2 = 4y$ is :
AIEEEAIEEE 2011Application of Integrals
Solution:
Area $= \int\limits^{4}_{0}\left(2\sqrt{x}-\frac{x^{2}}{4}\right)dx$
$= \left(2\left(\frac{x^{3/2}}{3/2}\right)-\frac{x^{3}}{12}\right)^{4}_{0}$
$= \frac{4}{3}\times8 - \frac{64}{12}$
$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$
