Q. The area bounded between the parabolas $x^{2} = \frac{y}{4}$ and $x^{2} = 9y$ and the straight line $y = 2$ is :
AIEEEAIEEE 2012Application of Integrals
Solution:
$y = 4x^{2} $
$y = \frac{1}{9}x^{2}$
Area $= 2\int^{2}_{0}\left(3\sqrt{y}-\frac{\sqrt{y}}{2}\right)dy = 2\left[\frac{5}{2} \frac{y\sqrt{y}}{3 / 2}\right]^{2}_{0} = 2. \frac{5}{3} 2\sqrt{2} = \frac{20\sqrt{2}}{3}$
