The area between the parabolas x²=\;y/4 and x²=9 y and the straight line y=2 is: [AIEEE 2012]

Correct Answer is : 2023{20{2}}{3}3202

Correct Answer is : 2023{20{2}}{3}3202

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AIEEE 2012 Solved Paper

Section: Mathematics

Question : 77 of 89

Marks: +1, -0

x^2=\;\frac{y}{4}

x^2=9 y

y=2

Solution:

Given curves

x^2=\;\frac{y}{4}

and

x^2=9 y

are the parabolas whose equations can be written as

y=4 x^2

and

y=\;\frac{1}{9} x^2

Also, given

y=2

Now, shaded portion shows the required area which is symmetric.

\therefore

Area

=2 \int\limits_{0}^{2} \left(\sqrt{9y}-\sqrt{\;\frac{y}{4}}\right) d y

Area

=2 \int\limits_{0}^{2} \left(3\sqrt{y}-\;\frac{\sqrt{y}}{2}\right) d y

=2 \left[ \;\frac{2}{3} \times 3 \cdot y^{\;\frac{3}{2}} - \;\frac{1}{2} \times \;\frac{2}{3} \cdot y^{\;\frac{3}{2}} \right]_0^2

=2 \left[ 2 y^{\;\frac{3}{2}} - \;\frac{1}{3} y^{\;\frac{3}{2}} \right] =2 \times \;\frac{5}{3} y^{\;\frac{3}{2}} \left.\frac{}{}\right|_{0}^{2}

=2 \cdot \;\frac{5}{3} 2 \sqrt{2} = \;\frac{20 \sqrt{2}}{3}

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