The area between the parabolas x^2=y4 and x^2=9y, and the straight line y=2, is:

Question

The area between the parabolas $x^2=\dfrac{y}{4}$ and $x^2=9y$, and the straight line $y=2$, is:

Accepted Answer

Given curves $${x^2} = {y \over 4}$$ and ${x^2} = 9y$ are the parabolas  whose equations can be written as $y = 4{x^2}$ and $$y = {1 \over 9}{x^2}.$$

Also, given $y=2.$

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

$$\therefore$$ Area $$ = 2\int\limits_0^2 {\left( {\sqrt {9y}  - \sqrt {{y \over 4}} } \right)} dy$$

Area $$ = 2\int\limits_0^2 {\left( {3\sqrt y  - {{\sqrt y } \over 2}} \right)} dy$$

$$ = 2\left[ {{2 \over 3} \times 3.{y^{{3 \over 2}}} - {1 \over 2} \times {2 \over 3}.{y^{{3 \over 2}}}} \right]_0^2$$

$$ = 2\left[ {2{y^{{3 \over 2}}} - {1 \over 3}{y^{{3 \over 2}}}} \right] = \left. {2 \times {5 \over 3}{y^{{3 \over 2}}}} \right|_0^2$$

$$ = 2.{5 \over 3}2\sqrt 2  = {{20\sqrt 2 } \over 3}$$

Solutions · 1