Question

The Area bounded between the parabola x²=y/4 and x²=9y, y =2

a) 20√2 b) 10√2 c)20√2/3 d)10√2​

Answer 1

We have to find out area bounded between x²=y/4 , x²=9y and , y =2 .

So , first we will draw rough graph of parabola [refer attached picture ]

Now, we have to find out area of shaded portion which is bounded by both the parabolas and the line.

$\rm \rightarrow {x}^{2} = 9y \\ \\ \rm \rightarrow {x} = \sqrt{9y}\\ \\ \rm \rightarrow {x} = 3 \: \sqrt{y}$

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$\rm \rightarrow {x}^{2} = \dfrac{y}{4} \\ \\ \rm \rightarrow {x}= \sqrt{\dfrac{y}{4} } \\ \\ \rm \rightarrow {x}= {\dfrac{\sqrt{y}}{2} }$

$\rm \longrightarrow \displaystyle \rm Required \: Area \: (A) = 2\int_{0}^{2} \: (3 \sqrt{y} - \frac{ \sqrt{y} }{2} ) \rm dy$

$\rm \longrightarrow \displaystyle\rm A = 2\int_{0}^{2} \: (3 \sqrt{y} - \frac{ \sqrt{y} }{2} ) \rm dy$

$\rm \longrightarrow \displaystyle\rm A = 2 \bigg[(2 \times {y}^{ \frac{3}{2} }) - \dfrac{ 1 }{3}( {y})^{ \frac{3}{2} } \bigg]_{0}^{2}$

$\rm \longrightarrow \displaystyle\rm A = 2 (2 \times {2}^{ \frac{3}{2} }) - \dfrac{ 2}{3}( {2})^{ \frac{3}{2} } - (0 - 0)$

$\rm \longrightarrow \displaystyle\rm A = {(2)}^{ \frac{7}{2} } - \dfrac{ 1}{3}( {2})^{ \frac{5}{2} }$

$\rm \longrightarrow \displaystyle\rm A = {(2)}^{ \frac{5}{2} } {(2)}^{ \frac{2}{2} }- \dfrac{ 1}{3}( {2})^{ \frac{5}{2} }$

$\rm \longrightarrow \displaystyle\rm A = {(2)}^{ \frac{5}{2} } ( 2- \dfrac{ 1}{3})$

$\rm \longrightarrow \displaystyle\rm A = \sqrt{ {2}^{5} } \times ( \dfrac{ 6 - 1}{3})$

$\rm \longrightarrow \displaystyle\rm A = 4 \sqrt{ {2} } \: \times ( \dfrac{ 5}{3})$

$\rm \longrightarrow \displaystyle\rm A = \frac{20}{3} \sqrt{ {2} } \: \: \: units$

Therefore , option c)20√2/3 is correct.

Answer 2

To find :-

• The Area bounded between the parabola x² = y/4 and x² = 9y, y = 2.

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$\huge {AηsωeR} ✍$

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$\large\underline\blue{\bold{Formula \: used:- }}$

To find the area bounded by curve with respect to y axis between the lines y = a and y = b is given by

$\bf \:  ⟼ \int_{a}^{b} xdy$

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$\underline{\boxed{\star{\sf{\blue{For \: the \: first \: curve : {x}^{2} = \dfrac{y}{4} }}}}}$

Its a upper parabola having vertex at (0,0) and

$\sf \:  ⟼x = \dfrac{ \sqrt{y} }{2}$

So, the area bounded by the curve x² = y/4 between the lines y = 0 and y = 2 and x = 0 is evaluated as

$\sf \:  Area_1 = \int_{0}^{2} xdy$

$\sf \:  = \int_{0}^{2} \dfrac{ \sqrt{y} }{2}dy$

$\sf \:  = \dfrac{1}{2} \int_{0}^{2} \sqrt{y} dy$

$\sf \:  = \dfrac{1}{2} \int_{0}^{2} {y}^{ \frac{1}{2} } dy$

$\sf \:  = \dfrac{1}{2} \times \dfrac{2}{3} [ {y}^{ \frac{3}{2} } ]_0^2$

$\sf \:  = \dfrac{1}{3} {(2)}^{ \frac{3}{2} }$

$\sf \:  = \dfrac{1}{3} {( \sqrt{2} )}^{2 \times \frac{3}{2} } = \dfrac{2}{3} \sqrt{2} \: sq. \: units$

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$\underline{\boxed{\star{\sf{\blue{For \: the \: second \: curve : {x}^{2} = 9y}}}}}$

Its an upper parabola with vertex at (0,0) and

$\sf \:  ⟼x = 3 \sqrt{y}$

So, the area bounded by the curve x² = y/4 between the lines y = 0 and y = 2 and x = 0 is evaluated as

$\sf \:  Area_2 = \int_{0}^{2} xdy$

$\sf \:  = \int_{0}^{2} 3 \sqrt{y} dy$

$\sf \:  = 3\int_{0}^{2} \sqrt{y} dy$

$\sf \:  = 3 \int_{0}^{2} {y}^{ \frac{1}{2} } dy$

$\sf \:  = 3 \int_{0}^{2} {y}^{ \frac{1}{2} } dy$

$\sf \:  =3 \times \dfrac{2}{3} [ {y}^{ \frac{3}{2} } ]_0^2$

$\sf \:  = 2 \times {(2)}^{ \frac{3}{2} } = 2 \times 2 \times \sqrt{2} = 4 \sqrt{2} \: sq. \: units$

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$\bf\implies \:Required \: area = 2(Area_2 - Area_1)$

$\sf \:  =2( 4 \sqrt{2} - \dfrac{2}{3} \sqrt{2} )$

$= 2(\dfrac{12 \sqrt{2} - 2 \sqrt{2} }{3} )$

$\sf \:  = 2 \times \dfrac{10 \sqrt{2} }{3}$

$\sf \:  = \dfrac{20 \sqrt{2} }{3} \: sq. \: units$

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$\large{\boxed{\boxed{\bf{Option (c) \: is \: correct}}}}$

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