Question

Find the area bounded by the following curve.
${y}^{2} = ax \: \: \: \: \: and \\ {x}^{2} + {y}^{2} = 4ax$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given curves are

$\rm :\longmapsto\: {y}^{2} = ax - - - (1)$

and

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 4ax - - - (2)$

Step :- 1 Point of intersection of curves

Substituting (1) in (2), we get

$\rm :\longmapsto\: {x}^{2} + ax = 4ax$

$\rm :\longmapsto\: {x}^{2} + ax - 4ax = 0$

$\rm :\longmapsto\: {x}^{2}- 3ax = 0$

$\rm :\longmapsto\:x( x- 3a) = 0$

$\bf\implies \:x = 0 \: \: or \: \: x = 3a$

➢ Pair of points of intersection of given curves are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 3a & \sf \sqrt{3}a \\ \\ \sf 3a & \sf - \sqrt{3}a \end{array}} \\ \end{gathered}$

Step :- 2 Curve Sketching

$\rm :\longmapsto\: {y}^{2} = ax$

represents the right handed parabola with vertex (0, 0).

And

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 4ax$

$\rm :\longmapsto\: {x}^{2} - 4ax+ {y}^{2} = 0$

$\rm :\longmapsto\: {x}^{2} - 4ax + {4a}^{2} + {y}^{2} = 4 {a}^{2}$

$\rm :\longmapsto\: {(x - 2a)}^{2} + {y}^{2} = {(2a)}^{2}$

So, it represents a circle whose centre is (2a, 0) and radius = 2a.

Step :- 3 Required Area is

$\rm \:  =  \:  \:2\displaystyle\int_0^{3a}\sf (y_{circle} - y_{parabola})dx$

$\rm \: = \:2\displaystyle\int_0^{3a}\sf \bigg( \sqrt{ {(2a)}^{2} - {(x - 2a)}^{2} } - \sqrt{a} \sqrt{x}\bigg)dx$

$\rm \:=2\bigg(\dfrac{x - 2a}{2} \sqrt{ {(2a)}^{2} - {(x - 2a)}^{2}} + \dfrac{ {4a}^{2} }{2} {sin}^{ - 1}\dfrac{x - 2a}{2a} - \dfrac{2}{3} \sqrt{a} {x}^{ \frac{3}{2} } \bigg)_0^{3a}$

$\rm \:  =2\bigg(\dfrac{a}{2} \sqrt{3}a + {2a}^{2} {sin}^{ - 1}\dfrac{1}{2} - \dfrac{2 \sqrt{a} }{3}3 \sqrt{3}a \sqrt{a} - {2a}^{2} {sin}^{ - 1}( - 1)\bigg)$

$\rm \:  =2\bigg(\dfrac{ \sqrt{3} {a}^{2} }{2}+{2a}^{2}\dfrac{\pi}{6} - 2 \sqrt{3} {a}^{2} + {2a}^{2} \dfrac{\pi}{2} \bigg)$

$\rm \:  =2\bigg(\dfrac{ \sqrt{3} {a}^{2} }{2}+{a}^{2}\dfrac{\pi}{3} - 2 \sqrt{3} {a}^{2} + {a}^{2}\pi \bigg)$

$\rm \:  =2 {a}^{2} \bigg(\dfrac{ \sqrt{3} }{2}+\dfrac{\pi}{3} - 2 \sqrt{3} + \pi \bigg)$

$\rm \:  =2 {a}^{2} \bigg( - \dfrac{ 3\sqrt{3} }{2}+\dfrac{4\pi}{3} \bigg) \: square \: units$