Question

find the area of the region lying between the parabolas y^2=2x and x^2=2y​

Answer 1

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

Given curves are

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

and

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

Let first the point of intersection of two given curves

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

$\rm :\implies\:x = \dfrac{ {y}^{2} }{2}$

Substituting this value in equation (2), we get

$\rm :\longmapsto\:\dfrac{ {y}^{4} }{4} = 2y$

$\rm :\longmapsto\: {y}^{4} = 8y$

$\rm :\longmapsto\: {y}^{4} - 8y = 0$

$\rm :\longmapsto\:y( {y}^{3} - 8) = 0$

$\bf\implies \:y = 0 \: \: \: \: or \: \: \: \: y = 2$

Hence, points of intersection are

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 2 & \sf 2 \end{array}} \\ \end{gathered}$

Now, from given curves,

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

$\bf\implies \:y = \sqrt{2} \: \sqrt{x}$

Let assume that,

$\bf\implies \:y_1 = \sqrt{2} \: \sqrt{x}$

Also,

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

$\bf\implies \:\:y = \dfrac{ {x}^{2} }{2}$

Let assume that

$\bf\implies \:\:y_2= \dfrac{ {x}^{2} }{2}$

Now, See the graph in attachment, so we concluded from graph that

The required area is

$\rm \:  =  \:  \: \displaystyle\int_0^{2}\sf y_1 \: dx \: - \: \displaystyle\int_0^{2}\sf y_2 \: dx$

$\rm \:  =  \:  \: \displaystyle\int_0^{2}\sf \sqrt{2} \: \sqrt{x} \: dx \: - \: \displaystyle\int_0^{2}\sf \dfrac{ {x}^{2} }{2} \: dx$

We know ,

$\boxed{ \bf{ \: \displaystyle\int\sf {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c}}$

So,

$\boxed{ \bf{ \: \displaystyle\int\sf {x}^{2} \: dx \: = \: \frac{ {x}^{3} }{3} + c}}$

and

$\boxed{ \bf{ \: \displaystyle\int\sf \sqrt{x} \: dx \: = \: \frac{ { 2\bigg(x \bigg)}^{ \dfrac{3}{2} } }{3} + c}}$

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

$\rm \:  =  \:  \: \dfrac{2 \sqrt{2} }{3} {\bigg( 2\bigg) }^{\dfrac{3}{2} } - \dfrac{1}{6}( {2}^{3})$

$\rm \:  =  \:  \: \dfrac{2 \sqrt{2} }{3} {\bigg( {( \sqrt{2} )}^{2} \bigg) }^{\dfrac{3}{2} } - \dfrac{1}{6}(8)$

$\rm \:  =  \:  \: \dfrac{2 \sqrt{2} }{3} {\bigg( {( \sqrt{2} )}^{3} \bigg) } - \dfrac{4}{3}$

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

$\rm \:  =  \:  \: \dfrac{8}{3} - \dfrac{4}{3}$

$\rm \:  =  \:  \: \dfrac{4}{3} \: square \: units$

Answer 2

Step-by-step explanation:

Given,

$\sf \: {y}^{2} = 2x \: \: - - - (1) \: \: \: \: and \: \: \: \: \: {x}^{2} = 2y \\ \\ \bf \: let \: \\ \sf \: y_1 = {y}^{2} = 2x \implies \: y_1 = y = \sqrt{2x} \: \: \: \: \: \: and \: \\ \sf \: y_2 = {x}^{2} = 2y \implies \: y_2 = y = \frac{ {x}^{2} }{2} \\ \\ \bf now \\ \sf \: \: \: \: \: \: \: \: \: {y}^{2} = 2x \\ \sf \implies \: ( { \frac{ {x}^{2} }{2} })^{2} = 2x \\ \sf \implies \: \frac{ {x}^{4} }{4} = 2x \\ \sf \implies \: {x}^{3} = 8 \\ \sf \implies \: x = 2 \\ \small{ \pink{ \bf \: putting \: the \: value \: of \: x = 2 \: in \: eqn.(1)}} \\ \\ \sf \: {y}^{2} = 2 \times 2 = 4 \\ \sf \implies \: y = 2 \\ \\ \bf \: both \: parabola \:are\: passing \: through \\ \bf \: center \\ \\ \huge{ \bf \: now \: area} \\ \\ \displaystyle\int_0^{2}\sf (y_1 - y_2) \: dx \\ \\ = \displaystyle\int_0^{2}\sf( \sqrt{2x} \: - \frac{ {x}^{2} }{2} )dx \\ \\ ={ \huge{ [ }}\sqrt{2} \times \frac{ {x}^{ \frac{3}{2} } }{ \frac{3}{2} } - \frac{ {x}^{3} }{2 \times 3} { \huge{ ]}} _0^2 \\ \\ = { \huge{ [ }} 2 \sqrt{2} \times \frac{ {x}^{ \frac{3}{2} } }{3} - \frac{ {x}^{3} }{6} { \huge{ ]}} _0^2 \\ \\ = 2 \sqrt{2} \times \frac{ {2}^{ \frac{3}{2} } }{3} - \frac{ {2}^{3} }{6} - 0 - 0 \\ \\ = \frac{4}{3} \: \sf{unit}^{2}$

Short technique :

$\sf \: if \: {y}^{2} = 4ax \: \: \: \: and \: \: \: {x}^{2} = 4ay \: \: \: are \: two \: \\ \sf \: prabola \: then \: the \: area \: between \: them \: is \\ \\ \bf = \frac{16 {a}^{2} }{ 3} \: {unit}^{2}$