Question

Find the area bounded between the curves

${x}^{2} = y \: \: and \: \: {y}^{2} = x$
​

Answer 1

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

Given curves are

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

and

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

Let's first find the point of intersection of two curves.

Substituting the value of y from equation (2) to equation (1), we get

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

$\rm :\longmapsto\: {x}^{4} = x$

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

$\rm :\longmapsto\: x({x}^{3} - 1) = 0$

$\bf\implies \:x = 0 \: \: or \: \: x = 1$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf 1 \end{array}} \\ \end{gathered}$

Curve Sketching :-

➢ y² = x represented the right handed parabola having vertex at the origin and axis along x - axis.

and

➢ x² = y represented the upper parabola having vertex at the origin and axis along y - axis.

[ See the attachment ]

Required Area

So, required area bounded between the two curves with respect to x - axis is given by

$\purple{\rm \:  =  \:\displaystyle\int_{0}^{1}\rm (y_1 - y_2) \: dx}$

$\purple{\rm \:  =  \:\displaystyle\int_{0}^{1}\rm ( \sqrt{x} - {x}^{2} ) \: dx}$

$\rm \:  =  \: \dfrac{ {\bigg(x\bigg) }^{\dfrac{1}{2} + 1} }{\dfrac{1}{2} + 1} - \dfrac{ {x}^{2 + 1} }{2 + 1} \bigg]_{0}^{1}$

$\rm \:  =  \: \dfrac{ {\bigg(x\bigg) }^{\dfrac{3}{2}} }{\dfrac{3}{2}} - \dfrac{ {x}^{3} }{3} \bigg]_{0}^{1}$

$\rm \:  =  \: \dfrac{2}{3} - \dfrac{1}{3} - 0$

$\rm \:  =  \: \dfrac{2 - 1}{3}$

$\rm \:  =  \: \dfrac{1}{3}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Question:-

Find the area bounded between the curves x² = y and y² = x.

Given:-

• The curves are x² = y and y² = x.

To Find:-

• The area bounded between the curves.

Solution:-

Curves are x² = y and y² = x. i.e. y = √x.

Clearly the given curves represents parabola in standard form.

The rough sketch of the parabolas is shown in figure:

For points of intersection y² = x.

$\sf \large ⇒y = \sqrt{x}$

$\sf \large \therefore \: x {}^{2} = \sqrt{x}$

$\sf \large ⇒x {}^{4} = x$

$\sf \large ⇒x(x {}^{3} - 1) = 0$

$\large \red {⇒} \sf \large { \boxed { \sf \red{x = 0,</p><p>1}}}$

$\large\red{⇒} \sf \large { \boxed{ \sf \red {y = 0,</p><p>1 }}}$

Point of intersection of these two parabolas are O(0,0) and A (1,1).

Now,

$\sf \large y {}^{2} = x \: or \: y = \sqrt{x } = f(x) \: (say) \\ \sf \large y = x {}^{2} = g(x) \: (say)$

$\sf \large where \: f(x) > g(x) \: in \: [0,1]$

Required area of shaded region:

$\sf \large \int ^1 _{0} \bigg[f(x) - g(x)dx = \int ^1_{0} \bigg( \sqrt{x} - x {}^{2} \bigg) dx$

$\sf \large = \bigg[ \frac{2}{3} \: x {}^{3/2 } - \frac{x {}^{3} }{3} \bigg]^1_</p><p>{0}$

$\sf \large = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \: sq.units$

Answer:-

$\sf \blue{ \therefore \: The \: Area \: of \: bounded \: between \: the} \\ \sf \blue{ \sf curves \: x² = y and \: y² = x = \frac{1}{3} \: sq.units. }$

[Refer to the above graph given]

Hope you have satisfied. ⚘