Question

Find the common area enclosed by the parabolas $y^2=x$ and $x^2=y$

Answer 1

Area required$=$Area $OABC$

To find point of intersection $B$:

Solving $y^2=x$ ....$\left(1\right)$

and $x^2=y$ ....$\left(2\right)$

Put $\left(2\right)$ in $\left(1\right)$

$y^2=x$

${\left(x^2\right)}^2=x$

$x^4-x=0$

$x(x^3-1)=0$

$\Rightarrow x=0,x=1$

For $x=0,y=x^2=0$.So, coordinates are $(0,0)$

For $x=1,y=1^2=1$.So, coordinates are $(1,1)$

Since point $B$ lies in first quadrant.

So, coordiantes of $B$ is $(1,1)$

Let us find the area of $OABC$

Area$OABC=$Area$OABD-$Area$OCBD$

${\int }_0^1{y}_{1}dx-{\int }_0^1{y}_{2}dx$

$y^2=x\Rightarrow y=\pm \sqrt{x}$

Since $OABD$ is in first quadrant $y_1=\sqrt{x}$

and $x^2=y\Rightarrow y_2=x^2$

Area Required$={\int }_0^1{y}_{1}dx-{\int }_0^1{y}_{2}dx$

$={\int }_0^1\sqrt{x}dx-{\int }_0^1{x}^{2}dx$

$={\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^1-{\left[\frac{x^{2+1}}{2+1}\right]}_0^1$

$={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1$

$=\frac{2}{3}[1-0]-\frac{1}{3}[1-0]$

$=\frac{1}{3}$

Hence, Area requires$=\frac{1}{3}$ sq.units.