Question

Prove that the curves $y^2=4x$ and $x^2=4y$ divide the area of the square bounded by $x=0,x=4,y=4$ and $y=0$ into three equal parts.

Answer 1

Given the curve $y=\cos x$ between $x=0$ and $x=2\pi$

We need to prove
Area OPQAO = Area OAQBO = Area OBQRO
Area OPQAO
$={\int }_0^{4}ydx$

$={\int }_0^4\frac{x^2}{4}dx=\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^4$

$=\frac{1}{12}\times [4^3-0^3]=\frac{1}{12}\times [64-0]$

$=\frac{64}{12}=\frac{16}{3}$

Area OBQRO= ${\int }_0^{4}xdy$

$={\int }_0^4\frac{y^2}{4}dy$

$=\frac{1}{4}{\left[\frac{y^3}{3}\right]}_0^4$

$=\frac{1}{12}\times [4^3-0^3]=\frac{1}{12}\times 64-0$

$=\frac{64}{12}=\frac{16}{3}$

Area OAQBO = Area OBQPO – Area OAQPO

$={\int }_0^4{y}_{1}dx-={\int }_0^4{y}_{2}dx$

Where $y_1=\sqrt{4x}\&y_2=\frac{x^2}{4}$

$={\int }_0^4\sqrt{4x}dx-={\int }_0^4\frac{x^2}{4}dx$

$=2{\int }_0^4\sqrt{x}‘dx-\frac{16}{3}$

$=2\times {\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^4-\frac{16}{3}$

$=2\times {\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^4-\frac{16}{3}$

$=\frac{4}{3}\left[\right(4{)}^{\frac{3}{2}}-\left(0{)}^{\frac{3}{2}}\right]-\frac{16}{3}$

$=\frac{4}{3}8-0-\frac{16}{3}$

$=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$

So, from step $2,3,4,$ we proved
Area OPQAO = Area OAQBO = Area OBQRO $=\frac{16}{3}$ square units