Question

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

Answer 1

We have,

$y^2=4x......\left(1\right)$

$x^2=4y......\left(2\right)$

According to figure,

Prove that:- $AreaOPQA=AreaOAQB=AreaOBQR$

Proof:-

So,

Area of $OPQA$ $={\int }_0^{4}ydx$

Here,

$4y=x^2$

$\Rightarrow y=\frac{x^2}{4}$

So,

Area of $OPQA={\int }_0^4\frac{x^2}{4}dx$

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

$=\frac{1}{12}\times (4^3-0^3)$

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

$=\frac{1}{12}\times 64$

$=\frac{16}{3}$

Area of $OBQR$

Since, Area is on Y-axis,

We can use formula,

$\int xdy$

So, Area of

$OBQR={\int }_0^{4}xdy$

$Here,y^2=4x$

$\Rightarrow x=\frac{y^2}{4}$

So, Area of $OBQR$ $={\int }_0^4\frac{y^2}{4}dy$

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

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

$=\frac{1}{12}\times 64$

$=\frac{16}{3}$

Area of $OBQP={\int }_0^{4}ydx$

Here,

$y^2=4x$

$y=\pm \sqrt{4x}$

As OBQP is in Ist quadrant.

We can use positive

$y=\sqrt{4x}$

So,

Area of $OBQP$ $={\int }_0^4\sqrt{4x}dx$

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

$=2{\int }_0^4{x}^{\frac{1}{2}}dx$

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

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

$=\frac{4}{3}[4^{\frac{3}{2}}-0^{\frac{3}{2}}]$

$=\frac{4}{3}\times 8$

$=\frac{32}{3}$

Now,

Area of $OAQP={\int }_0^{4}ydx$

Here

$x^2=4y$

$y=\frac{x^2}{4}$

Area of $OAQP$ $={\int }_0^4\frac{x^2}{4}dx$

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

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

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

$=\frac{1}{12}\times 64$

$=\frac{16}{3}$

So,

$AreaOPQA=AreaOAQB=AreaOBQR$

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

Hence, proved.