Question

Prove that line $x=y$ divides the intersecting area of the curves $y^2=4x$ and $x^2=4y$ into two equal parts.

Answer 1

$y^2=4x...\left(1\right),x^2=4y...\left(2\right),x=y...\left(3\right)$
Point of intersection of these curves is : $(4,4)$

Now, we have to prove that :
$\text{Area of}OACO=\text{Area of}OBCO$

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

$\text{Area of}OACO={\int }_0^4\sqrt{4x}dx-{\int }_0^{4}xdx={\left[\frac{4}{3}{x}^{3/2}\right]}_0^4-{\left[\frac{x^2}{2}\right]}_0^4=\frac{32}{3}-8=\frac{8}{3}\text{sq. units}$

$\text{Area of}OBCO={\int }_0^{4}xdx-{\int }_0^4\frac{x^2}{4}dx={\left[\frac{x^2}{2}\right]}_0^4-{\left[\frac{x^3}{12}\right]}_0^4=8-\frac{16}{3}=\frac{8}{3}\text{sq. units}$

$\Rightarrow \text{Area of}OACO=\text{Area of}OBCO$
Hence, it proves that line $y=x$ divides the intersecting area of the curves $y^2=4x$ and $x^2=4y$ into two equal parts.