Question

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

Answer 1

The curves $y^2=4x$ and $x^2=4y$ intersect at the point $(4,4).$

Area of the square = $4\times 4=16$

Area between the second curve and the x-axis is ${\int }_0^4\frac{x^2}{4}dx={\left(\frac{x^3}{12}\right)}_0^4=\frac{16}{3}$

Area between the first curve and the y-axis is ${\int }_0^4\frac{y^2}{4}dy={\left(\frac{y^3}{12}\right)}_0^4=\frac{16}{3}$

Therefore, area between the curves must be $16-2\times \frac{16}{3}=\frac{16}{3}$.

Hence, the curves divide the square region into three equal parts.