The parabolas${\mathrm{y}}^2 = 4\mathrm{x}, {\mathrm{x}}^2 = 4\mathrm{y}$ divide the square region bounded by the lines x = 4, y = 4 and the co-ordinate axes. If ${\mathrm{S}}_1, {\mathrm{S}}_2, {\mathrm{S}}_3$ are respectively the areas of these parts numbered from top to bottom then ${\mathrm{S}}_1: {\mathrm{S}}_2: {\mathrm{S}}_3$ is

$\text{ For point of intersection of the given parabolas }{x}^2=4y\text{ and }{y}^2=4x\text{, we have }$

$x^4=16y^2\Rightarrow x^4=16.4x$

$\Rightarrow x\left(x^3-64\right)=0\Rightarrow x=0\text{ or }x=4$

$\therefore \text{ Points of intersection are }O(0,0)\text{ and }B(4,4)$

$\text{ Area of }{S}_3={\int }_0^4 \frac{x^2}{4}dx={\left[\frac{x^3}{12}\right]}_0^4=\frac{64}{12}=\frac{16}{3}\text{ square units. }$

$\text{ Area of }\begin{array}{r}{S}_2={\int }_0^4 2\sqrt{x}dx-\frac{16}{3}=\frac{4}{3}{\left[x^{3/2}\right]}_0^4-\frac{16}{3}=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}\text{ square units. }\end{array}$

$\text{ Area of }{S}_1=\text{ Area of square - Area of }{S}_1-\text{ Area of }{S}_2$$=16-\frac{16}{3}-\frac{16}{3}=\frac{16}{3}\text{ square units. }$

$S_1:S_2:S_3=1:1:1$