The parabolas $y^2 = 4x$ and $x^2 = 4y$ divide the square region bounded by the lines $x = 4, y=4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom, then $S_1: S_2: S_3$, is

1 : 1 : 1

Clearly, $y^2 = 4x$ and $x^2 = 4y$ intersect at (4, 4).

$∴S_2+S_3=\int\limits_0^4\sqrt{4x}dx$

$⇒S_2+S_3=2\int\limits_0^4\sqrt{x}dx=\frac{4}{3}\left[x^{3/2}\right]_0^4=\frac{32}{3}$ sq. units

and, $S_3=\int\limits_0^4\frac{x^2}{4}dy=\frac{1}{4}\left[\frac{x^3}{3}\right]_0^4=\frac{16}{3}$ sq. units

Thus, we have

$S_2+S_3=\frac{32}{3}$ and $S_3=\frac{16}{3}⇒S_2=\frac{16}{3}$ sq. units.

$∵S_1+S_2+S_3$ = Area of square OAPB = $4 × 4 = 16$ sq. units.

$∴S_1=\frac{16}{3}$ sq. units.   $[∵S_2=S_3=\frac{16}{3} sq. units.]$

Hence, $S_1: S_2: S_3=1 : 1 : 1$