The point of intersection of the Parabolas y² = 4x and x² = 4y are (0, 0) and (4, 4) Now, the area of the region OAQBO bounded by curves $y^2$ = 4x and $x^2$ = 4y $\int\limits_{0}^{4} \left(2\sqrt{x} \cdot \frac{x^2}{4}\right)$ dx = $\left[\frac{2x^{3/2}}{\frac{3}{2}} - \frac{x^3}{12}\right]_{0}^{4}$ = $\frac{32}{3} - \frac{16}{3}$ = $\frac{16}{3}$ sq units ………..(i) Again, the area of the region OPQAO bounded by the curves $x^2$ = 4y, x = 0, x = 4 and the x-axis, $\int\limits_{0}^{4} \frac{x^2}{4}$ dx = $\left[\frac{x^3}{12}\right]_{0}^{4}$ = $\left(\frac{64}{12}\right)$ = $\frac{16}{3}$ sq units ……….(ii) Similarly, the area of the region OBQRO bounded by the curve $y^2$ = 4x, the y-axis, y = 0 and y = 4 $\int\limits_{0}^{4} \frac{y^2}{4}$ dy = $\left[\frac{y^3}{12}\right]_{0}^{4}$ = $\frac{16}{3}$ sq units ... (iii) From (i), (ii), and (iii) it is concluded that the area of the region OAQBO = area of the region OPQAO = area of the region OBQRO, i.e., area bounded by parabolas $y^2$ = 4x and $x^2$ = 4y divides the area of the square into three equal parts.