The intersection points of the curves \( y^2 = 4x \) and \( x^2 = 4y \) are \( O(0,0) \) and \( A(4,4) \). Thus, the required area is: \[ \int_{0}^{4} (y_2 - y_1) \, dx \] Substituting \( y_2 = \sqrt{4x} \) from \( y^2 = 4x \) and \( y_1 = \frac{x^2}{4} \) from \( x^2 = 4y \): \[ = \int_{0}^{4} \left( \sqrt{4x} - \frac{x^2}{4} \right) dx \] \[ = \int_{0}^{4} \left( 2x^{\frac{1}{2}} - \frac{x^2}{4} \right) dx \] Evaluating the integral: \[ \left[ \frac{2x^{3/2}}{3/2} - \frac{x^3}{12} \right]_{0}^{4} \] \[ = \left[ \frac{4x^{3/2}}{3} - \frac{x^3}{12} \right]_{0}^{4} \] Substituting \( x = 4 \): \[ = \frac{4(4^{3/2})}{3} - \frac{4^3}{12} \] \[ = \frac{4 \times 8}{3} - \frac{64}{12} \] \[ = \frac{32}{3} - \frac{16}{3} \] \[ = \frac{16}{3} \text{ square units} \] Thus, the required area is: \[ \boxed{\frac{16}{3} \text{ square units}} \]