Find the area of the region lying between the parabolas \(y^2 = 4x\) and \(x^2 = 4y\).

Answer: \(\frac{16}{3}\) sq. units

Points of intersection: from \(x^2 = 4y\), \(y = \frac{x^2}{4}\). Substituting in \(y^2 = 4x\): \(\frac{x^4}{16} = 4x \Rightarrow x(x^3 - 64) = 0 \Rightarrow x = 0\) or \(x = 4\). The points are O(0, 0) and A(4, 4).

[Figure: the parabola \(y^2 = 4x\) opening to the right and \(x^2 = 4y\) opening upwards meet at O(0, 0) and A(4, 4); the required region lies between them, with \(y^2 = 4x\) as the upper curve for \(0 \lt x \lt 4\).]

For \(0 \le x \le 4\), the upper curve is \(y = 2\sqrt{x}\) and the lower curve is \(y = \frac{x^2}{4}\).

Area \(= \int_0^4 \left(2\sqrt{x} - \frac{x^2}{4}\right)dx = \left[\frac{4}{3}x^{3/2} - \frac{x^3}{12}\right]_0^4 = \frac{4}{3}(8) - \frac{64}{12} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}\) sq. units.