The area included between the parabolas y2 = 4x and x2 = 4y is (in square units)

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$\frac{16}{3}$ We have, x = y^$\frac{2}{4}$ . . . . . \left( 1 \right) x^2 = 4y . . . . . \left( 2 \right) Points of intersection of two parabola is given by, \left( y^$\frac{2}{4}$ \right)^2 = 4y \Rightarrow y^4 - 64y = 0 \Rightarrow y\left( y^3 - 64 \right) = 0 \Rightarrow y = 0, 4 \Rightarrow x = 0, 4 Therefore, the points of intersection are A(0, 0) and C(4, 4). Therefore, the area of the required region ABCD, = \int_0^4 \left( y_1 - y_2 \right) d x ..............\left(\text{where, }y_1 = 2$\sqrt$(x)\text{ and }y_2 = x^$\frac{2}{4}$ \right) = \int_0^4 \left( 2$\sqrt$(x) - x^$\frac{2}{4}$ \right) d x = \left[ 2 $\times$ 2 x^3/2/3 - x^$\frac{3}{12}$ \right]_0^4 = \left( 2 $\times$ \frac{2 \left( 4 \right)^$\frac{3}{2}$}{3} - \frac{\left( 4 \right)^3}{12} \right) - \left( 2 $\times$ \frac{2 \left( 0 \right)^$\frac{3}{2}$}{3} - \frac{\left( 0 \right)^3}{12} \right) = \left( $\frac{32}{3}$ - $\frac{16}{3}$ \right) - 0 = $\frac{16}{3}$\text{ square units } [Solution diagram: 43925_solution_1.png]

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