• # question_answer The area enclosed between the parabolas ${{y}^{2}}=4x$ and ${{x}^{2}}=4y$ is              [Karnataka CET 1999, 2003] A)            $\frac{14}{3}$ sq. unit    B)            $\frac{3}{4}$ sq. unit C)            $\frac{3}{16}$ sq. unit    D)            $\frac{16}{3}$ sq. unit

Equations of curves ${{y}^{2}}=4x$ and ${{x}^{2}}=4y.$The given equations may be written as $y=2\sqrt{x}$ and $y=\frac{{{x}^{2}}}{4}.$            We know that area enclosed by the parabolas $=\int_{\,0}^{\,4}{2\sqrt{x}\,dx-}\int_{\,0}^{\,4}{\frac{{{x}^{2}}}{4}dx=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}}$sq. unit.