Question

The area included between the parabolas y² = 4x and x² = 4y is (in square units)
(a) 4/3
(b) 1/3
(c) 16/3
(d) 8/3

Answer 1

(c) 16/3





$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x=\frac{y^2}{4}.....\left(1\right) \\ {x}^2=4y.....\left(2\right) \end{gathered}$$
Points of intersection of two parabola is given by,
$$\begin{gathered} {\left(\frac{y^2}{4}\right)}^2=4y \\ \Rightarrow y^4-64y=0 \\ \Rightarrow y\left(y^3-64\right)=0 \\ \Rightarrow y=0,4 \end{gathered}$$$\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,

$$\begin{gathered} ={\int }_0^4\left(y_1-y_2\right)\mathit{d}x\left(\mathrm{where},y_1=2\sqrt{x}\mathrm{and}{y}_2=\frac{x^2}{4}\right) \\ ={\int }_0^4\left(2\sqrt{x}-\frac{x^2}{4}\right)\mathit{d}x \\ ={\left[2\times \frac{2x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}-\frac{x^3}{12}\right]}_0^4 \\ =\left(2\times \frac{2{\left(4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}-\frac{{\left(4\right)}^3}{12}\right)-\left(2\times \frac{2{\left(0\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}-\frac{{\left(0\right)}^3}{12}\right) \\ =\left(\frac{32}{3}-\frac{16}{3}\right)-0 \\ =\frac{16}{3}\mathrm{square}\mathrm{units} \end{gathered}$$