Question

Draw a rough sketch and find the area of the region bounded by the two parabolas y² = 4x and x² = 4y by using methods of integration.

Answer 1

To find the points of intersection between two parabola let us substitute $x=\frac{y^2}{4}$ in $x^2=4y$.

$$\begin{gathered} {\left(\frac{y^2}{4}\right)}^2=4y \\ \Rightarrow y^4-64y=0 \\ \Rightarrow y\left(y^3-64\right)=0 \\ \Rightarrow y=0,4 \end{gathered}$$
$\Rightarrow x=0,4$

Therefore, the points of intersection are $A(4,4)$ and $C\left(0,0\right)$.

Therefore, the area of the required region ABCD =${\int }_0^4{y}_{1}dx-{\int }_0^4{y}_{2}dx$ where $y_1=2\sqrt{x}$ and $y_2=\frac{x^2}{4}$

Required Area
$$\begin{gathered} ={\int }_0^4\left(2\sqrt{x}\right)dx-{\int }_0^4\left(\frac{x^2}{4}\right)dx \\ ={\left[2\times \frac{2x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}-\frac{x^3}{12}\right]}_0^4 \\ =\left[\left(2\times \frac{2{\left(4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}-\frac{{\left(4\right)}^3}{12}\right)-\left(2\times \frac{2{\left(0\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}-\frac{{\left(0\right)}^3}{12}\right)\right] \end{gathered}$$

After simplifying we get,

$=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$ square units