#### Find the area of the region bounded by the curve $y^2 = 4x, x^2 = 4y.$

The equations $y\textsuperscript{2 }= 4x$ is a parabola and no negative values of x are seen, therefore, this parabola lies to the right of the Y axis passing through (0, 0)

Similarly, for $x\textsuperscript{2} = 4y$ which is a parabola, not defined negative values of y lies above X axis and passing through (0, 0)

For finding point of interaction, solve simultaneously.
\begin{aligned} &\text { Put } y=\frac{x^{2}}{4} \text { in } y^{2}=4 x\\ &\Rightarrow\left(\frac{x^{2}}{4}\right)^{2}=4 x\\ &\Rightarrow \frac{x^{4}}{16}=4 x\\ &\Rightarrow x^{4}=64 x\\ &\Rightarrow x^{3}=64\\ &\Rightarrow x=4\\ &\text { Put } x=4 \text { in } y^{2}=4 x\\ &\Rightarrow y^{2}=4(4)\\ &\Rightarrow y=4 \end{aligned} \\

The point of interaction is therefore (4, 4)

For finding the area between two parabolas

Area between two parabolas = area under parabola $y\textsuperscript{2 }= 4x -\text{area under parabola } x\textsuperscript{2} = 4y \ldots . (1) \\$

Let us find area under parabola $y\textsuperscript{2}= 4x\\$

$= y = 2 \sqrt{x}$

Integrate from 0 to 4

Let us find area under parabola $\mathrm{y}^{2}=4 \mathrm{x}$

$\Rightarrow y=2 \sqrt{x}$

Integrate from 0 to 4

\Rightarrow \int_{0}^{4} \mathrm{ydx}=2 \int_{0}^{4} \sqrt{\mathrm{x}} \mathrm{dx} \\ \begin{aligned} &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2 \int_{0}^{4} \mathrm{x} \frac{1}{2} \mathrm{dx}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2\left[\frac{\mathrm{x} \frac{3}{2}}{\frac{3}{2}}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left[4^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left(2^{2}\right)^{\frac{3}{2}} \end{aligned} \\

$\\ \Rightarrow \int_{0}^{4} y d x=\frac{4}{3}(2)^{3} \\ \Rightarrow \int_{0}^{4} y d x=\frac{32}{3} \\ \\$

To find area under parabola $x\textsuperscript{2}= 4y\\$

$= x\textsuperscript{2}= 4 y\\$

$\\\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}(2)^{3}\\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{32}{3}\\ \text{Now let us find area under parabola} x^{2}=4 y\\ \Rightarrow x^{2}=4 y\\ \Rightarrow \mathrm{y}=\frac{\mathrm{x}^{2}}{4} \\ \text{Integrate from 0 to 4}\\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\int_{0}^{4} \frac{\mathrm{x}^{2}}{4} \mathrm{~d} \mathrm{x} \\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{4} \\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{1}{4}\left[\frac{4^{3}}{3}\right] \\ \Rightarrow \int_{0}^{4} y d x=\frac{4^{2}}{3} \\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{16}{3} \\ \text{Using (i)} \\ \Rightarrow \text{area bounded by two parabolas given} =\frac{32}{3}-\frac{16}{3} \\ \Rightarrow \text{area bounded by two parabolas given} =\frac{16}{3} \\$

Therefore, the area is found to be