#### Find the area of the region bounded by the parabola $y^2 = 2px, x^2 = 2py$

The equation of the parabola is $y\textsuperscript{2 }= 2px$ which shows no negative values for x. Therefore, the graph will be on right of Y-axis will be passing through (0, 0)

Similarly, for equations $x\textsuperscript{2} = 2py$ no negative values for y, therefore, the graph will be above the X axis passing through (0, 0).

For finding point of interaction, let us solve the equations simultaneously.

$\\ \text { Put } y=\frac{x^{2}}{2 p} \text { in } y^{2}=2 p x \\ \Rightarrow\left(\frac{x^{2}}{2 p}\right)^{2}=2 p x \\ \Rightarrow \frac{x^{4}}{4 p^{2}}=2 p x \\ \Rightarrow x^{4}=8 p^{3} x \\ \Rightarrow x^{3}=8 p^{3} \\$

$\\ \Rightarrow x=2 p \\ \text { Put } x=2 p \text { in } y^{2}=2 p x \\ \Rightarrow y^{2}=2 p(2 p) \\ \Rightarrow y=2 p$

The interaction point was found to be (2p, 2p)

Now, we have to find out the areas between the two parabolas.

The equation can be written as

Area between the two parabola = area under parabola

$y\textsuperscript{2 }= 2px - \text{area under the parabola } x\textsuperscript{2 }= 2py \ldots (1) \\$

For finding the area under$y\textsuperscript{2 }= 2px\: \: parabola \\$

$\item y = \sqrt 2p \sqrt x \\$

Integrating the equation from 0 to 2p

$\\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} \sqrt{x} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \frac{2}{3}\left[(2 p)^{\frac{3}{2}}-0\right]$

$\\\Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}}(2 p)^{\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}+\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{2} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{8 p^{2}}{3}$

For finding the area under $x\textsuperscript{2 }= 2py \text{ parabola} \\$

$\item y = x\textsuperscript{2}/ 2p \\$

Integrating from 0 to 2p

$\\\Rightarrow \int_{0}^{2 p} y d x=\int_{0}^{2 p} \frac{x^{2}}{2 p} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{x^{3}}{3}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{(2 p)^{3}}{3}\right] \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{(2 p)^{2}}{3} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{4 p^{2}}{3}$

Using
(i)

$\\\Rightarrow area bounded by two parabolas given =\frac{8 \mathrm{p}^{2}}{3}-\frac{4 \mathrm{p}^{2}}{3} \\\Rightarrow area bounded by two parabolas given =\frac{4 \mathrm{p}^{2}}{3}\\ Hence area is \frac{4 \mathrm{p}^{2}}{3} unit { }^{2}$