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

Toolbox:
• Let $f(x)$ be a continuous function defined on $[a,b]$.Then the area bounded by the curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $x=b$ is given by
• $\int_a^bf(x)dx$ or $\int_a^b ydx.$
Step 1:
$y^2=2px$ and $x^2=2py$
Clearly $y^2=2px$ is a parabola which is open rightward.
Similarly $x^2=2py$ is a parabola which is open upward.
The area of the required is the shaded portion shown in the fig.
$x^2=2py$
$\Rightarrow x^4=4p^2y^2$
$\Rightarrow y^2=\large\frac{x^4}{4p^2}$
$2px=\large\frac{x^4}{4p^2}$
$\Rightarrow x^3=8p^3$
$x=2p$
Step 2:
Clearly the point of intersection is the focus of the two curves.
The points of intersection are $(2p,0)$
Hence the required area is $A=\int_0^{2p}(y_2-y_1)dx.$
Where $y_1=\large\frac{\sqrt x^2}{2p}$ and $y_2=\sqrt{2px}$
The required area=$\int_0^{2p}(\large\frac{x^2}{2p}$$-\sqrt{2px})dx. \qquad\qquad\;\;\;\;=\int_0^{2p}\large\frac{1}{2p}$$x^2-\sqrt{2p}\int_0^{2p}\sqrt {x}dx.$
Step 3:
On integrating we get,
$\;\;\;=\large\frac{1}{2p}\begin{bmatrix}\large\frac{x^3}{3}\end{bmatrix}_0^{2p}$$-\sqrt{2p}\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_0^{2p} \;\;\;=\large\frac{1}{6p}\begin{bmatrix}x^3\end{bmatrix}_0^{2p}$$-\large\frac{2\sqrt{2p}}{3}\begin{bmatrix}x^{\Large\frac{3}{2}}\end{bmatrix}_0^{2p}$
On applying limits we get,
$\;\;\;=\large\frac{1}{6p}$$[(2p)^3-0]-\large\frac{2\sqrt{2p}}{3}$$[(2p)^{\Large\frac{3}{2}}-\normalsize 0]$
$\;\;\;=\large\frac{8p^3}{6}-\large\frac{2\sqrt{2p}}{3}$$\times 2\sqrt 2\times p\sqrt p. \;\;\;=\large\frac{4}{3}$$p^2-\large\frac{8p^2}{3}$
$\;\;\;=\large\frac{4}{3}$$p^2$sq.units.
answered Apr 26, 2013
edited Apr 26, 2013