Question

Draw a rough sketch of the region {(x, y) : y² ≤ 5x, 5x² + 5y² ≤ 36} and find the area enclosed by the region using method of integration.

Answer 1

The given region is intersection of
$y^2\le 5x\mathrm{and}5x^2+5y^2\le 36$
Clearly, $y^2\le 5x$ is a parabola with vertex at origin and the axis is along the x-axis opening in the positive direction.Also $5x^2+5y^2\le 36$ is a circle with centre at the origin and has a radius $\sqrt{\frac{36}{5}}\mathrm{or}\frac{6}{\sqrt{5}}$.
Corresponding equations of given inequations are
$$\begin{gathered} y^2=5x.....\left(1\right) \\ 5x^2+5y^2=36.....\left(2\right) \end{gathered}$$
Substituting the value of y² from (1) into (2), we get

$5x^2+25x=36$
$\Rightarrow 5x^2+25x-36=0$
$\Rightarrow x=\frac{-25\pm \sqrt{625+720}}{10}$
$\Rightarrow x=\frac{-25\pm \sqrt{1345}}{10}$
From the figure we see that x-coordinate of intersecting point can not be negative.
$\therefore x=\frac{-25+\sqrt{1345}}{10}$
Now assume that x-coordinate of intersecting point, $a=\frac{-25+\sqrt{1345}}{10}$
The Required area,
A = 2(Area of OACO + Area of CABC)
Approximating the area of OACO the length =$\left|y_1\right|$ and a width = dx=|y1=dx
$\mathrm{Area}\mathrm{of}OACO={\int }_0^a\left|y_1\right|dx$
$={\int }_0^a{y}_{1}dx$
$={\int }_0^a\sqrt{5x}dx\left(\because {y_1}^2=5x\Rightarrow y_1=\sqrt{5x}\right)$
= $\sqrt{5}{{\left[\frac{2x^{{\displaystyle \frac{3}{2}}}}{3}\right]}^a}_0$
Therefore, Area of OACO $=\frac{2\sqrt{5}{a}^{{\displaystyle \frac{3}{2}}}}{3}$
Similarly approximating the area of CABC the length $=\left|y_2\right|$ and the width = dx
$\mathrm{Area}\mathrm{of}CABC={\int }_a^{\frac{6}{\sqrt{5}}}\left|y_2\right|dx$
$={\int }_a^{\frac{6}{\sqrt{5}}}{y}_{2}dx$
$={\int }_a^{\frac{6}{\sqrt{5}}}\sqrt{\frac{36}{5}-x^2}dx$
$={{\left[\frac{x}{2}\sqrt{\frac{36}{5}-x^2}+\frac{18}{5}{\sin }^{-1}\left(\frac{x\sqrt{5}}{6}\right)\right]}_a}^{\frac{6}{\sqrt{5}}}$
$=\frac{6}{2\sqrt{5}}\sqrt{\frac{36}{5}-\frac{36}{5}}+\frac{18}{5}{\sin }^{-1}\left(1\right)-\frac{a}{2}\sqrt{\frac{36}{5}-a^2}-\frac{18}{5}{\sin }^{-1}\left(\frac{a\sqrt{5}}{{\displaystyle 6}}\right)$
$=0+\frac{18}{5}{\sin }^{-1}\left(1\right)-\frac{a}{2}\sqrt{\frac{36}{5}-a^2}-\frac{18}{5}{\sin }^{-1}\left(\frac{a\sqrt{5}}{6}\right)$
$=\frac{18}{5}\times \frac{\mathrm{\pi }}{2}-\frac{a}{2}\sqrt{\frac{36}{5}-a^2}-\frac{18}{5}{\sin }^{-1}\left(\frac{a\sqrt{5}}{6}\right)$
Area of CABC $=\frac{9\mathrm{\pi }}{5}-\frac{a}{2}\sqrt{\frac{36}{5}-a^2}-\frac{18}{5}{\sin }^{-1}\left(\frac{a\sqrt{5}}{6}\right)$
Thus the Required area, A = 2(Area of OACO + Area of CABC)
$A=2\left[\frac{2\sqrt{5}{a}^{{\displaystyle \frac{3}{2}}}}{3}+\frac{9\mathrm{\pi }}{5}-\frac{a}{2}\sqrt{\frac{36}{5}-a^2}-\frac{18}{5}{\sin }^{-1}\left(\frac{a\sqrt{5}}{6}\right)\right]$
$=\frac{4\sqrt{5}{a}^{{\displaystyle \frac{3}{2}}}}{3}+\frac{18\mathrm{\pi }}{5}-a\sqrt{\frac{36}{5}-a^2}-\frac{36}{5}{\sin }^{-1}\left(\frac{a\sqrt{5}}{6}\right)$ , $\mathrm{Where},a=\frac{-25+\sqrt{1345}}{10}$ .