Question

Draw a rough sketch of the region {(x, y) : y² ≤ 3x, 3x² + 3y² ≤ 16} and find the area enclosed by the region using method of integration.

Answer 1

The given region is the intersection of
$y^2\le 3x\mathrm{and}3x^2+3y^2\le 16$
Clearly ,y² = 3x is a parabola with vertex at (0, 0) axis is along the x-axis opening in the positive direction.
Also 3x² + 3y² = 16 is a circle with centre at origin and has a radius $\sqrt{\frac{16}{3}}$.
Corresponding equations of the given inequations are
$$\begin{gathered} y^2=3x.....\left(1\right) \\ \mathrm{and} \\ 3{\mathrm{x}}^2+3{\mathrm{y}}^2=16.....\left(2\right) \end{gathered}$$
Substituting the value of y² from (1) into (2)
$3x^2+9x=16$
$\Rightarrow 3x^2+9x-16=0$
$\Rightarrow x=\frac{-9\pm \sqrt{81+192}}{6}$
$\Rightarrow x=\frac{-9\pm \sqrt{273}}{6}$
By figure we see that the value of x will be non-negative.
$\therefore x=\frac{-9+\sqrt{273}}{6}$
Now assume that x-coordinate of the intersecting point, $a=\frac{-9+\sqrt{273}}{6}$
The Required area A = 2(Area of OACO + Area of CABC)
Approximating the area of OACO the length $=\left|y_1\right|$ width = dx
Area of OACO $={\int }_0^a\left|y_1\right|\mathit{d}x$
$={\int }_0^a{y}_1\mathit{d}x$
$={\int }_0^a\sqrt{3x}\mathit{d}x\left(\because {y^2}_1=\sqrt{3x}\Rightarrow y_1=\sqrt{3x}\right)$
$={{\left[\frac{2\sqrt{3}{x}^{{\displaystyle \frac{3}{2}}}}{3}\right]}^a}_0$
$=\frac{2\sqrt{3}{a}^{{\displaystyle \frac{3}{2}}}}{3}$
Therefore, Area of OACO $=\frac{2\sqrt{3}{a}^{{\displaystyle \frac{3}{2}}}}{3}$
Similarly approximating the are of CABC the length $=\left|y_2\right|$ and the width = dx
Area of CABC $={\int }_a^{\frac{4}{\sqrt{3}}}\left|y_2\right|d\mathrm{x}$
$={\int }_a^{\frac{4}{\sqrt{3}}}{y}_{2}dx$
$={\int }_a^{\frac{4}{\sqrt{3}}}\sqrt{\frac{16}{3}-x^2}\mathit{d}x\mathit{}\left(\because 3x^2+3{y_2}^2=16\Rightarrow y_2=\sqrt{\frac{16}{3}-x^2}\right)$
$={{\left[\frac{x}{2}\sqrt{\frac{16}{3}-x^2}+\frac{8}{3}{\sin }^{-1}\left(\frac{x\sqrt{3}}{4}\right)\right]}_a}^{\frac{4}{\sqrt{3}}}$
$=\left[\frac{4}{2\sqrt{3}}\sqrt{\frac{16}{3}-\frac{16}{3}}+\frac{8}{3}{\sin }^{-1}\left(\frac{4\sqrt{3}}{{\displaystyle 4\sqrt{3}}}\right)-\frac{a}{2}\sqrt{\frac{16}{3}-a^2}-\frac{8}{3}{\sin }^{-1}\left(\frac{a\sqrt{3}}{4}\right)\right]$

$=-\frac{a}{2}\sqrt{\frac{16}{3}-a^2}+\frac{8}{3}{\sin }^{-1}\left(1\right)-\frac{8}{3}{\sin }^{-1}\left(\frac{a\sqrt{3}}{4}\right)$
Area of CABC $=-\frac{a}{2}\sqrt{\frac{16}{3}-a^2}+\frac{4\mathrm{\pi }}{3}-\frac{8}{3}{\sin }^{-1}\left(\frac{a\sqrt{3}}{4}\right)$
Thus the required area A = 2(Area of OACO + Area of CABC)
$=2\left[\frac{2\sqrt{3}{a}^{{\displaystyle \frac{3}{2}}}}{3}-\frac{a}{2}\sqrt{\frac{16}{3}-a^2}+\frac{4\mathrm{\pi }}{3}-\frac{8}{3}{\sin }^{-1}\left(\frac{a\sqrt{3}}{4}\right)\right]$
$=\frac{4a^{{\displaystyle \frac{3}{2}}}}{\sqrt{3}}-a\sqrt{\frac{16}{3}-a^2}+\frac{8\mathrm{\pi }}{3}-\frac{16}{3}{\sin }^{-1}\left(\frac{a\sqrt{3}}{4}\right)$
$$\begin{gathered} \mathrm{Hence},\mathrm{the}\mathrm{reguired}\mathrm{area}\mathrm{is}, \\ \frac{4a^{{\displaystyle \frac{3}{2}}}}{\sqrt{3}}-a\sqrt{\frac{16}{3}-a^2}+\frac{8\mathrm{\pi }}{3}-\frac{16}{3}{\sin }^{-1}\left(\frac{a\sqrt{3}}{4}\right)\mathrm{square}\mathrm{units},\mathrm{where}a=\frac{-9+\sqrt{273}}{6} \end{gathered}$$