Question

Find the area of the region enclosed between the two curves x² + y² = 9 and (x − 3)² + y² = 9.

Answer 1

Let the two curves be named as y₁ and y₂ where
$$\begin{gathered} y_1:{\left(x-3\right)}^2+y^2=9.....\left(1\right) \\ {y}_2:x^2+y^2=9.....\left(2\right) \end{gathered}$$
The curve x² + y² = 9 represents a circle with centre (0, 0) and the radius is 3.
The curve (x − 3)² + y² = 9 represents a circle with centre (3, 0) and has a radius 3.
To find the intersection points of two curves equate them.
On solving (1) and (2) we get
$x=\frac{3}{2}\mathrm{and}y=\pm \frac{3\sqrt{3}}{2}$
Therefore, intersection points are $\left(\frac{3}{2},\frac{3\sqrt{3}}{2}\right)\mathrm{and}\left(\frac{3}{2},-\frac{3\sqrt{3}}{2}\right)$ .
Now, the required area(OABO) =2[area(OACO) +area(CABC)]
Here,
$\mathrm{Area}\left(OACO\right)={\int }_0^{\frac{3}{2}}{Y}_{1}dx$
$={\int }_0^{\frac{3}{2}}\sqrt{9-{\left(x-3\right)}^2}dx$
And
$\mathrm{Area}\left(CABC\right)={\int }_{\frac{3}{2}}^3\left|Y_2\right|dx$
$={\int }_{\frac{3}{2}}^3\sqrt{9-x^2}dx$
Thus the required area is given by,
A = 2[area(OACO) +area(CABC)]
$2\left({\int }_0^{\frac{3}{2}}\sqrt{9-{\left(x-3\right)}^2}dx+{\int }_{\frac{3}{2}}^3\sqrt{9-x^2}dx\right)$
$=2{{\left[\frac{\left(x-3\right)}{2}\sqrt{9-{\left(x-3\right)}^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{x-3}{3}\right)\right]}_0}^{\frac{3}{2}}+2{{\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{x}{3}\right)\right]}^3}_{\frac{3}{2}}$
$=2\left[\frac{{\displaystyle \frac{3}{2}-3}}{2}\sqrt{9-{\left(\frac{3}{3}-3\right)}^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{{\displaystyle \frac{3}{2}-3}}{3}\right)-\frac{0-3}{2}\sqrt{9-{\left(0-3\right)}^2}-\frac{9}{2}{\sin }^{-1}\left(\frac{0-3}{3}\right)\right]+2\left[\frac{3}{2}\sqrt{9-3^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{3}{3}\right)-\left(\frac{3}{4}\sqrt{9-\frac{9}{4}}\right)-\frac{9}{2}{\sin }^{-1}\left(\frac{{\displaystyle \frac{3}{2}}}{3}\right)\right]$

$=2\left[-\frac{9\sqrt{3}}{8}-\frac{9\mathrm{\pi }}{12}+\frac{9\mathrm{\pi }}{4}\right]+2\left[\frac{9\mathrm{\pi }}{4}-\frac{9\sqrt{3}}{8}-\frac{9\mathrm{\pi }}{12}\right]$

$=-\frac{18\sqrt{3}}{8}-\frac{18\mathrm{\pi }}{12}+\frac{18\mathrm{\pi }}{4}+\frac{18\mathrm{\pi }}{4}-\frac{18\sqrt{3}}{8}-\frac{18\mathrm{\pi }}{12}$

$=\frac{-36\sqrt{3}}{8}-\frac{36\mathrm{\pi }}{12}+\frac{36\mathrm{\pi }}{4}$

$=-\frac{9\sqrt{3}}{2}-3\mathrm{\pi }+9\mathrm{\pi }$
$=6\mathrm{\pi }-\frac{9\sqrt{3}}{2}$
Hence the required area is $6\mathrm{\pi }-\frac{9\sqrt{3}}{2}$ square units.