Question

Using integration, find the area enclosed between the two circles x² + y² = 4 and (x − 2)² + y² = 4.

Answer 1

$x^2+y^2=4$ .......(1) represents a circle with centre O(0,0) and radius 2
${\left(x-2\right)}^2+y^2=4$......(2) represents a circle with centre A(2 ,0) and radius 2

Points of intersection of two circles is given by solving the equations

$$\begin{gathered} {\left(x-2\right)}^2=x^2 \\ \Rightarrow x^2-4x+4=x^2 \\ \Rightarrow x=1 \\ \therefore y^2=3 \\ \Rightarrow y=\pm \sqrt{3} \\ \mathrm{Now},B\left(1,\sqrt{3}\right)\mathrm{and}B\text{'}\left(1,-\sqrt{3}\right)\mathrm{are}\mathrm{two}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{circles} \\ \mathrm{We}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{shaded}\mathrm{area}=2\times \mathrm{area}\left(OBAO\right)...\left(1\right) \\ \mathrm{Area}\left(OBAO\right)=\mathrm{area}\left(OBPO\right)\hspace{0.17em}+\mathrm{area}\left(PBAP\right)\mathit{} \\ ={\int }_0^1\left|y_1\right|dx+{\int }_1^2\left|y_2\right|dx\left\{\because y_1>0\Rightarrow \left|y_1\right|=y_1\mathrm{and}{y}_2>0\Rightarrow \left|y_2\right|=y_2\right\} \\ ={\int }_0^1{y}_{1}dx+{\int }_1^2{y}_{2}dx \\ ={\int }_0^1\sqrt{4-{\left(x-2\right)}^2}dx+{\int }_1^2\sqrt{4-x^2}dx \\ ={\left[\frac{1}{2}\left(x-2\right)\sqrt{4-{\left(x-2\right)}^2}+\frac{1}{2}\times 4\times {\sin }^{-1}\left(\frac{x-2}{2}\right)\right]}_0^1+{\left[\frac{1}{2}x\sqrt{4-x^2}+\frac{1}{2}\times 4\times {\sin }^{-1}\left(\frac{x}{2}\right)\right]}_1^2 \\ =\left[\frac{-\sqrt{3}}{2}+2{\sin }^{-1}\left(\frac{-1}{2}\right)\right]-\left[0+2{\sin }^{-1}\left(-1\right)\right]+\left(0-\frac{1}{2}\sqrt{3}\right)+2\left\{\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(1\right)-\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{1}{2}\right)\right\} \\ =\frac{-\sqrt{3}}{2}+2{\sin }^{-1}\left(\frac{-1}{2}\right)-0-2{\sin }^{-1}\left(-1\right)+0-\frac{1}{2}\sqrt{3}+2\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(1\right)-2\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{1}{2}\right) \\ =-\sqrt{3}-4{\sin }^{-1}\left(\frac{1}{2}\right)+4{\sin }^{-1}\left(1\right) \\ =-\sqrt{3}-4\times \frac{\mathrm{\pi }}{6}+4\times \frac{\mathrm{\pi }}{2} \\ =-\sqrt{3}-\frac{2\mathrm{\pi }}{3}+2\mathrm{\pi } \\ =\frac{4\mathrm{\pi }}{3}-\sqrt{3} \\ \mathrm{Now},\mathrm{From}\mathrm{equation}\left(1\right) \\ \mathrm{Shaded}\mathrm{area}=2\times \mathrm{area}\left(\mathrm{O}\mathrm{B}\mathrm{A}\mathrm{O}\right)=2\left(\frac{4\mathrm{\pi }}{3}-\sqrt{3}\right)=\left(\frac{8\mathrm{\pi }}{3}-2\sqrt{3}\right)\mathrm{sq}\mathrm{units} \end{gathered}$$