question_answer

Using integration, find the area of the region enclosed between the circles \[{{x}^{2}}+{{y}^{2}}=4\] and \[{{(x-2)}^{2}}+{{y}^{2}}=4.\]

Answer:

Equations of the given circles are \[{{x}^{2}}+{{y}^{2}}=4\]                            ...(i) and       \[{{(x-2)}^{2}}+{{y}^{2}}=4\]                                    ?(ii) \[\because \] Circle (i) has centre (0, 0), radius 2 and circle (ii) has centre (2, 0), radius 2.                           On putting the value of \[{{y}^{2}}\] from Eq. (i) in Eq. (ii), we get             \[{{(x-2)}^{2}}+{{x}^{2}}+4=4\] \[\Rightarrow \]   \[{{x}^{2}}+4-4x-{{x}^{2}}+4=4\]                                 \[[\because \,\,{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab]\] \[\Rightarrow \]   \[4x-4=0\] \[\Rightarrow \] x = 1 On putting x = 1 in Eq. (i), we get             \[{{1}^{2}}+{{y}^{2}}=4\] \[\Rightarrow \]   \[y=\pm \,\sqrt{3}\]             So, the given circles intersect each other at \[B(1,\,\,\sqrt{3})\] and \[D(1,\,\,-\sqrt{3}).\] Now, required area = 2 (Area of shaded region OABO) \[=2\left[ \int_{0}^{1}{circle}\,\,(ii)dx+\int_{1}^{2}{circle\,\,(i)dx} \right]\] \[=2\left[ \int_{0}^{1}{\sqrt{4-{{(x-2)}^{2}}}}dx+\int_{1}^{2}{\sqrt{4-{{x}^{2}}}dx} \right]\] \[=2\left[ \frac{x-2}{2}\sqrt{4-{{(x-2)}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\left( \frac{x-2}{2} \right) \right]_{0}^{1}\]                         \[+\,2\left[ \frac{1}{2}\times \sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{1}^{2}\] \[\left[ \because \,\,\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}+C} \right]\] \[=\left[ (x-2)\sqrt{4-{{(x-2)}^{2}}}+4{{\sin }^{-1}}\left( \frac{x-2}{2} \right) \right]_{0}^{1}\]                         \[+\left[ x\sqrt{4-{{x}^{2}}}+4{{\sin }^{-1}}\frac{x}{2} \right]_{1}^{2}\] \[=\left[ -\,\sqrt{3}+4{{\sin }^{-1}}\left( -\frac{1}{2} \right)-4{{\sin }^{-1}}(-\,1) \right]\]             \[+\left[ 0+4{{\sin }^{-1}}(1)-\sqrt{3}-4{{\sin }^{-1}}\left( \frac{1}{2} \right) \right]\] \[=\left[ \sqrt{3}+4\left( -\frac{\pi }{6} \right)-4\left( \frac{\pi }{2} \right) \right]+\left[ 4\cdot \frac{\pi }{2}-\sqrt{3}-4\times \frac{\pi }{6} \right]\] \[=-\,\sqrt{3}-\frac{2\pi }{3}+2\pi +2\pi -\sqrt{3}-\frac{2\pi }{3}\] \[=\frac{-2\pi +6\pi +6\pi -2\pi }{3}-2\sqrt{3}\] \[=\left( \frac{8\pi }{3}-2\sqrt{3} \right)\,\,\text{sq}\,\,\text{units}\]