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Question Number 167690 by peter frank last updated on 22/Mar/22

Answered by som(math1967) last updated on 23/Mar/22

$$bysolving{x}^2+y^2=4and{(x-2)}^2+y^2=4$$$$x=1,y=\pm \sqrt{3}$$$$areaofOABCO$$$$=2\times areaOMBAO$$$$=2\times 2MBAM$$$$=4{\int }_1^2\sqrt{4-x^2}dx$$$$=4{[\frac{x\sqrt{4-x^2}}{2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_1^2$$$$=4[{2\sin }^{-1}1-\frac{\sqrt{3}}{2}-{2\sin }^{-1}\frac{1}{2}]$$$$=4[\pi -\frac{\pi }{3}-\frac{\sqrt{3}}{2}]$$$$=(\frac{8\pi }{3}-2\sqrt{3})squnit$$

Commented by som(math1967) last updated on 23/Mar/22

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