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Question Number 87146 by john santu last updated on 03/Apr/20

$$\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{region}$$$$\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{polar}\mathrm{curve}$$$$\mathrm{r}=4+2\cos \theta ?$$

Answered by jagoll last updated on 03/Apr/20

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{area}=\underset{0}{\stackrel{2\pi }{\int }}\frac{1}{2}{\mathrm{r}}^2\mathrm{d}\theta$$$$=\frac{1}{2}\int \underset{0}{\stackrel{2\pi }{}}{(4+2\cos \theta )}^2\mathrm{d}\theta$$$$=\frac{1}{2}\underset{0}{\stackrel{2\pi }{\int }}(16+16\cos \theta +4(\frac{1}{2}+\frac{1}{2}\cos 2\theta ))\mathrm{d}\theta$$

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