Question Number 87146 by john santu last updated on 03/Apr/20
$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\: \\ $$$$\mathrm{enclosed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{curve}\: \\ $$$$\mathrm{r}\:=\:\mathrm{4}\:+\:\mathrm{2}\:\mathrm{cos}\:\theta\:? \\ $$
Answered by jagoll last updated on 03/Apr/20
Commented by jagoll last updated on 03/Apr/20
$$\mathrm{area}\:=\:\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{r}^{\mathrm{2}} \:\mathrm{d}\theta \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\:}}\:\left(\mathrm{4}+\mathrm{2cos}\:\theta\right)^{\mathrm{2}} \:\mathrm{d}\theta \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\left(\mathrm{16}+\mathrm{16}\:\mathrm{cos}\:\theta\:+\:\mathrm{4}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta\right)\right)\:\mathrm{d}\theta \\ $$