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Question Number 10495 by ajfour last updated on 14/Feb/17

$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr}\mathrm{cos}\:\theta}\:{d}\theta \\$$

Answered by robocop last updated on 14/Feb/17

$${todo}\:{en}\:{funcion}\:{de}\:\theta \\$$$$\\$$$$\mathrm{0}\leqslant\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rrcos}\theta}\:\:\leqslant\mathrm{2}\pi \\$$$$\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rrcos}\theta\:\:\:}\:\leqslant\mathrm{2}\pi \\$$$${R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rrcos}\theta\leqslant\mathrm{4}\pi^{\mathrm{2}} \\$$$$−\mathrm{2}{Rrcos}\theta\leqslant\mathrm{4}\pi^{\mathrm{2}} −{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \\$$$$\mathrm{2}{Rrcos}\theta\geqslant{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{4}\pi^{\mathrm{2}} \\$$$${cos}\theta\geqslant\frac{{R}^{\mathrm{2}} +{r}^{\mathrm{2}^{} } −\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{2}{Rr}} \\$$$$\\$$$$\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rrcos}\theta}\:\:\geqslant\mathrm{0} \\$$$${R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rrcos}\theta\geqslant\mathrm{0} \\$$$${cos}\theta\leqslant\frac{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}{Rr}} \\$$$$\\$$$$\int_{\frac{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{4}\pi^{\mathrm{2}\:\:} }{\mathrm{2}{Rr}}} ^{\frac{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}{Rr}}} \sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rrcos}\theta}\:{d}\theta=\frac{\mathrm{2}}{\mathrm{3}}\left({R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{4}\pi^{\mathrm{2}} \right)+{C} \\$$$$\\$$$$\\$$$$\\$$$$\\$$

Commented by ajfour last updated on 16/Feb/17

$${not}\:{satisfied}. \\$$

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