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

∫_0 ^(2π) (√(R^2 +r^2 −2Rrcos θ)) dθ

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

Answered by robocop last updated on 14/Feb/17

todo en funcion de θ 0≤(√(R^2 +r^2 −2Rrcosθ)) ≤2π (√(R^2 +r^2 −2Rrcosθ )) ≤2π R^2 +r^2 −2Rrcosθ≤4π^2 −2Rrcosθ≤4π^2 −R^2 −r^2 2Rrcosθ≥R^2 +r^2 −4π^2 cosθ≥((R^2 +r^2^ −4π^2 )/(2Rr)) (√(R^2 +r^2 −2Rrcosθ)) ≥0 R^2 +r^2 −2Rrcosθ≥0 cosθ≤((R^2 +r^2 )/(2Rr)) ∫_((R^2 +r^2 −4π^(2 ) )/(2Rr)) ^((R^2 +r^2 )/(2Rr)) (√(R^2 +r^2 −2Rrcosθ)) dθ=(2/3)(R^2 +r^2 −4π^2 )+C

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

Commented byajfour last updated on 16/Feb/17

not satisfied.

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

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