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0-2pi-R-2-r-2-2Rrcos-d-




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 by ajfour last updated on 16/Feb/17
not satisfied.
$${not}\:{satisfied}. \\ $$

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