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Question Number 34713 by abdo mathsup 649 cc last updated on 10/May/18

let a>0  calculate ∫∫_(x^2  +y^2  ≤3)  (1/(2 +x^2  +y^2 ))dxdy.

$${let}\:{a}>\mathrm{0}\:\:{calculate}\:\int\int_{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{3}} \:\frac{\mathrm{1}}{\mathrm{2}\:+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy}. \\ $$

Commented bymath khazana by abdo last updated on 11/May/18

let ude?the diffeomorphisme x=r cosθ ,  y = r sinθ  with 0≤r≤(√3)   and  −π≤θ≤π  I  =∫∫_(0≤r≤(√(3 )) and  −π≤ θ≤π)    (1/(2 +r^2 )) rdr dθ  = ∫_0 ^(√3)   (r/(2+r^2 )) dr. ∫_(−π) ^π  dθ = 2π ∫_0 ^(√3)     (r/(2+r^2 ))dr  =π[ln(2+r^2 )]_0 ^(√3)    = π{ln(5) −ln(2)} .

$${let}\:{ude}?{the}\:{diffeomorphisme}\:{x}={r}\:{cos}\theta\:, \\ $$ $${y}\:=\:{r}\:{sin}\theta\:\:{with}\:\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{3}}\:\:\:{and}\:\:−\pi\leqslant\theta\leqslant\pi \\ $$ $${I}\:\:=\int\int_{\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{3}\:}\:{and}\:\:−\pi\leqslant\:\theta\leqslant\pi} \:\:\:\frac{\mathrm{1}}{\mathrm{2}\:+{r}^{\mathrm{2}} }\:{rdr}\:{d}\theta \\ $$ $$=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{r}}{\mathrm{2}+{r}^{\mathrm{2}} }\:{dr}.\:\int_{−\pi} ^{\pi} \:{d}\theta\:=\:\mathrm{2}\pi\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\:\:\frac{{r}}{\mathrm{2}+{r}^{\mathrm{2}} }{dr} \\ $$ $$=\pi\left[{ln}\left(\mathrm{2}+{r}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\:=\:\pi\left\{{ln}\left(\mathrm{5}\right)\:−{ln}\left(\mathrm{2}\right)\right\}\:. \\ $$

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