Question Number 37278 by abdo.msup.com last updated on 11/Jun/18
$$\:{calculate}\:\int\int_{{D}} \:{x}\:{cos}\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\right. \\ $$$$\left.\mathrm{1}\leqslant{y}\leqslant\mathrm{3}\right\} \\ $$
Commented by prof Abdo imad last updated on 16/Jun/18
![let consider the diffeomorphisme (r,θ)→(x,y)/ x=r cosθ and y=rsinθ we have 1≤x^2 +y^2 ≤10 ⇒1≤r≤(√(10)) also x≥0 and y≥ ⇒ 0≤θ≤(π/2) I =∫∫_(1≤r≤(√(10)) and 0≤θ≤(π/2)) r cosθ cos(r^2 )rdrdθ = ∫_1 ^(√(10)) r^2 cos(r^2 )dr.∫_0 ^(π/2) cosθ dθ but ∫_1 ^(√(10)) r(r cos(r^2 ))dr=[(r/2)sin(r^2 )]_1 ^(√(10)) −∫_1 ^(√(10)) (r/2)sin(r^2 )dr =((√(10))/2) sin(10) −(1/2)sin(1) +(1/4)[cosr^2 ]_1 ^(√(10)) =((√(10))/2)sin(10) −(1/2)sin(1) +(1/4)( cos(10) −cos(1)) ∫_0 ^(π/2) cosθ dθ=[sinθ]_0 ^(π/2) =1 ⇒ I =((√(10))/2)sin(10) +(1/4)cos(10) −(1/2)sin(1)−(1/4)cos(1)](https://www.tinkutara.com/question/Q37680.png)
$${let}\:{consider}\:{the}\:{diffeomorphisme} \\ $$$$\left({r},\theta\right)\rightarrow\left({x},{y}\right)/\:{x}={r}\:{cos}\theta\:{and}\:{y}={rsin}\theta \\ $$$${we}\:{have}\:\mathrm{1}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{10}\:\:\Rightarrow\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{10}} \\ $$$${also}\:{x}\geqslant\mathrm{0}\:{and}\:{y}\geqslant\:\Rightarrow\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${I}\:=\int\int_{\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{10}}\:{and}\:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:{r}\:{cos}\theta\:{cos}\left({r}^{\mathrm{2}} \right){rdrd}\theta \\ $$$$=\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{10}}} {r}^{\mathrm{2}} {cos}\left({r}^{\mathrm{2}} \right){dr}.\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta\:{d}\theta\:{but} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{10}}} {r}\left({r}\:{cos}\left({r}^{\mathrm{2}} \right)\right){dr}=\left[\frac{{r}}{\mathrm{2}}{sin}\left({r}^{\mathrm{2}} \right)\right]_{\mathrm{1}} ^{\sqrt{\mathrm{10}}} −\int_{\mathrm{1}} ^{\sqrt{\mathrm{10}}} \:\frac{{r}}{\mathrm{2}}{sin}\left({r}^{\mathrm{2}} \right){dr} \\ $$$$=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}\:{sin}\left(\mathrm{10}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{1}\right)\:\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{cosr}^{\mathrm{2}} \right]_{\mathrm{1}} ^{\sqrt{\mathrm{10}}} \\ $$$$=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}{sin}\left(\mathrm{10}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\:{cos}\left(\mathrm{10}\right)\:−{cos}\left(\mathrm{1}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta\:{d}\theta=\left[{sin}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{1}\:\Rightarrow \\ $$$${I}\:=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}{sin}\left(\mathrm{10}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}{cos}\left(\mathrm{10}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}{cos}\left(\mathrm{1}\right) \\ $$