Question Number 127161 by kaivan.ahmadi last updated on 27/Dec/20
$$\left.{R}=\left({x},{y}\right):{y}\geqslant\mathrm{0}\:,\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\mathrm{9}\right\} \\ $$$$\int\underset{{R}} {\int}{cos}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dydx}=? \\ $$
Answered by mathmax by abdo last updated on 27/Dec/20
![we use the diffeomorphism { ((x=rcosθ)),((y=rsinθ)) :} x^2 +y^2 ≤9 ⇒r^2 ≤9 ⇒0≤r≤3 ,y≥0 ⇒0≤θ ≤π ⇒ ∫∫_R cos(x^2 +y^2 )dxdy =∫_0 ^3 ∫_0 ^π cos(r^2 )rdrdθ =π ∫_0 ^3 rcos(r^2 )dr =π[(1/2)sin(r^2 )]_0 ^3 =(π/2)sin(9)](https://www.tinkutara.com/question/Q127177.png)
$$\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{diffeomorphism}\:\:\begin{cases}{\mathrm{x}=\mathrm{rcos}\theta}\\{\mathrm{y}=\mathrm{rsin}\theta}\end{cases} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \:\leqslant\mathrm{9}\:\Rightarrow\mathrm{r}^{\mathrm{2}} \:\leqslant\mathrm{9}\:\Rightarrow\mathrm{0}\leqslant\mathrm{r}\leqslant\mathrm{3}\:\:,\mathrm{y}\geqslant\mathrm{0}\:\Rightarrow\mathrm{0}\leqslant\theta\:\leqslant\pi\:\Rightarrow \\ $$$$\int\int_{\mathrm{R}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \int_{\mathrm{0}} ^{\pi} \mathrm{cos}\left(\mathrm{r}^{\mathrm{2}} \right)\mathrm{rdrd}\theta \\ $$$$=\pi\:\int_{\mathrm{0}} ^{\mathrm{3}} \:\mathrm{rcos}\left(\mathrm{r}^{\mathrm{2}} \right)\mathrm{dr}\:=\pi\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{r}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{3}} \:=\frac{\pi}{\mathrm{2}}\mathrm{sin}\left(\mathrm{9}\right) \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 27/Dec/20
$${thank}\:{sir} \\ $$