Question Number 62197 by maxmathsup by imad last updated on 17/Jun/19
![calculate ∫∫_([0,1]^2 ) (√(x^2 +y^2 ))sin((√(x^2 +y^2 )))dxdy](Q62197.png)
$${calculate}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{sin}\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){dxdy} \\ $$
Commented by mathmax by abdo last updated on 21/Jun/19
![let use the diffeomorphism x =rcosθ and y =rsinθ 0≤x≤1 and 0≤y≤1 ⇒0≤x^2 +y^2 ≤2 ⇒0≤r^2 ≤2 ⇒0≤r≤(√2) A=∫∫_([0,1]^2 ) (√(x^2 +y^2 ))sin((√(x^2 +y^2 )))dxdy =∫∫_(0≤r≤(√2) and 0≤θ≤(π/2)) r sin(r)rdrdθ =∫_0 ^(√2) r^2 sin(r)dr ∫_0 ^(π/2) dθ =(π/2) ∫_0 ^(√2) r^2 sin(r)dr by parts u=r^2 and v^′ =sinr ∫_0 ^(√2) r^2 sin(r)dr =[−r^2 cos(r)]_0 ^(√2) +∫_0 ^(√2) 2r cosrdr =−2cos((√2)) +2 { [r sin(r)]_0 ^(√2) −∫_0 ^(√2) sinr dr} =−2cos((√2)) +2{ (√2)sin((√2)) +[cosr]_0 ^(√2) } =−2cos((√2)) +2(√2)sin((√2)) +2(cos((√2))−1) ⇒ A =(π/2){ −2cos((√2))+2(√2)sin((√2)) +2 cos((√2))−2}=π{(√2)sin((√2))−1} .](Q62469.png)
$${let}\:{use}\:{the}\:{diffeomorphism}\:{x}\:={rcos}\theta\:\:{and}\:{y}\:={rsin}\theta \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}\:\Rightarrow\mathrm{0}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{2}\:\Rightarrow\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\: \\ $$$${A}=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{sin}\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){dxdy}\:=\int\int_{\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\:\:{and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:{r}\:{sin}\left({r}\right){rdrd}\theta \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:{r}^{\mathrm{2}} {sin}\left({r}\right){dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:{r}^{\mathrm{2}} {sin}\left({r}\right){dr}\:\:\:\:\:\:{by}\:{parts}\:\:{u}={r}^{\mathrm{2}} \:{and}\:{v}^{'} \:={sinr} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:{r}^{\mathrm{2}} \:{sin}\left({r}\right){dr}\:=\left[−{r}^{\mathrm{2}} {cos}\left({r}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:+\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \mathrm{2}{r}\:{cosrdr} \\ $$$$=−\mathrm{2}{cos}\left(\sqrt{\mathrm{2}}\right)\:+\mathrm{2}\:\left\{\:\:\left[{r}\:{sin}\left({r}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:{sinr}\:{dr}\right\} \\ $$$$=−\mathrm{2}{cos}\left(\sqrt{\mathrm{2}}\right)\:+\mathrm{2}\left\{\:\:\sqrt{\mathrm{2}}{sin}\left(\sqrt{\mathrm{2}}\right)\:+\left[{cosr}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \right\} \\ $$$$=−\mathrm{2}{cos}\left(\sqrt{\mathrm{2}}\right)\:+\mathrm{2}\sqrt{\mathrm{2}}{sin}\left(\sqrt{\mathrm{2}}\right)\:+\mathrm{2}\left({cos}\left(\sqrt{\mathrm{2}}\right)−\mathrm{1}\right)\:\Rightarrow \\ $$$${A}\:=\frac{\pi}{\mathrm{2}}\left\{\:−\mathrm{2}{cos}\left(\sqrt{\mathrm{2}}\right)+\mathrm{2}\sqrt{\mathrm{2}}{sin}\left(\sqrt{\mathrm{2}}\right)\:+\mathrm{2}\:{cos}\left(\sqrt{\mathrm{2}}\right)−\mathrm{2}\right\}=\pi\left\{\sqrt{\mathrm{2}}{sin}\left(\sqrt{\mathrm{2}}\right)−\mathrm{1}\right\}\:. \\ $$$$ \\ $$$$ \\ $$