Question Number 46848 by maxmathsup by imad last updated on 01/Nov/18
$${caculate}\:\:\int\int_{{D}} \:\:\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\:{e}^{−{x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} } {dxdy}\:\:{with} \\ $$$${D}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{4}\right\} \\ $$
Commented by maxmathsup by imad last updated on 03/Nov/18
![let consider the diffeomrphisme (r,θ) →ϕ(r,θ)=(x,y) / x =r cosθ and y =rsinθ we have x^2 +y^2 ≤4 ⇒0<r≤2 and 0≤θ≤2π ∫∫_D (x^2 −y^2 )e^(−(x^2 +y^2 )) dx =∫∫_W foϕ(r,θ)rdrdθ =∫∫_(0<r≤2) r^2 {cos^2 θ −sin^2 θ}e^(−r^2 ) r dr dθ =∫_0 ^2 r^3 e^(−r^2 ) dr .∫_0 ^(2π) cos2θ dθ but ∫_0 ^(2π) cos(2θ)dθ =[((sin(2θ))/2)]_0 ^(2π) =0 ⇒ I =0 .](https://www.tinkutara.com/question/Q46981.png)
$${let}\:{consider}\:{the}\:{diffeomrphisme}\:\left({r},\theta\right)\:\rightarrow\varphi\left({r},\theta\right)=\left({x},{y}\right)\:/ \\ $$$${x}\:={r}\:{cos}\theta\:{and}\:{y}\:={rsin}\theta\:\:{we}\:{have}\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{4}\:\Rightarrow\mathrm{0}<{r}\leqslant\mathrm{2}\:\:\:{and}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$$$\int\int_{{D}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \:{dx}\:=\int\int_{{W}} {fo}\varphi\left({r},\theta\right){rdrd}\theta \\ $$$$=\int\int_{\mathrm{0}<{r}\leqslant\mathrm{2}} {r}^{\mathrm{2}} \left\{{cos}^{\mathrm{2}} \theta\:−{sin}^{\mathrm{2}} \theta\right\}{e}^{−{r}^{\mathrm{2}} } {r}\:{dr}\:{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} \:{r}^{\mathrm{3}} \:{e}^{−{r}^{\mathrm{2}} } {dr}\:.\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{cos}\mathrm{2}\theta\:{d}\theta\:\:{but}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{cos}\left(\mathrm{2}\theta\right){d}\theta\:=\left[\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}\pi} =\mathrm{0}\:\Rightarrow \\ $$$${I}\:=\mathrm{0}\:. \\ $$