Menu Close

Evaluate-0-0-e-x-2-y-2-dydx-




Question Number 20612 by ajfour last updated on 29/Aug/17
Evaluate ∫_0 ^(  ∞) ∫_0 ^(  ∞) e^(−(x^2 +y^2 )) dydx .
$${Evaluate}\:\int_{\mathrm{0}} ^{\:\:\infty} \int_{\mathrm{0}} ^{\:\:\infty} {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dydx}\:. \\ $$
Answered by ajfour last updated on 29/Aug/17
let  x^2 +y^2 =r^2   that is  x=rcos θ  ;  y=rsin θ  dx=−rsin θdθ   ;  dy=rcos θdθ  ∫_0 ^(  ∞) ∫_0 ^(  ∞) e^(−(x^2 +y^2 )) dydx =I  I=∫_0 ^(  ∞) [∫_( 0) ^(  π/2) e^(−r^2 ) rdθ]dr    =(π/2)∫_0 ^(  ∞) e^(−r^2 ) rdr  =(π/4)∫_∞ ^(  0) e^(−r^2 ) (−2rdr)  I= (π/4)(e^(−r^2 ) )∣_∞ ^0  = (π/4) .
$${let}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${that}\:{is}\:\:{x}={r}\mathrm{cos}\:\theta\:\:;\:\:{y}={r}\mathrm{sin}\:\theta \\ $$$${dx}=−{r}\mathrm{sin}\:\theta{d}\theta\:\:\:;\:\:{dy}={r}\mathrm{cos}\:\theta{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\:\:\infty} \int_{\mathrm{0}} ^{\:\:\infty} {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dydx}\:={I} \\ $$$${I}=\int_{\mathrm{0}} ^{\:\:\infty} \left[\int_{\:\mathrm{0}} ^{\:\:\pi/\mathrm{2}} {e}^{−{r}^{\mathrm{2}} } {rd}\theta\right]{dr} \\ $$$$\:\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\infty} {e}^{−{r}^{\mathrm{2}} } {rdr}\:\:=\frac{\pi}{\mathrm{4}}\int_{\infty} ^{\:\:\mathrm{0}} {e}^{−{r}^{\mathrm{2}} } \left(−\mathrm{2}{rdr}\right) \\ $$$${I}=\:\frac{\pi}{\mathrm{4}}\left({e}^{−{r}^{\mathrm{2}} } \right)\mid_{\infty} ^{\mathrm{0}} \:=\:\frac{\pi}{\mathrm{4}}\:. \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *