Question Number 62213 by maxmathsup by imad last updated on 17/Jun/19
$${calculate}\:\int\int\int_{{D}} \:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }{dxdydz}\:{with} \\ $$$${D}\:=\left\{\left({x},{y},{z}\right)\in{R}^{\mathrm{3}} \:/\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:,\:\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\:\:{and}\:\:\:\mathrm{2}\leqslant{z}\leqslant\mathrm{3}\:\right\} \\ $$
Commented by prof Abdo imad last updated on 20/Jun/19
![let use the diffeomorphism x=rcosθ ,y=rsinθ 0≤x≤1 and 1≤y≤2 ⇒1 ≤x^2 +y^2 ≤5 ⇒1≤r≤(√5) ∫∫∫_D e^(−x^2 −y^2 ) (√(x^2 +y^2 +z^2 ))dxdydz =∫_2 ^3 ( ∫_1 ^(√5) (e^(−r^2 ) (√(r^2 +z^2 )))rdr∫_0 ^(π/2) dθ)dz =(π/2) ∫_2 ^3 ( ∫_1 ^(√5) (re^(−r^2 ) (√(r^2 +z^2 ))dr)dz but by parts ∫_1 ^(√5) (re^(−r^2 ) )(√(r^2 +z^2 ))dr =[−(1/2) e^(−r^2 ) (√(r^2 +z^2 ))]_1 ^(√5) +(1/2)∫_1 ^(√5) e^(−r^2 ) ((2r)/( (√(r^2 +z^2 )))) dr =(1/2)(e^(−1) (√(1+z^2 ))−e^(−5) (√(5+z^2 ))) +∫_1 ^(√5) ((re^(−r^2 ) )/( (√(r^2 +z^2 )))) dr r =zu ⇒ ∫_1 ^(√5) ((r e^(−r^2 ) )/( (√(r^2 +z^2 )))) dr =∫_(1/z) ^((√5)/z) ((zu e^(−z^2 u^2 ) )/(z(√(1+u^2 )))) zdu =z ∫_(1/z) ^((√5)/z) ((u e^(−z^2 u^2 ) )/( (√(1+u^2 )))) du .....be continued...](https://www.tinkutara.com/question/Q62385.png)
$${let}\:{use}\:{the}\:{diffeomorphism}\:{x}={rcos}\theta\:\:,{y}={rsin}\theta \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\:\Rightarrow\mathrm{1}\:\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{5}\:\Rightarrow\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{5}} \\ $$$$\int\int\int_{{D}} {e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }{dxdydz} \\ $$$$=\int_{\mathrm{2}} ^{\mathrm{3}} \left(\:\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \left({e}^{−{r}^{\mathrm{2}} } \sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }\right){rdr}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\right){dz} \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{2}} ^{\mathrm{3}} \:\left(\:\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \left({re}^{−{r}^{\mathrm{2}} } \sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }{dr}\right){dz}\:{but}\right. \\ $$$${by}\:{parts} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \:\left({re}^{−{r}^{\mathrm{2}} } \right)\sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }{dr}\: \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{r}^{\mathrm{2}} } \sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }\right]_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \:{e}^{−{r}^{\mathrm{2}} } \:\:\frac{\mathrm{2}{r}}{\:\sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }}\:{dr} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{−\mathrm{1}} \sqrt{\mathrm{1}+{z}^{\mathrm{2}} }−{e}^{−\mathrm{5}} \sqrt{\mathrm{5}+{z}^{\mathrm{2}} }\right)\:+\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \:\:\frac{{re}^{−{r}^{\mathrm{2}} } }{\:\sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }}\:{dr} \\ $$$${r}\:={zu}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \:\frac{{r}\:{e}^{−{r}^{\mathrm{2}} } }{\:\sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }}\:{dr}\:=\int_{\frac{\mathrm{1}}{{z}}} ^{\frac{\sqrt{\mathrm{5}}}{{z}}} \:\:\:\:\frac{{zu}\:{e}^{−{z}^{\mathrm{2}} {u}^{\mathrm{2}} } }{{z}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:{zdu} \\ $$$$={z}\:\int_{\frac{\mathrm{1}}{{z}}} ^{\frac{\sqrt{\mathrm{5}}}{{z}}} \:\:\:\frac{{u}\:{e}^{−{z}^{\mathrm{2}} {u}^{\mathrm{2}} } }{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:{du}\:…..{be}\:{continued}… \\ $$