Question Number 141219 by mathmax by abdo last updated on 16/May/21
![find ∫∫_([0,1]^2 ) e^(−(x^2 +y^2 )) (√(x^4 +y^4 ))dxdy](https://www.tinkutara.com/question/Q141219.png)
$$\mathrm{find}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\mathrm{e}^{−\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)} \sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} }\mathrm{dxdy} \\ $$
Answered by Ar Brandon last updated on 16/May/21
![I=∫_0 ^1 ∫_0 ^1 e^(−(x^2 +y^2 )) (√(x^4 +y^4 ))dxdy =∫_0 ^(π/2) ∫_0 ^1 re^(−r^2 ) (√(r^4 −2r^4 sin^2 θcos^2 θ))drdθ =∫_0 ^(π/2) ∫_0 ^1 r^3 e^(−r^2 ) (√(1−(1/2)sin^2 2θ))drdθ =−(1/2)∫_0 ^1 r^2 ∙(−2re^(−r^2 ) )dr×∫_0 ^(π/2) (√(1−(1/2)sin^2 2θ))dθ =−(1/2)[r^2 ∙(e^(−r^2 ) /2)−∫re^(−r^2 ) dr]_0 ^1 ×∫_0 ^(π/2) (√(1−(1/2)sin^2 2θ))dθ =−(1/2)[r^2 ∙(e^(−r^2 ) /2)+(e^(−r^2 ) /2)]_0 ^1 ×∫_0 ^(π/2) (√(1−(1/2)sin^2 2θ))dθ](https://www.tinkutara.com/question/Q141236.png)
$$\mathcal{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)} \sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} }\mathrm{dxdy} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \sqrt{\mathrm{r}^{\mathrm{4}} −\mathrm{2r}^{\mathrm{4}} \mathrm{sin}^{\mathrm{2}} \theta\mathrm{cos}^{\mathrm{2}} \theta}\mathrm{drd}\theta \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{r}^{\mathrm{3}} \mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta}\mathrm{drd}\theta \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{r}^{\mathrm{2}} \centerdot\left(−\mathrm{2re}^{−\mathrm{r}^{\mathrm{2}} } \right)\mathrm{dr}×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta}\mathrm{d}\theta \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{r}^{\mathrm{2}} \centerdot\frac{\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } }{\mathrm{2}}−\int\mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{dr}\right]_{\mathrm{0}} ^{\mathrm{1}} ×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta}\mathrm{d}\theta \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{r}^{\mathrm{2}} \centerdot\frac{\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } }{\mathrm{2}}+\frac{\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} ×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta}\mathrm{d}\theta \\ $$