Question Number 203349 by Mathspace last updated on 17/Jan/24
![calculate ∫∫_([0,a]^2 ) e^(−x^2 −y^2 ) dxdy can you find ∫_0 ^a e^(−x^2 ) dx ? a>0](https://www.tinkutara.com/question/Q203349.png)
$${calculate}\:\int\int_{\left[\mathrm{0},{a}\right]^{\mathrm{2}} } \:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy} \\ $$$${can}\:{you}\:{find}\:\int_{\mathrm{0}} ^{{a}} {e}^{−{x}^{\mathrm{2}} } {dx}\:\:\:\:? \\ $$$${a}>\mathrm{0} \\ $$
Answered by witcher3 last updated on 17/Jan/24
![the idee ∫∫_([0,∞]^2 ) f(x,y)dxdy=∫_0 ^(π/2) ∫_0 ^∞ f(rcos(a),rsin(a))rdrda can′t hellp ther for a>0 [0,a]^2 is a square if you try to fill it withe disc is not easy 0≤rcos(z)≤a 0≤rsin(z)≤a 0<r<(a/(cos(z))),0≤r<(a/(sin(z))) ⇒r<min((a/(cos(z))),(a/(sin(z)))) =∫_0 ^(π/4) ∫_0 ^(a/(cos(z))) re^(−r^2 ) drdz+∫_(π/4) ^(π/2) ∫_0 ^(a/(sin(z))) re^(−r^2 ) drdz =∫_0 ^(π/4) ∫_0 ^(a/(cos(z))) 2re^(−r^2 ) drdz =∫_0 ^(π/4) (−e^(−r^2 ) ]_0 ^(a/(cos(z))) dz =∫_0 ^(π/4) 1−e^(−(a^2 /(cos^2 (z)))) dz tg(z)=t =∫_0 ^1 (1−e^(−a^2 (1+t^2 )) )(dt/(1+t^2 )),not easy from heree we can use feyneman and givre answer withe “erf” ∫_0 ^a e^(−x^2 ) dx....should bee first Quation=((√π)/2)erf(a) erf(z)=(2/( (√π)))∫_0 ^z e^(−t^2 ) dt Than ∫∫e^(−x^2 −y^2 ) dxdy=∫_0 ^a e^(−x^2 ) dx.∫_0 ^a e^(−y^2 ) dy =(∫_0 ^a e^(−x^2 ) )^2 dx=(π/4).erf^2 (a)](https://www.tinkutara.com/question/Q203358.png)
$$\mathrm{the}\:\mathrm{idee}\:\int\int_{\left[\mathrm{0},\infty\right]^{\mathrm{2}} } \mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dxdy}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \mathrm{f}\left(\mathrm{rcos}\left(\mathrm{a}\right),\mathrm{rsin}\left(\mathrm{a}\right)\right)\mathrm{rdrda} \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{hellp}\:\mathrm{ther}\: \\ $$$$\mathrm{for}\:\mathrm{a}>\mathrm{0}\:\left[\mathrm{0},\mathrm{a}\right]^{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{if}\:\mathrm{you}\:\mathrm{try}\:\mathrm{to}\:\mathrm{fill}\:\mathrm{it}\:\mathrm{withe}\:\mathrm{disc} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{easy} \\ $$$$\:\mathrm{0}\leqslant\mathrm{rcos}\left(\mathrm{z}\right)\leqslant\mathrm{a} \\ $$$$\mathrm{0}\leqslant\mathrm{rsin}\left(\mathrm{z}\right)\leqslant\mathrm{a} \\ $$$$\mathrm{0}<\mathrm{r}<\frac{\mathrm{a}}{\mathrm{cos}\left(\mathrm{z}\right)},\mathrm{0}\leqslant\mathrm{r}<\frac{\mathrm{a}}{\mathrm{sin}\left(\mathrm{z}\right)} \\ $$$$\Rightarrow\mathrm{r}<\mathrm{min}\left(\frac{\mathrm{a}}{\mathrm{cos}\left(\mathrm{z}\right)},\frac{\mathrm{a}}{\mathrm{sin}\left(\mathrm{z}\right)}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\mathrm{cos}\left(\mathrm{z}\right)}} \mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{drdz}+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\mathrm{sin}\left(\mathrm{z}\right)}} \mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{drdz} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\mathrm{cos}\left(\mathrm{z}\right)}} \mathrm{2re}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{drdz} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(−\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\frac{\mathrm{a}}{\mathrm{cos}\left(\mathrm{z}\right)}} \mathrm{dz} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{1}−\mathrm{e}^{−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{z}\right)}} \mathrm{dz} \\ $$$$\mathrm{tg}\left(\mathrm{z}\right)=\mathrm{t} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{e}^{−\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)} \right)\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} },\mathrm{not}\:\mathrm{easy}\:\mathrm{from}\:\mathrm{heree} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{feyneman}\:\mathrm{and}\:\mathrm{givre}\:\mathrm{answer}\:\mathrm{withe}\:“\mathrm{erf}'' \\ $$$$\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}….\mathrm{should}\:\mathrm{bee}\:\mathrm{first}\:\mathrm{Quation}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left(\mathrm{a}\right) \\ $$$$\mathrm{erf}\left(\mathrm{z}\right)=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{z}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt} \\ $$$$\mathrm{Than}\:\int\int\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } \mathrm{dxdy}=\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}.\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \mathrm{dy} \\ $$$$=\left(\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \right)^{\mathrm{2}} \mathrm{dx}=\frac{\pi}{\mathrm{4}}.\mathrm{erf}^{\mathrm{2}} \left(\mathrm{a}\right) \\ $$$$ \\ $$