Question Number 146067 by gsk2684 last updated on 10/Jul/21
$${transform}\:{the}\:{cartesian}\:{inyegral}\: \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\:\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {\int}}{e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \:{dy}\:{dx}\:{into}\:{polar}\:{integral}\: \\ $$$${and}\:{evaluate}\:{it}. \\ $$
Answered by Olaf_Thorendsen last updated on 10/Jul/21
![A = ∫_0 ^1 ∫_0 ^(√(1−x^2 )) e^(−(x^2 +y^2 )) dydx x = rcosθ, y = rsinθ dS = dydx = rdrdθ A = ∫_0 ^(π/2) ∫_0 ^1 e^(−r^2 ) rdrdθ A = (π/2)[−e^(−r^2 ) ]_0 ^1 = (π/2)(1−(1/e))](https://www.tinkutara.com/question/Q146068.png)
$${A}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dydx} \\ $$$${x}\:=\:{r}\mathrm{cos}\theta,\:{y}\:=\:{r}\mathrm{sin}\theta \\ $$$${dS}\:=\:{dydx}\:=\:{rdrd}\theta \\ $$$${A}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{r}^{\mathrm{2}} } {rdrd}\theta \\ $$$${A}\:=\:\frac{\pi}{\mathrm{2}}\left[−{e}^{−{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right) \\ $$
Commented by gsk2684 last updated on 11/Jul/21
![thank you sir ∫_0 ^(Π/2) [((−e^(−r^2 ) )/2)]_0 ^1 dθ = ((−1)/2)∫_0 ^(Π/2) (e^(−1) −1)dθ =((−1)/2)((1/e)−1)((Π/2))=(Π/4)(1−(1/e))](https://www.tinkutara.com/question/Q146078.png)
$${thank}\:{you}\:{sir} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\Pi}{\mathrm{2}}} {\int}}\left[\frac{−{e}^{−{r}^{\mathrm{2}} } }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} {d}\theta\:=\:\frac{−\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\Pi}{\mathrm{2}}} {\int}}\left({e}^{−\mathrm{1}} −\mathrm{1}\right){d}\theta \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{e}}−\mathrm{1}\right)\left(\frac{\Pi}{\mathrm{2}}\right)=\frac{\Pi}{\mathrm{4}}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right) \\ $$