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Question Number 26601 by 01779506249 last updated on 27/Dec/17

Use polar co-ordinates to evaluate ∫∫_R e^(−(x^2 +y^2 )) dA, where the region R is enclosed by the circle x^2 +y^2 =1.

$${Use}\:{polar}\:{co}-{ordinates}\:{to}\:{evaluate}\:\int\underset{{R}} {\int}{e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dA},\:{where}\:{the}\:{region}\:{R}\:{is}\:{enclosed}\:{by}\:{the}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}. \\ $$

Answered by ajfour last updated on 27/Dec/17

=∫_0 ^(  2π) ∫_0 ^(  1) e^(−r^2 ) (rdr)dθ  =π∫_1 ^(  0) e^(−r^2 ) (−2rdr)=π(e^(−r^2 ) )∣_1 ^0    =π(1−(1/e)) .

$$=\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:\:\mathrm{1}} {e}^{−{r}^{\mathrm{2}} } \left({rdr}\right){d}\theta \\ $$$$=\pi\int_{\mathrm{1}} ^{\:\:\mathrm{0}} {e}^{−{r}^{\mathrm{2}} } \left(−\mathrm{2}{rdr}\right)=\pi\left({e}^{−{r}^{\mathrm{2}} } \right)\mid_{\mathrm{1}} ^{\mathrm{0}} \\ $$$$\:=\pi\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right)\:. \\ $$

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