Evaluation of double integration with polar coordinates

Asked May 25 '20 at 17:09

Active May 25 '20 at 17:19

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Evaluation of $$\iint_R e^{-(x^2+y^2)} \, dA$$

where $R$ is the region given by inequalities $x^2+y^2\leq 1$ and $0\leq y\leq\sqrt{3}x$

What i try:: drawing Circle and line. We get a sector which makes an angle of $\pi/3$ with $x$ axis.

Now put $x^2+y^2=r^2$ and $dx\,dy=r\,dr\,d\theta$

So integral is $$\int^{\frac{\pi}{3}}_0 \int^1_0 e^{-r^2}\,dr\,d\theta = \frac{\pi}{6} \bigg(e-1\bigg)$$

Can anyone please tell me is my solution is right? If not then tell me how do I solve it. Thanks.

edited Jun 12 '20 at 10:38

Community

asked May 25 '20 at 17:09

jacky

Yep... looks fine to me. – David G. Stork May 25 '20 at 17:17
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You need $\displaystyle \int^{\frac{\pi}{3}}_0 \int^1_0 e^{-r^2} r,dr,d\theta$ rather than $\displaystyle \int^{\frac{\pi}{3}}_0 \int^1_0 e^{-r^2},dr,d\theta. \qquad$ – Michael Hardy May 25 '20 at 17:21
But the value you get at the end is correct. – Michael Hardy May 25 '20 at 17:21
1

Isn't it rather $\int^{\frac{\pi}{3}}{0}\int^{1}{0}re^{-r^2}drd\theta=\frac{\pi}{6}\bigg(1-e^{-1}\bigg)?$ – user376343 May 25 '20 at 17:22
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I edited the question to change $\displaystyle \int\int_R e^{-(x^2+y^2)} , dA$ to $\displaystyle \iint_R e^{-(x^2+y^2)} , dA,$ coding it with \iint (with two consecutive "i"s). You can also write $\displaystyle \iint\limits_R e^{-(x^2+y^2)} , dA$ (with the subscript $R$ formatted differently). $ \qquad$ – Michael Hardy May 25 '20 at 17:23
Thanks so much Michael Hardy. – jacky May 25 '20 at 17:41

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