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 April 14th, 2017, 05:42 AM #1 Newbie   Joined: Apr 2017 From: India Posts: 20 Thanks: 0 Double integration While studying the plume rise Gaussian Model , I came across following Improper integral which I was unable to solve : ∫∫exp[-((y^2+z^2)/2)]dydz....Here Integral on the left side ranges from 0 to ∞ and the integral next to that varies from −∞ to ∞. The source I am referring says to apply the transformation y=r cosθ z=r sinθ I am unable to do solve this integral. Please help me out. Answer : 2π
 April 14th, 2017, 06:06 AM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,497 Thanks: 755 $\displaystyle \int_0^\infty \int_{-\infty}^{\infty}~e^{-\frac{(y^2+z^2)}{2}}~dy~dz$ $y = r\cos(\theta)$ $z = r \sin(\theta)$ This is just using polar coordinates so we know that $dy~dz \to r ~dr~d\theta$ $\displaystyle \int_{-\pi/2}^{\pi/2}\int_0^\infty~e^{-\frac{r^2}{2}}r~dr~d\theta$ now just let $u=\dfrac{ r^2}{2},~du = r~dr$ $\displaystyle \int_{-\pi/2}^{\pi/2}\int_0^\infty~e^{-u}du~d\theta$ $\pi\left(\left. e^{-u}\right|_{\infty}^0 \right) = \pi$ $\pi$ is the correct answer, not $2\pi$
 April 14th, 2017, 07:05 AM #3 Newbie   Joined: Apr 2017 From: India Posts: 20 Thanks: 0 I got it.Thanks a lot. But a quick question arose in mind that if in the same question , the limit of the left integral ranges from −∞ to ∞ (which was initially ranging from 0 to ∞) Does this cause any change in the the answer? If yes,then what is the answer in that case. Please elaborate.
 April 14th, 2017, 07:25 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707 Think of the yz-plane (a little peculiar but just like an xy-plane). With y going form negative infinity to infinity, z from 0 to infinity, we are dealing with the upper half plane. Using romsek's change of variable (polar coordinates as he says) r will still go from 0 to infinity but theta will go from 0 to $\pi$ rather than from 0 to $2\pi$ as it would if we were covering the entire plane. (Romsek, in going form $-pi/2$ to [/tex]\pi/2[/tex] covers the right half plane rather than the upper half plane. But because the function being integrated is symmetric in y and z, that will be the same thing. I have the y-axis horizontal, z-axis vertical. Romsek has the y-axis vertical, z-axis horizontal.)
 April 14th, 2017, 07:55 AM #5 Newbie   Joined: Apr 2017 From: India Posts: 20 Thanks: 0 Well, that concept of symmetry of y and z and choosing the limit -pi/2 to pi/2 instead of 0 to pi was interesting. When I solved the same question with modification like this: ∫∫exp[-((y^2+z^2)/2)]dydz Here the limit of the left integral is −∞ to ∞ and the limit of next integral is also −∞ to ∞. Then I reached the answer : 2pi I have used the same steps as suggested earlier in response to my question but would like to get it verified whether my answer is correct or not?
 April 14th, 2017, 12:35 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707 If both integrals are from negative infinity to infinity then you are covering the entire plane and must take r from 0 to infinity and theta from 0 to 2pi. That is $\int_0^\infty\_0^{2\pi} e^{-r^2) (rd\theta)dr$. That can be separated as the product $\left(\int_0^{2\pi}d\theta\right)\left(\int_0^\inf ty r dr\right)$. Integrating with respect to $\theta$first, we have $2\pi\int_0^\infty r e^{-r^2}dr$. To do that last integral, let $u= r^2$ so that $du= 2r dr$ so that becomes $2\pi\int_0^\infty e^{-u}du= \left[-e^{-u}\right]_0^\infty]= 2\pi$ since $e^{-u}$ goes to 0 as u goes to infinity.

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