April 14th, 2017, 05:42 AM  #1 
Newbie Joined: Apr 2017 From: India Posts: 16 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: CA Posts: 1,238 Thanks: 637 
$\displaystyle \int_0^\infty \int_{\infty}^{\infty}~e^{\frac{(y^2+z^2)}{2}}~dy~dz$ $y = r\cos(\theta)$ $z = r \sin(\theta)$ 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: 16 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: 2,484 Thanks: 628 
Think of the yzplane (a little peculiar but just like an xyplane). 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 rather than from 0 to as it would if we were covering the entire plane. (Romsek, in going form 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 yaxis horizontal, zaxis vertical. Romsek has the yaxis vertical, zaxis horizontal.)

April 14th, 2017, 07:55 AM  #5 
Newbie Joined: Apr 2017 From: India Posts: 16 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: 2,484 Thanks: 628 
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 . That can be separated as the product . Integrating with respect to first, we have . To do that last integral, let so that so that becomes since goes to 0 as u goes to infinity. 

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double, double integral, improper integral, integration 
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