My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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
shashank dwivedi is offline  
 
April 14th, 2017, 06:06 AM   #2
Senior Member
 
romsek's Avatar
 
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 $2\pi$
romsek is online now  
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.
shashank dwivedi is offline  
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 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 y-axis horizontal, z-axis vertical. Romsek has the y-axis vertical, z-axis horizontal.)
Country Boy is offline  
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?
shashank dwivedi is offline  
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
. 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.
Country Boy is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
double, double integral, improper integral, integration



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Double integration spottedtango Calculus 2 May 23rd, 2014 11:12 AM
double integration zell^ Calculus 6 April 16th, 2012 09:39 AM
double integration help!! skdnjswnd Calculus 1 April 15th, 2009 07:20 PM
Double integration (x/y)*e^y Calculus 10 March 17th, 2009 04:28 PM
Please Help With Double Integration sp1227 Calculus 2 March 22nd, 2008 06:23 AM





Copyright © 2017 My Math Forum. All rights reserved.