What is the correct integral for rexp(-ar^2) using cylindrical coordinates?

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Homework Help Overview

The discussion revolves around the integration of the function involving an exponential decay term, specifically rexp(-ar^2), in the context of cylindrical coordinates. Participants are exploring the correct formulation of the integral and the implications of using different coordinate systems.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the integral, questioning whether it should be indefinite or related to area integration in polar coordinates. There is mention of the error function and the need for substitutions in the context of polar coordinates.

Discussion Status

The conversation is ongoing, with participants providing insights into the nature of the integral and the coordinate systems involved. One participant has clarified their initial misunderstanding regarding the integral's formulation, indicating a shift in focus towards cylindrical coordinates.

Contextual Notes

There is a noted confusion between cylindrical and spherical coordinates, which may affect the setup of the integral. The discussion also touches on the use of non-elementary functions like the error function in relation to the integral.

francescopadormo
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Hi! Can someone help me with this integral? I can't seem to get my head around it!

Integrate r^2.exp(-ar^2) imperfectly.

Thanks!
 
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Do you mean indefinitely? If so, this will involve erf, a non-elementary function. That is, unless you are doing an area integral in polar coordinates, in which case the actual function you integrate will be r^3e^(-ar^2), and this can be solved by substitution.
 
I imagine it would be easy to integrate it "imperfectly"! :smile:

As StatusX said: [itex]\int x^2 e^{-x^2}dx[/itex], like [itex]\int e^{-x^2}dx[/itex] has no expression in terms of elementary functions, only in terms of the "error" function Erf(x).

HOWEVER, since your variable is "r", it is possible that you are really attempting to integrate over polar coordinates, in which case you need to remember that the "differential of area" in polar coordinates is [itex]rdrd\theta[/itex]. And the "r" in that let's you make the substitution u= r2.
 
Sorry everyone! I was mistaken in what the integral should actually be: it should be the integral of rexp(-ar^2). I should be using cylindrical co-ords, not spherical.

Thanks for the help though!
 

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