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

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In summary, the integral in question is r^2.exp(-ar^2) and it can be solved by substitution in polar coordinates or with the use of the error function. However, if the intended variable is "r", then the integral should be rexp(-ar^2) and cylindrical coordinates should be used instead of spherical.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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!
 

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

What is the meaning of r^2.exp(-ar^2) in the context of integral help?

In integral help, r^2.exp(-ar^2) represents a specific type of function that is commonly encountered when solving integrals. It is a Gaussian function, also known as a bell curve, that is often used to model various natural phenomena.

How do you solve integrals involving r^2.exp(-ar^2)?

To solve integrals involving r^2.exp(-ar^2), you can use various techniques such as integration by parts, substitution, or completing the square. The specific method used will depend on the specific integral and its complexity.

What is the significance of the "a" value in r^2.exp(-ar^2)?

The "a" value in r^2.exp(-ar^2) represents a constant that affects the shape and magnitude of the Gaussian function. It determines the width of the curve and how quickly it approaches zero as r increases.

Can you provide an example of a real-world application of r^2.exp(-ar^2)?

One example of a real-world application of r^2.exp(-ar^2) is in physics, specifically in the study of Brownian motion. This function can be used to model the probability distribution of the displacement of a particle undergoing random motion in a fluid.

How does changing the "a" value affect the integral of r^2.exp(-ar^2)?

Changing the "a" value in r^2.exp(-ar^2) can significantly affect the integral of the function. It can change the overall area under the curve, the location of the maximum value, and the symmetry of the curve. This is why the "a" value is an essential factor to consider when solving integrals involving this type of function.

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