What is the integral of a vector field with the divergence theorem?

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Homework Help Overview

The discussion revolves around evaluating an integral involving a vector field and the application of the divergence theorem. The integral in question includes an exponential term and is set over a volume that extends to infinity, raising questions about its implications and the nature of the divergence involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the divergence theorem and express confusion regarding the exponential term and the implications of integrating over an infinite volume. There are attempts to clarify the divergence of the vector field and its relation to delta functions.

Discussion Status

Some participants have offered insights into the nature of the divergence and the integral's components, while others are questioning their own reasoning and the relevance of certain mathematical steps. The discussion is ongoing with multiple interpretations being explored.

Contextual Notes

There is mention of the divergence theorem and its application to a triple integral involving delta functions, indicating a complex setup that may require careful consideration of the limits and definitions involved.

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Homework Statement


Evaluate the integral
\int\limits_{V=\infty} e^{-r} \left[ \nabla \cdot \frac {\widehat{r}} {r^2} \right] , d^3 x

Homework Equations


Divergence theorem:
\int\limits_{V} \left ( \nabla \cdot A \right ) \, d^3 x<br /> = \oint\limits_{S} A \cdot \, da}<br />

The Attempt at a Solution


I know that I have to apply the div theorem somewhere, but this e^{-r} is confusing and what does it mean if the lower limit V is infinity?
I haven't seen the integral of \frac{1}{e^r} before but I'm kinda guessing
\int \frac{1}{e^r} \, dr <br /> = \frac{1}{e^r} \int \frac{1}{u} \frac{du}{e^r}<br /> = ln(e^r)<br /> = r<br />
where I used a substitution u=e^r and du= e^r dr
 
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What is the divergence of vec(r)/r^2?

ehild
 
It's defined as
4 \pi \delta x \delta y \delta z

but then I don't know how to apply Stokes' (which I guess to use because of the d^3 x and V in the integral. Could I split it into a triple integral and \delta x dx at a time?
 
I know that I have to apply the div theorem somewhere, but this e^{-r} is confusing and what does it mean if the lower limit V is infinity?
I haven't seen the integral of \frac{1}{e^r} before but I'm kinda guessing
\int \frac{1}{e^r} \, dr <br /> = \frac{1}{e^r} \int \frac{1}{u} \frac{du}{e^r}<br /> = ln(e^r)<br /> = r<br />
where I used a substitution u=e^r and du= e^r dr

This is wrong.

\frac{1}{e^r} = e^{-r}

\int e^{-r} dr

u=-r, du=-dr, -du=dr

\int -e^u du = -e^u = -e^{-r} = \frac{-1}{e^r}

The integral of e^-r isn't r, as that would imply that e^-r is a constant number.
 
I agree, my above reasoning was useless

Ok so I can integrate the e^-r but I don't think that really matters when there's a delta in the integral... my main problem is how to solve a third order delta integral, probably using the Divergence theorem because of the third order and volume. So we have

4 pi \int_{V=\infty} e^{-r} {\delta}^3 x z d^3 x
 

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