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

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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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