# Integral with divergence thm

1. Aug 14, 2010

### SpY]

1. The problem statement, all variables and given/known data
Evaluate the integral
$$\int\limits_{V=\infty} e^{-r} \left[ \nabla \cdot \frac {\widehat{r}} {r^2} \right] , d^3 x$$

2. Relevant equations
Divergence theorem:
$$\int\limits_{V} \left ( \nabla \cdot A \right ) \, d^3 x = \oint\limits_{S} A \cdot \, da}$$

3. 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
$$\int \frac{1}{e^r} \, dr = \frac{1}{e^r} \int \frac{1}{u} \frac{du}{e^r} = ln(e^r) = r$$
where I used a substitution $$u=e^r$$ and $$du= e^r dr$$

Last edited: Aug 14, 2010
2. Aug 15, 2010

### ehild

What is the divergence of vec(r)/r^2?

ehild

3. Aug 17, 2010

### SpY]

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?

4. Aug 17, 2010

### Char. Limit

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.

5. Aug 17, 2010

### SpY]

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

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

6. Aug 19, 2010