# Proving Divergence Theorem using Gauss' Theorem

• latentcorpse
In summary, the conversation discusses using Gauss' Theorem (Divergence Theorem) and integration by parts to show that a certain integral is equal to another. The participants go through the steps of the derivation, including setting up the necessary vector identities and using spherical coordinates. In the end, they arrive at the solution and discuss the limits of the integral to show that it equals zero. The expert summarizes the conversation and provides a clear and concise explanation of the steps taken.
latentcorpse
I need to show that, using Gauss' Theorem (Divergence Theorem), i.e. integration by parts, that:

$\int_V dV e^{-r} \nabla \cdot (\frac{\vec{\hat{r}}}{r^2}) = \int_V dV \frac{e^{-r}}{r^2}$

any ideas on where to start?

Hint:The following vector identity should be useful:

$$\vec{\nabla}\cdot{f\vec{A}}=f(\vec{\nabla}\cdot\vec{A})+\vec{A}\cdot(\vec{\nabla}f)$$

thanks.
1, do i set $f=\frac{1}{r^2}$?
2, how did you know to do that so quickly?

1. No, set $$f=e^{-r}$$

2. It's a common theme in many EM derivations.

ok so it then equals
$\int_V dV \nabla \cdot (\frac{e^{-r} \vec{\hat{r}}}{r^2}) - \int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) = \int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS - \int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r})$
My instinct is to do the integral over the surface S using S = sphere of radius a as then $\vec{n}=\vec{\hat{r}}$ and i can get rid of that dot product. but then $dS=4 \pi a^2$ which messes stuff up and also what do I do on the right?

I'm assuming that the original integral was over all of space, correct?

If so, then your surface integral is over the boundary of all space which is at $$r\to\infty$$...what is the value of the integrand there?

For the second term; Take the gradient of $$e^{-r}$$ using spherical coords...

ok. might be getting somewhere now. on the right i took the grad of the exponential term and got $\nabla e^{-r} = -e^{-r} \vec{\hat{r}}$ is that right?
seems to be so because that causes the integral on the right to simplify substantially to $-\int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) = + \int_V dV \frac{e^{-r}}{r^2}$ which is what we're looking for - therefore it appears to just be a matter of showing the integral on the left vanishes?

indeed it is over all space. i would then imagine it would zero because $e^{-r} \rightarrow 0$ as $r \rightarrow \infty$. If this is the case, how owuld you recommend I write that in order to be an acceptable answer? Also, I took the grad of the exponential using Cartesians - does that matter?

latentcorpse said:
indeed it is over all space. i would then imagine it would zero because $e^{-r} \rightarrow 0$ as $r \rightarrow \infty$. If this is the case, how owuld you recommend I write that in order to be an acceptable answer? Also, I took the grad of the exponential using Cartesians - does that matter?

Formally, you could start by assuming that your volume is a sphere of radius $a$ and then take the limit as $a$ approaches infinity.

And it doesn't matter that you used Cartesian coords to take the gradient; although it would probably have been quicker to use spherical coords.

ok
but then
$\int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS = \int_S (\frac{e^{-r} \vec{\hat{r}} \cdot \vec{\hat{r}}}{r^2}) dS= \frac{e^{-r}}{r^2} 4 \pi a^2$
if i take the limit as $a \rightarrow \infty$ i get $\infty$

unless i take $dS=4 \pi r^2$ and then the intergal becomes $4 \pi e^{-r}$ and then the limit as $r \rightarrow \infty$ would be 0. Any advice? Cheers.

latentcorpse said:
ok
but then
$\int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS = \int_S (\frac{e^{-r} \vec{\hat{r}} \cdot \vec{\hat{r}}}{r^2}) dS= \frac{e^{-r}}{r^2} 4 \pi a^2$
if i take the limit as $a \rightarrow \infty$ i get $\infty$

Along the surface, $r=a$.

i don't understand what you mean there sorry

actually surely if i integrate over a sphere of radius a, $dS=a^2 \sin{\theta} d\theta d\phi$?

in which case this integral becomes $\int_S \frac{e^{-r} a^2}{r^2} \sin{\theta} d\theta d\phi = 4 \pi \frac{a^2}{r^2} e^{-r}$

would i then let a or r go to infinity, or is this all complete bollocks? HELP I am really confused now! arghhhh!

$r$ is the distance from the origin to the infinitesimal area element $dS$. Since the entire surface is a distance $a$ from the origin, that means that $r=a$ everywhere on the surface.

$$\implies \int_S \frac{e^{-r}}{r^2}dS=\int_S \frac{e^{-a}}{a^2} dS =\int_S \frac{e^{-a} a^2}{a^2} \sin{\theta} d\theta d\phi = 4 \pi e^{-a}$$

cheers buddy!

## 1. What is the Divergence Theorem and how does it relate to Gauss' Theorem?

The Divergence Theorem, also known as Gauss's Theorem or Ostrogradsky's Theorem, is a fundamental result in vector calculus that relates the flux of a vector field through a closed surface to the triple integral of the divergence of the field over the volume enclosed by the surface. In other words, it states that the flux of a vector field through a closed surface is equal to the volume integral of the vector field's divergence over the enclosed volume.

## 2. How is the Divergence Theorem used to prove Gauss' Theorem?

The Divergence Theorem can be used to prove Gauss' Theorem by showing that both sides of the equation are equivalent. This is done by converting the surface integral on the left side of the equation into a volume integral using the Divergence Theorem. Then, by comparing the integrands on both sides, we can show that they are equal and thus prove the theorem.

## 3. What are the prerequisites for understanding the proof of the Divergence Theorem using Gauss' Theorem?

A strong understanding of vector calculus, including vector fields, line integrals, and surface integrals, is necessary to understand the proof of the Divergence Theorem using Gauss' Theorem. It is also helpful to have a familiarity with multivariable calculus and the concepts of divergence and curl of a vector field.

## 4. Can the Divergence Theorem be used in any situation, or are there limitations?

The Divergence Theorem can only be applied to vector fields that are continuous and have continuous first-order partial derivatives over the enclosed volume. Additionally, the surface being integrated over must be smooth and closed. If these conditions are not met, the Divergence Theorem cannot be used.

## 5. Why is the Divergence Theorem important in the field of physics?

The Divergence Theorem is important in physics because it allows us to solve problems involving the flow of a vector field through a closed surface, which is a common scenario in many physical systems. It is also a powerful tool for simplifying complex integrals and making calculations more manageable. Additionally, the Divergence Theorem has applications in various fields of physics, such as electromagnetism, fluid dynamics, and thermodynamics.

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