# Gauss' Theorem - Divergence Theorem for Sphere

## Homework Statement

Using the fact that $\nabla \cdot r^3 \vec{r} = 6 r^2$ (where $\vec{F(\vec{r})} = r^3 \vec{r}$) where S is the surface of a sphere of radius R centred at the origin.

## Homework Equations

$$\int \int \int_V \nabla \cdot \vec{F} dV =\int \int_S \vec{F} \cdot d \vec{S}$$
That is meant to be a double surface integral but not sure how to do that in latex

## The Attempt at a Solution

Very new to this, so be kind :)

for the LHS
$$\int \int \int_V \nabla \cdot \vec{F} dV = 6 \int \int \int_V r^2 dV \\ = 6 \int_0^R r^2 r^2 dr \int_0^{2 \pi} d\phi \int_0^{\pi} \sin{\theta} d\theta \\ = 6 [\frac{1}{5}r^5]_0^R [\phi]_0^{\pi} [\cos{\theta}]_0^{2 \pi} \\ = 6(\frac{1}{5} R^5) (2 \pi ) (2) = \frac{24}{5} \pi R^5$$

for the RHS
$$\int \int_S \vec{F} \cdot d\vec{S} = \int \int_S r^3 \vec{r} \cdot d \vec{S} = \int \int_S r^3 \vec{r} \cdot \vec{\hat{n}} dS \\$$

Then I calc $r^3 \vec{r} \cdot \vec{\hat{n}}$

$$r^3 \vec{r} \cdot \vec{\hat{n}} = r^3 \vec{r} \cdot \frac{r^3 \vec{r}}{r^3 r} = \frac{r^2 r^3}{r} = r^4$$

so...

$$\int \int_S r^3 \vec{r} \cdot \vec{\hat{n}} dS = \int \int_S r^4 dS \\ = \int_0^{2 \pi} d \phi \int_0^{\pi} R^4 R^2 \sin{\theta} d \theta \\ = R^6 [\phi]_0^{2 \pi} [\cos{\theta}]_0^{\pi} = 4 \pi R^6$$

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## Answers and Replies

Geofleur
Science Advisor
Gold Member
Well, for starters, I don't think that ## \nabla \cdot r^3 \mathbf{r} = 6r^2 ##.

Well, for starters, I don't think that ## \nabla \cdot r^3 \mathbf{r} = 6r^2 ##.

That is a given though, it is part of the question. It says, "given that ## \nabla \cdot r^3 \mathbf{r} = 6r^2 ## ..."

I am assuming that the vector r=(x,y,z) as that is what is used throughout the coursework.

Geofleur
Science Advisor
Gold Member
First, note that, if ## r ## has the dimension of length (##L##), then ##r^3 \mathbf{r} ## must have the dimension of ## L^4 ##. But the derivative has the dimension of ## 1 / L ##, so the result must have the dimension of ## L^3 ##, not ## L^2 ##. Second, you can just calculate it. Hint: Write ## r^3 \mathbf{r}## as ##r^3(x\mathbf{i} + y \mathbf{j} + z\mathbf{k}) ## and note that, for example, ## \frac{\partial r}{\partial x} = \frac{x}{r} ##.

First, note that, if ## r ## has the dimension of length (##L##), then ##r^3 \mathbf{r} ## must have the dimension of ## L^4 ##. But the derivative has the dimension of ## 1 / L ##, so the result must have the dimension of ## L^3 ##, not ## L^2 ##. Second, you can just calculate it. Hint: Write ## r^3 \mathbf{r}## as ##r^3(x\mathbf{i} + y \mathbf{j} + z\mathbf{k}) ## and note that, for example, ## \frac{\partial r}{\partial x} = \frac{x}{r} ##.

EDIT: Yes I have just calculated it, and i do get $6r^3$!!

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Geofleur
Science Advisor
Gold Member
Even the greatest books have been known to contain errors. Besides, if you do the problem with the correct value of ## \nabla \cdot \mathbf{F} ##, everything works out the way it's supposed to.

Even the greatest books have been known to contain errors. Besides, if you do the problem with the correct value of ## \nabla \cdot \mathbf{F} ##, everything works out the way it's supposed to.
Yes, see my edit above. I get [itex[6r^3[/itex] ! Haha. My lecturer may have told us about the error, but if he did, I was either not there or didnt hear it! haha

Thanks!!

Even the greatest books have been known to contain errors. Besides, if you do the problem with the correct value of ## \nabla \cdot \mathbf{F} ##, everything works out the way it's supposed to.

Yes I can see now that the LHS would also equal $4 \pi R^6$ when using the correct value for the divergence of F. No wonder they were not equal to eachother! I should have realised the mistake myself to be fair, but this was my first actual problem using Gauss' Theorem, thanks for your help.