Gauss' Theorem - Divergence Theorem for Sphere

1. Oct 21, 2015

FaraDazed

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

3. 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...

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

Last edited: Oct 21, 2015
2. Oct 21, 2015

Geofleur

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

3. Oct 21, 2015

FaraDazed

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.

4. Oct 21, 2015

Geofleur

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, $\frac{\partial r}{\partial x} = \frac{x}{r}$.

5. Oct 21, 2015

FaraDazed

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

Last edited: Oct 21, 2015
6. Oct 21, 2015

Geofleur

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.

7. Oct 21, 2015

FaraDazed

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!!

8. Oct 21, 2015

FaraDazed

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted