Gauss' Theorem - Divergence Theorem for Sphere

[FaraDazed]

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

<br /> \int \int \int_V \nabla \cdot \vec{F} dV =\int \int_S \vec{F} \cdot d \vec{S}<br />
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
<br /> \int \int \int_V \nabla \cdot \vec{F} dV = 6 \int \int \int_V r^2 dV \\<br /> = 6 \int_0^R r^2 r^2 dr \int_0^{2 \pi} d\phi \int_0^{\pi} \sin{\theta} d\theta \\<br /> = 6 [\frac{1}{5}r^5]_0^R [\phi]_0^{\pi} [\cos{\theta}]_0^{2 \pi} \\<br /> = 6(\frac{1}{5} R^5) (2 \pi ) (2) = \frac{24}{5} \pi R^5<br />

for the RHS
<br /> \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 \\<br />

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

<br /> 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<br />

so...

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

 

[Geofleur]

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

 

[FaraDazed]

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]

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

 

[FaraDazed]

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!

 

[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.

 

[FaraDazed]

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!

 

[FaraDazed]

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 each other! I should have realized the mistake myself to be fair, but this was my first actual problem using Gauss' Theorem, thanks for your help.