# Gauss' Theorem - Divergence Theorem for Sphere

In summary, the conversation discusses a problem involving Gauss' Theorem and the surface integral of a vector field. The incorrect value of the divergence of the vector field is given, resulting in incorrect calculations. However, the conversation concludes with the correct value being used and the equations balancing out.

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

Geofleur said:
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.

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} ##.

Geofleur said:
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 $6r^3$!

Last edited:
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.

Geofleur said:
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!

Geofleur said:
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.

## 1. What is Gauss' Theorem - Divergence Theorem for Sphere?

Gauss' Theorem, also known as the Divergence Theorem for Sphere, is a mathematical principle that relates the flux of a vector field through a closed surface to the divergence of the vector field within the enclosed volume. It is a fundamental theorem in electromagnetism and fluid dynamics, and is often used in the calculation of electric and magnetic fields.

## 2. How is Gauss' Theorem - Divergence Theorem for Sphere used?

Gauss' Theorem is used to calculate the flux of a vector field through a closed surface. It can also be used to determine the enclosed volume of a sphere, as well as the total surface area of a sphere. Additionally, it is used in various applications in physics and engineering, such as in the calculation of electric and magnetic fields.

## 3. What is the formula for Gauss' Theorem - Divergence Theorem for Sphere?

The formula for Gauss' Theorem is: ∫∫∫S F · dA = ∫∫D &nabla · F dV, where F is the vector field, S is a closed surface, D is the enclosed volume, and &nabla · F is the divergence of the vector field F.

## 4. What are the assumptions behind Gauss' Theorem - Divergence Theorem for Sphere?

There are three main assumptions behind Gauss' Theorem: 1) the vector field must be continuous and differentiable within the enclosed volume, 2) the surface must be smooth and have a well-defined normal vector at every point, and 3) the surface must be closed, meaning it has no boundary.

## 5. Can Gauss' Theorem - Divergence Theorem for Sphere be applied to any shape besides a sphere?

Gauss' Theorem can be applied to any closed surface, not just a sphere. As long as the three assumptions mentioned in the previous question are met, the theorem can be used to calculate flux and other quantities for any shape. However, the theorem is particularly useful for spheres due to their symmetry and simplicity in calculations.

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