Does the divergence theorem work for a specific vector field?

[Tom McCurdy]

Homework Statement

Show divergence theorem works
For the vector field $$E = \hat{r}10e^{-r}-\hat{z}3z$$

Homework Equations

$$\int_{v}\nabla \cdot E dv = \oint_{s} E \cdot ds$$

The Attempt at a Solution

$$\nabla \cdot E = 1/r \frac{d}{dr}(rAr)+1/r\frac{dA\phi}{d\phi}+\frac{dAz}{dz}$$
Ar=10e^(-r)
Aphi=0
Az=-3z

$$\nabla \cdot E = \frac{1}{r}(10e^{-r}-10re^{-r})+3$$

$$\int_{0}^{2\pi}\int_{0}^{4}\int_{0}^{2} (r)(\frac{1}{r}(10e^{-r}-10re^{-r})+3) dr dz d\phi = -82.77$$

$$\oint_{s} E \cdot ds= \int_{0}^{2} \int_{0}^{4} (\hat{r}10e^{-r}-\hat{z}3z)\cdot(16 \pi \hat{r}+4\pi\hat{z}) = 2341.7$$

wow that took awhile to type

 

[jhicks]

The surface form you chose to integrate over wasn't my first choice to use. The volume integral looks correct (barring calculator errors), but could you explain your thought process for the surface geometry?

 

[HallsofIvy]

Homework Statement

Show divergence theorem works
For the vector field $$E = \hat{r}10e^{-r}-\hat{z}3z$$

You haven't given the full problem. What are s and v? What is the surface to be integrated over and its boundary?

Homework Equations

$$\int_{v}\nabla \cdot E dv = \oint_{s} E \cdot ds$$

The Attempt at a Solution

$$\nabla \cdot E = 1/r \frac{d}{dr}(rAr)+1/r\frac{dA\phi}{d\phi}+\frac{dAz}{dz}$$
Ar=10e^(-r)
Aphi=0
Az=-3z

$$\nabla \cdot E = \frac{1}{r}(10e^{-r}-10re^{-r})+3$$

$$\int_{0}^{2\pi}\int_{0}^{4}\int_{0}^{2} (r)(\frac{1}{r}(10e^{-r}-10re^{-r})+3) dr dz d\phi = -82.77$$

$$\oint_{s} E \cdot ds= \int_{0}^{2} \int_{0}^{4} (\hat{r}10e^{-r}-\hat{z}3z)\cdot(16 \pi \hat{r}+4\pi\hat{z}) = 2341.7$$

wow that took awhile to type

 

[Tom McCurdy]

oh shoot i forgot to say it is bounded by the cylinder z=0 z=4 r=2

my thought process from the surface integral was
i found ds in r direction or the surface area normal to r 16 pi
and surface area normal to z and doted it with E

 

[Dick]

The surface integral has two parts. There's the cylinder surface and then the two disks at each end. Looking at the 'normal' part of your integral, I think you know this, but you can't combine them both into a single integral. Split them up. It also looks like you are including some sort of area factor into the normal and then also integrating.