Showing Divergence Theorem Equivalence

[Eruditee]

Homework Statement

The problem states that a cube encloses charge. This cube is given in three space by <0,0,0> and <a,a,a>. The electric field is given by:
$\hat{E}=\frac{4e}{a^{2}e_{0}}[\frac{xy}{a^{2}}\hat{i}+\frac{(y-x)}{a}\hat{j}+\frac{xyz}{a^{2}}\hat{k}]$. I am to find the total charge enclosed using both methods or "sides" of the divergence theorem equivalence.

Homework Equations

$\displaystyle \iiint \limits_U \left({\nabla \cdot \mathbf F} \right) d V = \iint \limits_{\partial U} \mathbf F \cdot \mathbf n \ d S$
$\nabla=\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}$

The Attempt at a Solution

The first half worked out fine and makes sense, so I think this part is right. Do note; I do get a bit sloppy with notation; this is a rough draft:
$\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}\nabla\bullet EdV$
$\nabla\bullet\hat{E}=\frac{4e}{a^{2}e_{0}}[ \frac{y}{a^{2}}\hat{i}+\frac{1}{a}\hat{j}+\frac{xy}{a^{2}}\hat{k}]$

$\frac{4e}{a^{2}e_{0}}\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}(\frac{y}{a^{2}}+\frac{1}{a}+\frac{xy}{a^{2}})dV$
[itex]=\frac{4e}{a^{2}e_{0}}\int(\frac{y}{a^{2}}+\frac{1}{a}+\frac{xy}{a^{2}})dx(dA)=\frac{4e}{a^{2}e_{0}}(\int(\frac{y}{a^{2}})dx+\int\frac{1}{a}dx)+\int\frac{xy}{a^{2}}dx)dA=\frac{yx}{a^{2}}+\frac{x}{a}+\frac{x^{2}y}{2a^{2}}|_{0}^{a}=\frac{y}{a}+1+\frac{y}{2}
[\itex]
$=\frac{4e}{a^{2}e_{0}}\int_{0}^{a}(\frac{y}{a}+1+\frac{y}{2}dy)dz=\frac{y^{2}}{2a}+y+\frac{y^{2}}{4a}|_{0}^{a}=\frac{a}{2}+a+\frac{a}{4}$
$=\int\frac{a}{2}+a+\frac{a}{4}dz=\frac{az}{2}+az+\frac{az}{4}=\frac{a^{2}}{2}+a^{2}+\frac{a^{2}}{4}=a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}=a^{2}(\frac{1}{2}+\frac{1}{4})=\frac{3a^{2}}{4}*\frac{4e}{a^{2}e_{0}}=\frac{3e}{e_{0}}$
which makes sense, as there are 3 faces in which flux can go by?

However, the surface integral is a mess:
when z=a ; Sz(surface raised from xy by a)=a
$\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet-kdxdy =-\int_{0}^{a}\frac{xyz}{a^{2}}dxdy=-z\int_{0}^{a}\frac{xy}{a^{2}}dxdy=-\int\frac{x^{2}y}{2a^{2}}dy=\frac{-x^{2}z}{2a^{2}}\int ydy=\frac{-x^{2}zy^{2}}{4a^{2}}=-\frac{a^{3}}{4}$
likewise:
$\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet jdxdy =\frac{1}{a}\int_{0}^{a}(y-x)dxdz=\frac{1}{a}\int yx-\frac{x^{2}}{2}dz=yxz-\frac{x^{2}z}{2}=a^{2}-\frac{a^{2}}{2}$
$\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet idydz =\int \frac{xy}{a^{2}}dydz=\frac{x}{a^{2}}\int ydydz=\int\frac{xy^{2}}{2}dz=\frac{xy^{2}z}{a^{2}}=\frac{a^{2}}{2}$
And the total:
$a^{2}-\frac{a^{3}}{4}=\frac{4a^{2}-a^{3}}{4}=\frac{a^{2}(4-a)}{4}$

which I know is wrong because the flux is zero at a=4.

Where am I going wrong? I haven't done surface/volume integrals in about 4-5 years, so bear with me. I'm trying to relearn through Boas.

 

[vela]

Homework Statement

The problem states that a cube encloses charge. This cube is given in three space by <0,0,0> and <a,a,a>. The electric field is given by:
$\hat{E}=\frac{4e}{a^{2}e_{0}}[\frac{xy}{a^{2}}\hat{i}+\frac{(y-x)}{a}\hat{j}+\frac{xyz}{a^{2}}\hat{k}]$. I am to find the total charge enclosed using both methods or "sides" of the divergence theorem equivalence.

Homework Equations

$\displaystyle \iiint \limits_U \left({\nabla \cdot \mathbf F} \right) d V = \iint \limits_{\partial U} \mathbf F \cdot \mathbf n \ d S$
$\nabla=\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}$

The Attempt at a Solution

The first half worked out fine and makes sense, so I think this part is right. Do note; I do get a bit sloppy with notation; this is a rough draft:
$\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}\nabla\bullet EdV$
$\nabla\bullet\hat{E}=\frac{4e}{a^{2}e_{0}}[ \frac{y}{a^{2}}\hat{i}+\frac{1}{a}\hat{j}+\frac{xy}{a^{2}}\hat{k}]$

$\frac{4e}{a^{2}e_{0}}\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}(\frac{y}{a^{2}}+\frac{1}{a}+\frac{xy}{a^{2}})dV$
[itex]=\frac{4e}{a^{2}e_{0}}\int(\frac{y}{a^{2}}+\frac{1}{a}+\frac{xy}{a^{2}})dx(dA)=\frac{4e}{a^{2}e_{0}}(\int(\frac{y}{a^{2}})dx+\int\frac{1}{a}dx)+\int\frac{xy}{a^{2}}dx)dA=\frac{yx}{a^{2}}+\frac{x}{a}+\frac{x^{2}y}{2a^{2}}|_{0}^{a}=\frac{y}{a}+1+\frac{y}{2}
[\itex]
$=\frac{4e}{a^{2}e_{0}}\int_{0}^{a}(\frac{y}{a}+1+\frac{y}{2}dy)dz=\frac{y^{2}}{2a}+y+\frac{y^{2}}{4a}|_{0}^{a}=\frac{a}{2}+a+\frac{a}{4}$
$=\int\frac{a}{2}+a+\frac{a}{4}dz=\frac{az}{2}+az+\frac{az}{4}=\frac{a^{2}}{2}+a^{2}+\frac{a^{2}}{4}=a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}=a^{2}(\frac{1}{2}+\frac{1}{4})=\frac{3a^{2}}{4}*\frac{4e}{a^{2}e_{0}}=\frac{3e}{e_{0}}$
which makes sense, as there are 3 faces in which flux can go by?

However, the surface integral is a mess:
when z=a ; Sz(surface raised from xy by a)=a
$\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet-kdxdy =-\int_{0}^{a}\frac{xyz}{a^{2}}dxdy=-z\int_{0}^{a}\frac{xy}{a^{2}}dxdy=-\int\frac{x^{2}y}{2a^{2}}dy=\frac{-x^{2}z}{2a^{2}}\int ydy=\frac{-x^{2}zy^{2}}{4a^{2}}=-\frac{a^{3}}{4}$

Your normal is pointing in the wrong direction. It should point outward from the volume, so on the z=a face, it should point in the $+\hat{k}$ direction.

likewise:
$\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet jdxdy =\frac{1}{a}\int_{0}^{a}(y-x)dxdz=\frac{1}{a}\int yx-\frac{x^{2}}{2}dz=yxz-\frac{x^{2}z}{2}=a^{2}-\frac{a^{2}}{2}$
$\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet idydz =\int \frac{xy}{a^{2}}dydz=\frac{x}{a^{2}}\int ydydz=\int\frac{xy^{2}}{2}dz=\frac{xy^{2}z}{a^{2}}=\frac{a^{2}}{2}$
And the total:
$a^{2}-\frac{a^{3}}{4}=\frac{4a^{2}-a^{3}}{4}=\frac{a^{2}(4-a)}{4}$

which I know is wrong because the flux is zero at a=4.

Where am I going wrong? I haven't done surface/volume integrals in about 4-5 years, so bear with me. I'm trying to relearn through Boas.

You need to calculate the flux through all six faces and sum them.

Also, your expression for $\vec{E}$ looks wrong. Should the z-component be $xyz/a^3$ rather than $xyz/a^2$? It doesn't work out dimensionally otherwise.

Finally, I got a different result for the integral of the divergence, so you probably made an error integrating somewhere. Your setup looked fine.

 

[Eruditee]

Corrected

Your normal is pointing in the wrong direction. It should point outward from the volume, so on the z=a face, it should point in the $+\hat{k}$ direction.


You need to calculate the flux through all six faces and sum them.

Also, your expression for $\vec{E}$ looks wrong. Should the z-component be $xyz/a^3$ rather than $xyz/a^2$? It doesn't work out dimensionally otherwise.

Finally, I got a different result for the integral of the divergence, so you probably made an error integrating somewhere. Your setup looked fine.

Thankx, it was a cubed!

I actually made a mistake in the addition; it should be:
$=\int\frac{a}{2}+a+\frac{a}{4}dz=\frac{az}{2}+az+\frac{az}{4}=\frac{a^{2}}{2}+a^{2}+\frac{a^{2}}{4}=a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}=a^{2}(\frac{1}{2}+\frac{1}{4})+a^{2}=\frac{7a^{2}}{4}*\frac{7e}{4a^{2}e_{0}}=\frac{7e}{4e_{0}}$
I'm a bit lost with the surface integral.

 

[vela]

Thankx, it was a cubed!

I redid the integral. Somehow, I decided to add an a in just the right place to make it consistent with the a cubed variety. Also, it comes out pretty neatly to be 3 times the charge through a plane, which makes sense because there's three flux possibilities? The professor I have seems to set up most of his problems to have a usually simple answer.

I'm not sure what you mean by three flux possibilities, but the integral should evaluate to $\frac{7 e}{\epsilon_0}$.

It does look like that's the correct integral of divergence, but I'm not integrating properly for the surfaces. If I hold a surface constant, doesn't that mean the volume is just the area*the height of the surface, which is what I'm trying to do?

There's no volume involved in doing the surface integrals. I'm not sure what you're getting at.

 

[Eruditee]

I'm not sure what you mean by three flux possibilities, but the integral should evaluate to $\frac{7 e}{\epsilon_0}$.


There's no volume involved in doing the surface integrals. I'm not sure what you're getting at.

Sorry, I edited it before seeing your response. I'm getting 7/4ths. I'll try it again. thank you.

 

[Eruditee]

Right Answer

Sorry, I edited it before seeing your response. I'm getting 7/4ths. I'll try it again. thank you.

Yes, same answer youu got, forgot the constant coeff. I should do this much more neatly in lyx; the whole issue was adding improperly there.

Ok, so I finally got it done:
$=\frac{4e}{a^{2}e_{0}}\int(\frac{y}{a^{2}}+\frac{1}{a}+\frac{xy}{a^{3}})dx(dA)=\frac{4e}{a^{2}e_{0}}(\int(\frac{y}{a^{2}})dx+\int\frac{1}{a}dx)+\int\frac{xy}{a^{3}}dx)dA=\frac{yx}{a^{2}}+\frac{x}{a}+\frac{x^{2}y}{2a^{3}}|_{0}^{a}=\frac{y}{a}+1+\frac{y}{2a}$
$=\frac{4e}{a^{2}e_{0}}\int_{0}^{a}(\frac{y}{a}+1+\frac{y}{2a}dy)dz=\frac{y^{2}}{2a}+y+\frac{y^{2}}{4a}|_{0}^{a}=\frac{a}{2}+a+\frac{a}{4}$
$=\int\frac{a}{2}+a+\frac{a}{4}dz=\frac{az}{2}+az+\frac{az}{4}=\frac{a^{2}}{2}+a^{2}+\frac{a^{2}}{4}=a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}=a^{2}(\frac{1}{2}+\frac{1}{4})+a^{2}=\frac{7a^{2}}{4}*\frac{4e}{a^{2}e_{0}}=$
$\frac{7e}{e_{0}}$

And by the surface integral:
$First at x=0 (the yz axis) \int\int F\bullet-idydx=0$
$x=a \int\int F\bullet idydx=\frac{x}{a^{2}}\int\int ydydz=\frac{y^{2}zx}{2a^{2}}=\frac{a^{4}}{2a^{2}}=\frac{a^{2}}{2}$
$at y=0 \int\int F\bullet-jdxdz=\frac{1}{a}\int\int xdxdy=\frac{x^{2}z}{2a}=\frac{a^{2}}{2}$
This is where I went wrong; I automatically assumed this plane to be of zero when there was a y-x term not y-xy term or so.
$at y = a \int\int F\bullet jdxdz=\frac{1}{a}\int(y-x)dxdz=yxz-\frac{x^{2}z}{2}=\frac{a^{3}-\frac{a^{3}}{2}}{a}=a^{2}-\frac{a^{2}}{2}$
$at z = 0 \int\int F\bullet-kdxdz=0$
$at z=a \int\int F\bullet kdxdz=\frac{z}{a^{3}}\int xydxdy=\frac{x^{2}y^{2}z}{4a^{3}}=\frac{a^{5}}{4a^{3}}=\frac{a^{2}}{4}$

$\frac{a^{2}}{2}+\frac{a^{2}}{2}+a^{2}-\frac{a^{2}}{2}+\frac{a}{4}=a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}=a^{2}(\frac{1}{2}+\frac{1}{4})+a^{2}=\frac{7a^{2}}{4}*\frac{4e}{a^{2}e_{0}}=\frac{7e}{e_{0}}$

Thank you!

Aside: On a side note, I'm using lyx. I'm actually a MSCS student; I just want to work on scientific computation. Is there a real urgency in learning LaTex itself outside of WYSIG?

 

[vela]

Good work!

I don't know if there's an urgency to learn LaTeX for your field, but it's not very hard to get decent at it.