Calculate Divergence using limit definition

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Saladsamurai
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Homework Statement



Evaluate div v at P = (0, 0, 0) by actually evaluating [itex](\int_S\mathbf{\hat{n}}\cdot \mathbf{v}\,dA)/V[/itex] and taking the limit as B-->0. Take B to be the cube [itex]|x|\le\epsilon,|y|\le\epsilon,|z|\le\epsilon[/itex]. Let [itex]\mathbf{v} = x\mathbf{\hat{i}} + 2y\mathbf{\hat{j}} - 4z^3\mathbf{\hat{k}}[/itex]

Homework Equations


The Attempt at a Solution

So what I need to do is to first find the integral [itex]\int_S\mathbf{\hat{n}}\cdot \mathbf{v}\,dA[/itex] and to do so, I will break it up into 6 integrals, one for each face of the cube.

First I have a question: the way that the bounds of the cube are given suggest that the cube is [itex]2\epsilon[/itex] in length in each direction. I am wondering how I am to position my coordinate system. Should it be centered in the cube? Should it be at a corner? Does it matter? I would like to think that it does not matter, but I cannot figure out how to justify that assumption.

I have more questions, but I would like to clarify this one first. I started the problem by positioning the origin at the center of the cube, but I want to confirm that's ok before typing my work in.
 
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Hi Saladsamurai! :smile:
Saladsamurai said:
First I have a question: the way that the bounds of the cube are given suggest that the cube is [itex]2\epsilon[/itex] in length in each direction. I am wondering how I am to position my coordinate system. Should it be centered in the cube? Should it be at a corner? Does it matter? I would like to think that it does not matter, but I cannot figure out how to justify that assumption.

… I started the problem by positioning the origin at the center of the cube, but I want to confirm that's ok before typing my work in.

Yes, that's fine …

in fact, you really don't have a choice, since v depends on x y and z, and you'll get really confused if you try to change variables (especially if you keep the names x y and z ! :wink:).
 
tiny-tim said:
Hi Saladsamurai! :smile:


Yes, that's fine …

in fact, you really don't have a choice, since v depends on x y and z, and you'll get really confused if you try to change variables (especially if you keep the names x y and z ! :wink:).

Ok great tiny - tim, thanks! So let me show what I have done so far so I can show where I am confused. I am calculating the integral for the two faces of the cube that lie in the xy-plane first:

[tex]I_{xy-plane} = \int_y\int_x \mathbf{\hat{k}}<br /> \cdot<br /> (\mathbf{\hat{i}} + 2y\mathbf{\hat{j}} - 4z^3\mathbf{\hat{k}}) \,dx\ \,dy \, - <br /> \int_y\int_x \mathbf{\hat{k}} <br /> \cdot<br /> (\mathbf{\hat{i}} + 2y\mathbf{\hat{j}} - 4z^3\mathbf{\hat{k}})\,dx\,dy \,<br /> \qquad(1)[/tex]

or

[tex]I_{xy-plane} = \int_y\int_x (-4z^3) \,dx\,dy \,<br /> - <br /> \int_y\int_x (-4z^3)\,dx\,dy \,<br /> \qquad(2)[/tex]



Now, I feel like this is missing something or else the integrals are just going to cancel. Also, I don't see how I am including the point P = (0,0,0) anywhere. Can someone calrify these two things for me?