https://picasaweb.google.com/lh/photo/Nzi23gp0euuA-nSKY81Yk3S3VwNJ5qi-az28rYAARew?feat=directlink

My intention was to upload the picture/problem above, and I am well aware that this problem would be better placed on a mathematical webside. But circuits is almost all about math, or what? The p in front of the materials Ω pr meter should be a ρ, but I guess you understood that I have come to the expression R=ρ*(dx/(phi*r^2)) using the hint, but I'm not even sure that is correct. So if there are any math-genius with some time to spare out there I would appreciate some help

 

Feel free to move this to the math section, I'm new here and not sure where it belongs.

 

Where are the terminals of this resistor? On the two ends? If so, I don't find their hint to be particularly helpful (nor can I think of a terminal configuration for which it would be helpful).

Instead, imagine breaking the thing into a bunch of circular disks. What is the resistance of one of these disks? What is the total resistance in terms of the resistances of the individual disks?

Note also that rho (the bulk resistivity) has units of resistance * distance, not resistance per distance.

 

The terminals is at each end of the cone/resistor as you mentioned. I don't know how the hint is supposed to be helpful, but I have a feeling that it has to do with the format dx/du you use when you integreate, but that is when x is the variable. Here I guess that r must be the variable... Exuse me if I'm far out here. But the problem is presented in the link, as we got it presented in our task at school. Sorry for my bad English, but hope you understand. And one more thing, when you say that it would be easier to solve it in small discs, that is exactly what you do when you integrate. Ask if this became a mess:-\

 

There are many ways to divide things up when you integrate. As you say, you can integrate along x or you can integrate along r, in this case. But in order to set up the integral properly (and thereby get the right answer), you have to have one foot in reality, as well.

To integrate with a bunch of cones, you have to assume that the current is spread out evenly across the interface from one differential cone to the next. If the current is injected at one end of the cone, this is an unreasonable assumption. But if you integrage along x using small disks, then it is reasonable to assume that the current is evenly spread out because it is injected into (and taken out of) the end disks evenly spread out. Unless you the terminals are small compared to the radius or unless the cone is really squat and flat, then this assumption should yield a reasonable answre.

The integral itself is very tame.

 

How to upload an attachment.

To attach a document or picture to a post

Click "Go Advanced" at the bottom of your post,

One of several options will pop up,

Click "Manage Attachments" much further at the bottom of your post,

To upload a file from your computer click "Browse", then select the file.

Schematics should be .gif or .png format, pictures should be .jpg. .jpg formats will fuzz out schematics, and should not be used for that purpose.

---------------

If you want to display this file there is an old thread I made,

How to Display Attachments Full Size

 

I believe the hint is telling you to picture this as a tube, not a cone.

This would result in you assigning an average value of diameter for the tube, and therefor your 'thin disc' as well, This would also be true at the point found an equal distance from either end of the cone.

 

I believe the hint is telling you to picture this as a tube, not a cone.

This would result in you assigning an average value of diameter for the tube, and therefor your 'thin disc' as well, This would also be true at the point found an equal distance from either end of the cone.

There's several problems with that. First, the problem specifically calls it a cone. Second, there is no information given regarding the inner diameter and there is no basis upon which to make any kind of a reasonable assumption regarding it. Third, we still have the same problem if we try to integrate using cylindral shells, namely that there is no basis upon which to assume a uniform current density or voltage at the interface between differential elements -- in fact we absolutely know the latter is an impossibility.

If I had to guess, I would say that the hint was intended for a different problem, or perhaps was for this problem previously but then the shape go changed and the hint didn't get updated.