Resistance Variable Cross Section

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The discussion focuses on calculating the resistance of a disk with a variable cross-sectional area, where connections are made along the cylindrical sides rather than the flat ends. The resistance formula R = ρ(L/A) is referenced, with L being the length of the disk and A representing the cross-sectional area. The user contemplates using integration to account for the changing area as it relates to the radius, suggesting a method involving the equation of a circle. There is uncertainty regarding the setup of the cross-sectional area and how the connections to the resistor are made, particularly concerning the deposition of conductive material. Integration is confirmed as necessary to accurately determine the resistance in this configuration.
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Homework Statement



The question asks for resistance R of a disk with radius r and fixed width w, whose cross sectional area is variable. Unlike in the picture below, the resistor is not connected to the circuit on the flat ends, but on the cylindrical sides.


Homework Equations



2_3_4_1_eng.png


The Attempt at a Solution



With R = \rho \frac{L}{A}L = 2r, ρ is given, but I'm having trouble setting up an expression for the cross sectional area. Normally, for cross sectional area of this type of problem it's just the area of a circle, and some integration will be required with the bounds being the radius at each end.

But here my initial thought is to set up an expression for a one dimensional line along the top of the disk, then multiple that by the width of the disk, w. The area will change with the radius though, so I'm not sure if integration is required.

From the equation for the circle of radius r, I get

y = \sqrt{r^2-x^2}

which then needs to be multiplied by 2, and then by w to get a sort of cross sectional area:

R = \rho \frac{2r}{2w*\sqrt{r^2-x^2}}

Am I on the right track?
 
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How are the connections made to the resistor? Is the curved area divided into two with conductor deposited along the entire curved surface except for two thin cracks, or ? A picture would be a good thing to have here.
 
Attached is a picture. The disk has radius 2r, I'm assuming the wires connect somewhere on the cylindrical sides.
 

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  • disk.png
    disk.png
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The picture still does not show exactly how the metalization is deposited. For example, were the flat ends connected instead, the metalization would be across the entire flat surfaces.

So let's assume two diametrically opposed and arbitrarily thin stripes running the length of the cylinder. Then you can integrate dR from o to r and multiply by two. You definitely need to integrate.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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