Resistance of spheroid like object

[pugtm]

Homework Statement

a resistor of a solid sphere of radius R=.75 cm and is made of a material of resistivity\rho=6.8e-8\Omegam. the sphere is sliced on either side vertically so that the horizontal diameter is equal to R. if the current is flowing from left to right what is the objects resistance?
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Homework Equations

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The Attempt at a Solution

well i can appreciate that i need to multiply \rho by a meter amount so
\intof \Pir^2 of the cut off portion from -r/2 to r/2 would give me an area. to be honest I'm not even sure how to set this integral up.



any help would be greatly appreciated

 

[da_nang]

The formula for resistance is R = ρ*L/A, where L is length and A is the area of the cross section. Consider the length as dL and the cross section as a function of L. How would you set up the integral for this case?

Spoiler

_-r/2∫^r/2{ρdL/A(L)}. A(L) = π[√(1-L^2)]^2 = π(1-L^2), since it's a sphere with a circular cross section.

So _-r/2∫^r/2{ρdL/[π(1-L^2)]}

 

[pugtm]

sorry i don't understand how you achieved the A(L)
also is L the length of the width of the sphere or the radius of the cross sections?

 

[da_nang]

I admit, I mixed it up with the unit circle there. Sorry.

But say you put the object centered at origo in a x,y,z-coordinate system, then look at it on the xy-plane. It's a bounded circle, which when you rotate it gives you the object. Its equation as a circle would be

x²+y² = R². Solving this for y² gives you:

y² = R² - x², where -R ≤ x ≤ R. However, to get the object in your assignment, you bound the x-domain so that -R/2 ≤ x ≤ R/2.

When you rotate this curve, you get the volume. In calculus, when you rotate a function around the x-axis, the formula is the definite integral ∫πf(x)²dx from a to b, where f(x) is a function of x to give the radius of the disks that you sum. This is basically adding up many cylinders with height dx. However, where only interested in the cross section. You can get this from that formula, since the object is a sphere cutoff for certain x-values. The cross section A(x) = πf(x)². f(x) in this case would be y². So, A(x) = πy² = π(R² - x²). This is what you'll use for to get in the resistance integral. We merely swap x for L as they're both length.

Basically, _-R/2∫^R/2{ρdL/[πA(x)]} = _-R/2∫^R/2{ρdL/[π(R² - x²)]}, where R is a constant.

To simplify, you can move out some of the constants.

(ρ/π)* _-R/2∫^R/2[dL/(R² - x²)]

 

[pugtm]

i'm not sure how you can rotate the circle and get the spheroid. it's sphere with semi spheres cut off of the edges... so i fail to see how rotating the bounded circle would produce this shape. so like slicing off both the ends of an onion. maybe I'm not understanding what you mean.

 

[da_nang]

This is the kind of figure I imagine being on the xy-plane if look at the object from the sides. I'm basing the integral on this. Is it the right shape you have in mind? (Assume that the curve is connected.)

 

[pugtm]

yes, essentially
so don't i need to slice it in order to get all the surface area's
so i don't get where you are getting r in regards to X from when the integral is dL

 

[da_nang]

You're not getting the surface area, you're getting the area of the cross section. That's what the A in R = ρ*L/A is. As for the other thing, I seem to have forgotten to swap the x as well.

(ρ/π)* _-R/2∫^R/2[dL/(R² - L²)]

Remember that the L and x are the same thing, as they both can be used for the distance that the current travels through the resistor. I could just as easily have made the equation of the circle as L²+y² = R².

You should be able to solve it now.

 

[pugtm]

my thanks, did you get 3.170605653e-6 as the final answer by any chance?

 

[da_nang]

Indeed I did. It sounds reasonable too, given the size of the cross section compared to the distance.

 

[CarlosSoli]

Electrical resistance is, in general, not a localized property. Depends on the whole circuit because it is a relationship between voltage and currents, which in general are nonuniform. Look at
http://independent.academia.edu/Csoliverez/Papers/1101798/The_notion_of_electrical_resistance