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Surface integral - I don't understand it

by Woopydalan
Tags: integral, surface
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Woopydalan
#1
Feb28-13, 12:50 AM
P: 746
Hello,

So I am trying to understand surface integrals so I can can more insight to understand Gauss's Law.

I am reading a book about it, and the example that is used to explain a surface integral is to have a flat surface that has a mass density that changes as a function of position in the x & y position σ(x,y)

The author then goes on to say that to find the total mass of the surface, take the area density and multiply by the area, and to make a summation of small areas.

The author says ''the smaller you make the area segments, the closer this gets to the true mass, since your approximation of constant σ is more accurate for smaller segments.


I don't understand this statement, if the area density changes with x & y, how can I say that the area in some corner that is very dense is approximately equal to another corner that is much less dense.
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dx
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Feb28-13, 04:24 AM
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When you multiply the density by a small area, you take the density to be constant over that area, i.e. take σ = σ(x, y) for some particular (x, y) in that area.

This will be close to the actual mass because of the fact that σ is continuous, i.e. by choosing sufficiently small area the difference between σ at two points in that area can be made arbitrarily small.
HallsofIvy
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Feb28-13, 09:53 AM
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Quote Quote by Woopydalan View Post
Hello,

So I am trying to understand surface integrals so I can can more insight to understand Gauss's Law.

I am reading a book about it, and the example that is used to explain a surface integral is to have a flat surface that has a mass density that changes as a function of position in the x & y position σ(x,y)

The author then goes on to say that to find the total mass of the surface, take the area density and multiply by the area, and to make a summation of small areas.

The author says ''the smaller you make the area segments, the closer this gets to the true mass, since your approximation of constant σ is more accurate for smaller segments.


I don't understand this statement, if the area density changes with x & y, how can I say that the area in some corner that is very dense is approximately equal to another corner that is much less dense.
The statement you quote doesn't mean "that the area in some corner that is very dense is approximately equal to another corner that is much less dense"
It is saying that the sum of masses of such small regions is approximately the total mass. (And it is talking about mass not area.)

Woopydalan
#4
Feb28-13, 10:14 AM
P: 746
Surface integral - I don't understand it

why is it that if you make smaller segments, the area density approaches a constant? If it becomes constant over the entire surface, then you would be saying that the less dense regions, if made appropriately small enough, are about the same as the more dense regions, if the segments are made small enough?

I'm still not convinced, help!!
WannabeNewton
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Feb28-13, 11:03 AM
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Quote Quote by Woopydalan View Post
I'm still not convinced, help!!
If you take a sufficiently small region, the continuity of sigma will allow you to say that the variations in sigma over the region are extremely small to the point of being negligible. Then when you take limits, this "approximation" becomes exact. This is pretty much what dx said above.
Whovian
#6
Feb28-13, 11:09 AM
P: 642
Okay, the point here is the same concept behind Riemann integrals. In Riemann integration, over a "very small" region, for the purposes of approximation, we can consider a function to be constant (the idea behind the rectangles.) However, in the next "very small" region, we consider it (again, for the purposes of approximation) to be another constant. Does it make sense that, as our small regions get small, the sum of the rectangles' area gets close to the area under the curve?

The same idea applies here. Over a "very small" part of the surface, for approximation, we can call the function a constant, though over the other "very small" regions, the function is considered to be a different constant (well, not necessarily different, but not necessarily the same) over each region.
Woopydalan
#7
Mar1-13, 11:12 AM
P: 746
Quote Quote by Whovian View Post
Okay, the point here is the same concept behind Riemann integrals. In Riemann integration, over a "very small" region, for the purposes of approximation, we can consider a function to be constant (the idea behind the rectangles.) However, in the next "very small" region, we consider it (again, for the purposes of approximation) to be another constant. Does it make sense that, as our small regions get small, the sum of the rectangles' area gets close to the area under the curve?

The same idea applies here. Over a "very small" part of the surface, for approximation, we can call the function a constant, though over the other "very small" regions, the function is considered to be a different constant (well, not necessarily different, but not necessarily the same) over each region.
So I get the part about in one region of the surface the function is constant if made sufficiently small, but why can I say that in another region sufficiently small it is the same constant as in another region? That is how I am interpreting the statement
Robert1986
#8
Mar1-13, 11:24 AM
P: 828
Quote Quote by Woopydalan View Post
So I get the part about in one region of the surface the function is constant if made sufficiently small, but why can I say that in another region sufficiently small it is the same constant as in another region? That is how I am interpreting the statement
It isn't the same constant. Each small region has its own constant.
Chestermiller
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Mar1-13, 08:08 PM
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This is the two dimensional version of approximating the integral of a function f(x) with respect to x by subdividing the region into finite rectangles of height f and width Δx.


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