Surface integral - I don't understand it

As Δx gets smaller, the sum of the areas of the rectangles approaches the integral. In this case, the role of Δx is played by the area of the small region, which is being made smaller and smaller to approximate the integral. So in summary, the smaller the area segments, the closer the approximation of the total mass becomes, as the function sigma is considered to be constant over each small region and the sum of these regions gets closer to the actual total mass.
  • #1
member 392791
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
Woopydalan said:
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.)
 
  • #4
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!
 
  • #5
Woopydalan said:
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.
 
  • #6
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.
 
  • #7
Whovian said:
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
 
  • #8
Woopydalan said:
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.
 
  • #9
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.
 

1. What is a surface integral?

A surface integral is a mathematical concept used to calculate the total value of a function over a two-dimensional surface. It is similar to a regular integral, but instead of integrating over a one-dimensional interval, it integrates over a two-dimensional surface.

2. How is a surface integral different from a regular integral?

A surface integral is different from a regular integral in that it calculates the value of a function over a two-dimensional surface instead of a one-dimensional interval. It also requires a different type of integration method, such as parametric or double integration.

3. When is a surface integral used?

A surface integral is commonly used in physics and engineering to calculate quantities such as electric flux, fluid flow, and surface area. It is also used in geometry to find the area of a curved surface.

4. How do I calculate a surface integral?

The process for calculating a surface integral depends on the specific problem and the type of surface being integrated over. In general, it involves determining the limits of integration, setting up the appropriate integral equation, and solving for the value using integration techniques.

5. Can you provide an example of a surface integral?

One example of a surface integral is calculating the surface area of a sphere with radius r. The equation for the surface area of a sphere is A = 4πr^2. To calculate this using a surface integral, we would set up a double integral with limits of integration for the angles theta and phi, and integrate the function 4πr^2 over the surface of the sphere.

Similar threads

Replies
4
Views
1K
Replies
2
Views
1K
Replies
33
Views
3K
Replies
2
Views
264
  • Calculus
Replies
1
Views
2K
Replies
46
Views
1K
Replies
1
Views
2K
  • Calculus
Replies
5
Views
2K
Replies
3
Views
624
Back
Top