Surface integral to Lateral integral?

In summary, Guass's theorem can be used to simplify an angle-independent surface integral if the relative angle and field strength don't change.
  • #1
KingBigness
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Homework Statement


I have some working out my lecture gave me to a problem and I don't think I understand part of it. Hoping you could help me.


It's using Gauss' Law to find the capacitance of a cylindrical capacitor of length L but this information shouldn't matter for my question.

[itex]\lambda=\frac{∂q}{∂l}[/itex]

[itex]\oint E \bullet \Delta A = \frac{\sum q}{\epsilon}[/itex]

He then jumps to.

[itex]E \int \Delta A = \frac{\lambda L}{\epsilon}[/itex]

First question, how does he go from the surface integral to the normal integral? Has it got anything to do with removing the dot product?

He then changes
[itex]E \int \Delta A[/itex] to E(2πrL)

I understand this is taking the integral of dA which becomes A, which is the area of a circle, hence 2πr, but how does he then bring the L into play...would this not be the Volume, not the area?

Thanks for any help.
 
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  • #2
could it be that E is constant in the integral and always makes the same angle with the surface normal?

This is generally the best way to make use of Guass's theorem - exploit the symmetry of the problem to simplify any integrals
 
  • #3
lanedance said:
could it be that E is constant in the integral and always makes the same angle with the surface normal?

This is generally the best way to make use of Guass's theorem - exploit the symmetry of the problem to simplify any integrals

So when an angle doesn't change a surface integral can be turned into a lateral integral?
Sorry I haven't done much study on surface integrals so I don't know a lot about them. I'll look into them.
 
  • #4
so if the relative angle, and field strength both don't change the scalar result of the vector dot product is just EdA, everywhere over the surface. As E is constant, the integral is just E.A
 
  • #5
lanedance said:
so if the relative angle, and field strength both don't change the scalar result of the vector dot product is just EdA, everywhere over the surface. As E is constant, the integral is just E.A

Perfect sense, thanks so much for that
 

1. What is a surface integral and how is it different from a lateral integral?

A surface integral is a mathematical concept used in multivariable calculus to find the flux (flow) of a vector field over a curved surface. It is similar to a double integral, but with an additional parameter for the surface orientation. A lateral integral, on the other hand, is used to find the area under a curve on a flat plane. The main difference is that a surface integral is performed over a 3-dimensional surface, while a lateral integral is performed over a 2-dimensional plane.

2. What is the formula for calculating a surface integral?

The formula for calculating a surface integral is:
∫∫S F(x,y,z) dS = ∫∫D F(x(u,v),y(u,v),z(u,v)) ||∂(x,y,z)/∂(u,v)|| dA
Where F(x,y,z) is the vector field, S is the surface, D is the parameter domain of the surface, x(u,v), y(u,v), and z(u,v) are the parametric equations of the surface, and ||∂(x,y,z)/∂(u,v)|| is the magnitude of the partial derivatives of the parametric equations.

3. What are some real-world applications of surface integrals?

Surface integrals have many real-world applications, such as calculating the flow of fluid through a pipe, the amount of heat transfer over a curved surface, and the electric flux through a closed surface. They are also used in computer graphics to calculate the amount of light that hits a curved surface, and in engineering to analyze stress and strain on a 3-dimensional object.

4. How is the orientation of the surface important in a surface integral?

The orientation of the surface is important in a surface integral because it determines the direction of the flow or flux being calculated. The surface can be oriented in two ways - inward or outward - depending on the direction of the normal vector at each point. The orientation must be specified in the surface integral formula to get an accurate result.

5. Can a surface integral be calculated without using parametric equations?

Yes, a surface integral can be calculated without using parametric equations by using a special type of surface called a graph surface. A graph surface can be defined by a function of two variables, such as z = f(x,y). In this case, the surface integral formula simplifies to:
∫∫S F(x,y,z) dS = ∫∫D F(x,y,f(x,y)) √(1+∂f/∂x)^2 + (∂f/∂y)^2 + 1) dA
Where D is the parameter domain of the graph surface and ∂f/∂x and ∂f/∂y are the partial derivatives of the function.

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