Double integrals (line vs. Area)

  • #1
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Hey! So we were doing double integrals in electricity and magnetism for vectors dA and A (for electric flux).

I’m a little confused. Doing a double integral of vectors dx and dy gave an area (vector) dA and A.

Thinking back to calc 1, when we had FUNCTIONS (not vectors) they gave the area under a curve.


Does this mean doing a double integral for vectors (line integral) gives an area spanned by those two vectors in 2D or 3D space.

A double integral for regular functions, therefore two variables, gives the volume of that function in 3D space?

Am I getting this right?

I tried asking someone the first part (about vectors) and they said that’s not necessarily true and now I’m even more confused :)).


I’m doing calc 3 at the same time as electricity and magnetism so I don’t know too much, our physics prof is teaching us what he thinks we need to know for our lessons.
 

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  • #2
.Scott
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Let's start with 0D.
Say I have one photodetector sitting in sunlight.
If I wanted the total power from that cell it will be a function of the intensity of the sunlight on that cell, the angle that the light is hitting it, and perhaps other factors.
There is no integration here.

Now 1D:
I have a whole line of these photo cells. The intensity will be the sum of the power generated by each cell. I need to integrate across those length of this line of cells to get the total power. If I plotted this with cells along the bottom and power along the side, the "area under the curve" would be my total power.

Now 2D:
I have an arrays of cells. To get the total power, I need to do two integrals. First along one axis, then across the other axis using the result from the first integration.
I could set up a 3-d plot with power as the vertical axis and the cell coordinates along the floor of the plot. The total volume under that curve would be my total power.
 
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  • #3
Charles Link
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There are two theorems that change the dimensionality of the integrals that are commonly encountered in E&M (electricity and magnetism): One is Gauss' law: ## \int \nabla \cdot E \, d^3 x=\int E \cdot \hat{n} \, dA ## , and the other is Stokes' theorem: ## \int \nabla \times E \cdot \hat{n} \, dA=\oint E \cdot dl ##. ## \\ ## These two integral theorems see extensive use in even the semi-elementary E&M courses. Gauss' law is usually first taught as ## \int_V E \cdot \, \hat{n} \, dA=\frac{Q}{\epsilon_o } ##, and Ampere's law is taught as ## \oint B \cdot dl=\mu_o I ##. Later, they teach the differential forms of Maxwell's equations, that with these integral theorems, gives the integral forms of Maxwell's equations. Usually they start off with the integral form of the equation, even though its origins are in the differential form.
 
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  • #4
Charles Link
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I need to add to the above, Faraday's law, that in differential form reads ## \nabla \times E=-\frac{\partial{B}}{\partial{t}} ## when integrated over ## dA ## and with Stokes' theorem reads ## \mathcal{E}=\int E \cdot dl=-\frac{d \Phi_m}{dt} ##. ## \\ ## I should also give the differential forms of the other Maxwell equations I mentioned: ## \\ ## ## \nabla \cdot E=\frac{\rho}{\epsilon_o} ##. Integrated over ## d^3 x ## and with Gauss' law becomes ## \int E \cdot \, \hat {n} \, dA =\frac{Q}{\epsilon_o} ##. ## \\ ## and ## \\ ## ## \nabla \times B=\mu_o J +\mu_o \epsilon_o \dot{E} ##. For steady state case ## \dot{E}=\frac{dE}{dt}=0 ## . Integrated over ## dA ## and with Stokes' theorem it becomes ## \oint B \cdot dl=\mu_o I ##. ## \\ ## The fourth Maxwell equation ## \nabla \cdot B=0 ## also has an integral form upon integrating over ## d^3 x ## and using Gauss' law: ## \int B \cdot \, \hat{n} \, dA=0 ##.
 
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