Surface Integrals: Calculating dxdy, dxdz, dydz

coverband
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When calculating surface integrals do you have to calculate double integrals for dxdy, dxdz and dydz and add up or what?
 
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It depends on how you parametrize the surface, really. If you use any 2 of the three variables x,y,z then you have to set up the double integrals with respect to them. Otherwise you perform a double integral dudv if your parameters are u,v.
 
Defennder said:
If you use any 2 of the three variables x,y,z then you have to set up the double integrals with respect to them.

But this is sufficient? If you integrate say with respect to dxdy and have correct and appropriate limits then this is the question answered so to speak. You don't have to go on and do dxdz and dydz?
 
The thing here is that the surface integral is assumed to be done with respect to x,y and that the surface may be parametrized in x,y. If so, then yes.
 
coverband said:
But this is sufficient? If you integrate say with respect to dxdy and have correct and appropriate limits then this is the question answered so to speak. You don't have to go on and do dxdz and dydz?

Hi coverband! :smile:

Just draw a squareish grid across the surface.

If dxdy describes the grid, where does dz come into it? :smile:
 
I appreciate the attention but is the answer to do you just integrate with respect to dxdy in all cases yes !?
 
If the grid is described by dxdy, then yes. :smile:
 
coverband said:
I appreciate the attention but is the answer to do you just integrate with respect to dxdy in all cases yes !?

I don't think that's true. x, y and z are just the names you give to the axis, you can call them whatever you want. You integrate with respect to surface area i.e. dS I believe. Have you read this: http://planetmath.org/encyclopedia/IntegrationWithRespectToSurfaceArea.html
 
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Thanks no more exams. So say if you have surface z=x+y+1 then you integrate with respect to dxdy and that's it finished problem solved

If you have x=z-y-1 then you integrate with respect to dzdy and that's it finished

You don't do all three and add them up or anything !?
 
  • #10
coverband said:
Thanks no more exams. So say if you have surface z=x+y+1 then you integrate with respect to dxdy and that's it finished problem solved

If you have x=z-y-1 then you integrate with respect to dzdy and that's it finished

You don't do all three and add them up or anything !?

Maybe I should have also pointed you here: http://en.wikipedia.org/wiki/Surface_integral,
note where it says: "A natural question is then whether the definition of the surface integral depends on the chosen parametrization. For integrals of scalar fields, the answer to this question is simple, the value of the surface integral will be the same no matter what parametrization one uses." So yes, whichever way you want to parametrize your surface, that's what will affect what your dS = . You definitely would not do it 3 different ways and add them up. Better yet, do it 3 different ways and see what answers you get.
 
  • #11
Thanks nomoreexams
 
  • #12
can someone help me with the solution to ∫ex² dx.i need to apply the solution for something am working on.
 
  • #13
for the question i posted, the x² is a power of e. thanx.
 
  • #14
nsama said:
for the question i posted, the x² is a power of e. thanx.

You should probably start a new thread instead of asking it in one that is not related. In any case \int e^{x^{2}} \, dx has no closed form but it is related to http://en.wikipedia.org/wiki/Gaussian_integral I believe.
 
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