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#1 2006-10-10 20:56:39

otto
Member
Registered: 2006-10-10
Posts: 3

Integral

I need to solve integral (2x-z+2/3y)ds where ds is x/1+y/2+z/3=1(in first octant).
I wuld be happy if anyone could help solving this...
Thnx!

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#2 2006-10-11 12:37:52

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: Integral

I'm not sure what this question is asking. You say it's a line integral and you use the conventional ds notation, but then you say ds is is x/1+y/2+z/3=1, which is the equation for a plane, not a line. If you can communicate to me what exactly it is you're trying to integrate over then I might be able to help.

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#3 2006-10-11 20:05:00

otto
Member
Registered: 2006-10-10
Posts: 3

Re: Integral

It is plane integral.
The assignment is this:
Solve SS(2x-z+2/3y)ds if s is part of the plane x/1 + y/2 + z/3 =1 in the first octant. (SS is double integral)...
I hope i translated all correctly...

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#4 2006-10-13 03:19:34

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: Integral

Ah, I see. What you want to solve is the integral:

[align=center]

[/align]

Where the integration is over S, the surface of the plane given by the equation you listed (in the first octant). Basically, since we don't know how to directly integrate over funny shapes, the trick is to project them onto a surface we do know how to integrate over - like the xy plane (xz or yz planes would work just as well). The formula you want to use here is this:

[align=center]

[/align]

Where D is the projection of the surface onto the xy-plane. So, you can just evaluate the right hand side in the same way you would solve other integrals you've seen in previous calculus classes. It's just a double integral over x and y, and anywhere you see a z you just plug in what that is in terms of x and y given by the plane equation (that's what the z(x,y) means). Just remember to limit D to the projection of the region of the plane you want to integrate over - that is, the part in the first octant.

You can find a number of websites that explain this in more detail if you google "surface integrals". The formula is pretty strange, so if you want to understand why it works, you can look here

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/surface/surface.html#derivation

for a derivation. Also, it certainly won't be an issue on any school problems you're asked to solve, but the derivation for the surface integral formula relies on an assumption that you can approximate the surface at any given point by the tangent plane at that point. So I think this forumla would fail if you ever tried to integrate over a surface that had a sharp kink in it (meaning there is no tangent plane). Could possibly be an issue in certain real life applications.

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#5 2006-10-13 09:01:37

otto
Member
Registered: 2006-10-10
Posts: 3

Re: Integral

Thnx!

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