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Flux integral

  • Thread starter mathman44
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  • #1
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Homework Statement



Directly compute the flux integral [tex]\int\int(G*n dS)[/tex] where D is the part of the plane x+y+z=1 in the first octant, oriented upwards.

G = -y-z-x

My attempt at solution:

normal vector, normalized, is [tex]1/sqrt(3) * (1,1,1)[/tex] and since z=1-y-x, the integral simplifies to

[tex]1/sqrt(3)*\int\int(-1 dS)[/tex]

How do I evaluate this, without parametrization of the plane?
 

Answers and Replies

  • #2
tiny-tim
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Hi mathman44! :smile:

(have a square-root: √ and an integral: ∫ :wink:)

∫∫ dS is the area, so you need the area of an equilateral triangle of side … ? :smile:
 
  • #3
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Doh!

Of side √2. Meaning the area is 1, and the flux is -1/√3... is that okay?
 
  • #4
tiny-tim
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Doh!

Of side √2. Meaning the area is 1 …
sin60º ≠ 1 :redface:
 
  • #5
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Ha, oops... so area is √3/2 and the flux is then -1/2.
 
  • #6
tiny-tim
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Ha, oops... so area is √3/2
Yup! :biggrin:
and the flux is then -1/2.
ah … now I didn't actually understand the original question :redface:

G is presumably a vector, but what are its coordinates? :confused:
 
  • #7
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G is -y(i) - z(j) - x(k)

so G dotted with n

= 1/√(3)*(1,1,1) * G = 1/√(3) * (-y-z-x)

But because x+y+z=1, z=1-x-y and so -y-z-x = -1, becoming:
1/√(3) * (-y-z-x) = 1/√(3) * (-1)

So then flux should be -1/√3 * dS, which is -1/√3 * √3/2 = -1/2
 
  • #8
tiny-tim
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G is -y(i) - z(j) - x(k)
ah! all is now clear! :smile:

Yes, -1/2 looks good. :wink:
 

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