Find the flux of the vector field F ( r ) = π 1 ∣ ∣ r ∣ ∣ 3 r through the triangular surface S with its vertices at ( 1 , ϕ , 0 ) , ( 0 , 1 , ϕ ) and ( ϕ , 0 , 1 ) , where ϕ = 2 1 + 5 is the golden ratio.
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Yes, exactly! I just stole your analogous problem about the cube and dressed it up a bit ;)
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Yeah, that's a nice expansion which increases the challenge level a bit
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The triangular surface is one face of a convex regular icosahedron surrounding the origin. The total outward flux of 4 , divided by 2 0 faces gives a flux of 0 . 2 for each face.