The flux of the vector-field F = 1i + 1j + 2k across the surface S is 2. We find out the flux of the vector-field using Green's Theorem.
Flux form of Green's Theorem for the given vector-field
φ = ∫ F.n ds
= ∫∫ F. divG.dA
Here G is equivalent to the part of the plane = 2x+4y+z = 2.
and given F = 1i + 1j + 2k
divG = div(2x+4y+z = 2) = 2i + 4j + k
Flux = ∫(1i + 1j + 2k) (2i + 4j + k) dA
φ = ∫ (2 + 4 + 2)dA
= 8∫dA
A = 1/2 XY (on the given x-y plane)
2x+4y =2
at x = 0, y = 1/2
y = 0, x = 1
1/2 (1*1/2) = 1/4
Therefore flux = 8*1/4 = 2
φ = 2.
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