The flux of the vector-field F = 4i + 4j + 1k across the surface S is 63/4. We find out the flux of the vector-field using Green's Theorem.
<h3>Define Green's Theorem.</h3>
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 = 3x+2y+z = 3.
and given F = 4i + 4j + 1k
divG = div(3x+2y+z = 3) = 3i + 2j + k
Flux = ∫(4i + 4j + 1k) (3i + 2j + k) dA
φ = ∫ (12 + 8 + 1)dA
= 21∫dA
A = 1/2 XY (on the given x-y plane)
3x+2y =3
at x = 0, y = 3/2
y = 0, x = 1
1/2 (1*3/2) = 3/4
Therefore flux = 21*3/4 = 63/4
φ = 63/4.
To know more about Green's theorem visit:
brainly.com/question/27549150
#SPJ4
