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Let S be the part of the plane 2x + 1y + z = 2 which lies in the first octant, oriented upward. Find the flux of the vector field F = 1i + 1j + 3k across the surface S. F = 1i + 1j + 3k across the surface s2x + 1y + z = 2 which lies in the first octant, oriented upward. Find the flux of the vector field.

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rafikiedu08

The flux of the vector-field F = 1i + 1j + 3k 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 + 1y + z = 2.

and given F = 1i + 1j + 2k

div G = div(2x + 1y + z = 2) = 2i + 4j + k

Flux = ∫(1i + 1j + 3k) (2x + 1y + z) 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.

Green's theorem has numerous applications. One method is to solve two-dimensional flow integrals, which state that the sum of fluid outflows from a volume equals the total outflow summed around an enclosing area.

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