flux of vector field | Physics Library

Let

⃗U = Ux⃗i + Uy⃗j + Uz⃗k

be a vector field in ℝ³ and let a be a portion of some surface in the vector field. Define one side of a to be positive; if a is a closed surface, then the positive side must be the outer surface of it. For any surface element da of a, the corresponding vectoral surface element is

d⃗a = ⃗nda,

where n is the unit normal vector on the positive side of da.

The flux of the vector U through the surface a is the surface integral

∫ ⃗U ⋅ d⃗a. a

Remark. One can imagine that U represents the velocity vector of a flowing liquid; suppose that the flow is stationary, i.e. the velocity U depends only on the location, not on the time. Then the scalar product U ⋅ da is the volume of the liquid flown per time-unit through the surface element da; it is positive or negative depending on whether the flow is from the negative side to the positive side or contrarily.

Example. Let U = xi + 2yj + 3zk and a be the portion of the plane x + y + x = 1 in the first octant (x ≧ 0, y ≧ 0, z ≧ 0) with the positive normal away from the origin.

One has the constant unit normal vector:

-1-⃗ -1-⃗ -1-⃗ ⃗n = √3-i + √3-j + √3-k.

The flux of U through a is

∫ ∫ -1-- φ = ⃗U ⋅ d⃗a = √ -- (x + 2y + 3z)da. a 3 a

However, this surface integral may be converted to one in which a is replaced by its projection A on the xy-plane, and da is then similarly replaced by its projection dA;

dA = cosα da

where α is the angle between the normals of both surface elements, i.e. the angle between n and k:

⃗ -1-- cosα = ⃗n ⋅k = √--. 3

Then we also express z on a with the coordinates x and y:

1 ∫ √ -- ∫ 1 (∫ 1− x ) φ = √--- (x + 2y + 3(1 − x − y)) 3 dA = (3 − 2x − y)dy dx = 1 3 A 0 0

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