Leibniz Integral Rule - Proofs - Three-dimensional, Time-dependent Form

Three-dimensional, Time-dependent Form

At time t the surface Σ in Figure 1 contains a set of points arranged about a centroid R ( t ) and function F ( r, t) can be written as F ( R ( t ) + r − R(t), t ) = F ( R ( t ) + ρ, t ), with ρ independent of time. Variables are shifted to a new frame of reference attached to the moving surface, with origin at R ( t ). For a rigidly translating surface, the limits of integration are then independent of time, so:

where the limits of integration confining the integral to the region Σ no longer are time dependent so differentiation passes through the integration to act on the integrand only:

with the velocity of motion of the surface defined by:

This equation expresses the material derivative of the field, that is, the derivative with respect to a coordinate system attached to the moving surface. Having found the derivative, variables can be switched back to the original frame of reference. We notice that (see article on curl ):

and that Stokes theorem allows the surface integral of the curl over Σ to be made a line integral over ∂Σ:

The sign of the line integral is based on the right-hand rule for the choice of direction of line element ds. To establish this sign, for example, suppose the field F points in the positive z-direction, and the surface Σ is a portion of the xy-plane with perimeter ∂Σ. We adopt the normal to Σ to be in the positive z-direction. Positive traversal of ∂Σ is then counterclockwise (right-hand rule with thumb along z-axis). Then the integral on the left-hand side determines a positive flux of F through Σ. Suppose Σ translates in the positive x-direction at velocity v. An element of the boundary of Σ parallel to the y-axis, say ds, sweeps out an area vt × ds in time t. If we integrate around the boundary ∂Σ in a counterclockwise sense, vt × ds points in the negative z-direction on the left side of ∂Σ (where ds points downward), and in the positive z-direction on the right side of ∂Σ (where ds points upward), which makes sense because Σ is moving to the right, adding area on the right and losing it on the left. On that basis, the flux of F is increasing on the right of ∂Σ and decreasing on the left. However, the dot-product v × F • ds = −F × v• ds = −F • v × ds. Consequently, the sign of the line integral is taken as negative.

If v is a constant,

which is the quoted result. This proof does not consider the possibility of the surface deforming as it moves.