**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 *d***s**. 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 *d***s**, sweeps out an area **v***t* **×** *d***s** in time *t*. If we integrate around the boundary ∂Σ in a counterclockwise sense, **v***t* **×** *d***s** points in the negative *z*-direction on the left side of ∂Σ (where *d***s** points downward), and in the positive *z*-direction on the right side of ∂Σ (where *d***s** 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 •** *d***s** = −**F × v•** *d***s** = −**F • v ×** *d***s**. 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.

Read more about this topic: Leibniz Integral Rule, Proofs

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