We notice that (see article on curl) and that Stokes theorem equates the surface integral of the curl over Σ with a line integral over : 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 . 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. Alternative derivation Lemma. One has: Proof. From the proof of the fundamental theorem of calculus, and Suppose a and b are constant, and that f(x) involves a parameter α which is constant in the integration but may vary to form different integrals. Assume that f(x, α) is a continuous function of x and α in the compact set {(x, α) : α0 ≤ α ≤ α1 and a ≤ x ≤ b}, and that the partial derivative fα(x, α) exists and is continuous. If one defines: then may be differentiated with respect to α by differentiating under the integral sign, i.e., By the Heine–Cantor theorem it is uniformly continuous in that set. In other words, for any ε > 0 there exists Δα such that for all values