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We want our workforce to reflect the diversity of our work. com 47 Calculus 2b Line integrals Special cases: In the following special cases we reduce 2) in the procedure by inserting the given area element dS. 1) Integral over a graph for z = f (x, y), rectangular: dS = 1+ 2 ∂f ∂x + 2 ∂f ∂y dx dy. ) 2) Integral over a cylindric surface r(t, z) = (X(t), Y (t), z), rectangular: dS = dX dt 2 dY dt + 2 dt dz = ds dz, where ds is the curve element for ˜r(t) = (X(t), Y (t)) in the plane, cf. Chapter 15.

In principle this can always be applied, but it is often very cumbersome. 1) Divide if necessary F into convenient sub-surfaces F1 , . . , Fk each having its own unit normal vector field n1 , . . , nk . 2) Check, whether F (or Fj ) is “flat”, and if it is not too difficult to calculate ordinary plane integral, because F is lying in a plane set. F V · n dS as an 3) If F is not flat, we calculate the normal vector corresponding to the specific parametric representation in the variables (u, v), N(u, v) = ex ey ez ∂x ∂u ∂y ∂u ∂z ∂u ∂x ∂v ∂y ∂v ∂z ∂v .

If so, then K V · dx = F (b) − F (a) + K U · dx. g. used in Thermodynamics, where the vector field usually is not a gradient field. In these reductions one can take advantage of the well-known rules of calculus for differentials: α df + dg = d(α f + g), α constant f dg + g df = d(f g), f dg − g df = f 2 d g f , f = 0, F (f ) df = d(F ◦ f ). wanted: ambitious people Please click the advert At NNE Pharmaplan we need ambitious people to help us achieve the challenging goals which have been laid down for the company.

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