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Extra resources for Calculus 2b, Guidelines for Solutions of Some Types of Problems
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.