By Mejlbro L.
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Additional resources for Calculus 2b, Guidelines for Solutions of Some Types of Problems
Cf. also the remarks to section A. 2) Calculate the Jacobian ∂(x, y, z) = ∂(u, v, w) ∂x ∂u ∂x ∂v ∂x ∂w ∂y ∂u ∂y ∂v ∂y ∂w ∂z ∂u ∂z ∂v ∂z ∂w . Then the volume element is dΩ = dx dy dz = ∂(x, y, z) du dv dw, ∂(u, v, w) where one must be careful not to forget the numerical signs of the Jacobian. 3) Insert and calculate the right hand side by means of one of the methods from the chapter 12–14 in the formula f (x, y, z) dx dy dz = B f (r(u, v, w)) D ∂(x, y, z) du dv dw. 1 Calculate the integral f (x) dS, A where the domain A is bounded, and where either A is not closed, or the integrand f (x) is not deﬁned for all x ∈ A.
3) Insert the result and calculate the right hand side by using the methods from Chapter 10 or Chapter 11 in the expression f (x, y) dx dy = B f (r(u, v)) D ∂(x, y) du dv. 1 Usually u and v are given as functions in x and y instead of the form we shall use: (1) u = U (x, y) and v = V (x, y). Then one has to solve these equations with respect to x and y. If we in this way obtain a unique solution, then we have at the same time implicitly proved that the map is injective. Apply furthermore (1) to ﬁnd the images of the boundary curves of B, thereby ﬁnding the boundary of D.
Com 45 Calculus 2b 16 Line integrals Surface integrals We consider 2-dimensional surfaces imbedded in R3 . The idea is to pull the integration over the surface F back to a plane integral over the parameter domain E, where we can use one of the methods from Chapter 10 or Chapter 11. This procedure has its price because we must add some weight function as a factor to the integrand. 8 1 Figure 12: Example of a surface F with a corresponding parameter domain E in the (x, y)-plane. Procedure: 1) Write down a rectangular parameter representation of the surface F: (u, v) ∈ E.