- You Are Attempting To Perform An Integral 1 Sis 5 2 1 2 Dr Dy Dz For A Given Co Function Ds I F Over A Region D In 1 (131.21 KiB) Viewed 25 times
You are attempting to perform an integral 1 = SIS, 5(2,1, 2) dr dy dz for a given Cº function ds I f, over a region D in three-dimensional space. Suppose you use a change-of-variables to write this integral in terms of new variables (u, v, w), such that =u , y=vw and z=wu, and assume that this mapping between (x, y, z) and (u, v, w) spaces is invertible (a bijection) in the region which is described by D in (x, y, z)-space. If D* is the region in (u, v, w)-space corresponding to D in (x, y, z)-space, then I is equal to O 2 SIT. |uw|f\uv, vw, wu) du dvdw , O SIL, 5(u, v, w) dwdv du D* О f(uv, vw, wu) du du dw D* SIL.. ) SIL.5(u,0, u)V42 + x2 + wa du dv dw O D* O SIL.5: f(u, v, w) dx dydz D