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20e-lecture18 - Lecture 18 Monday May 11th [email protected]

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Lecture 18 - Monday May 11th [email protected] Key words: Change of variables, Jacobian, polar, cylindrical and spherical co-ordinates Key concepts: Know how to change variables in integrals 18.1 Change of variables In this section we give the change of variables theorem for multiple integrals. This is the analog of the substitution rule for integrals of functions of one variable. Suppose that we want to make the substitution x = x ( u ) in the integral of a function f from a to b , and that for a x b , we have c u d . Then Z b a f ( x ) dx = Z d c f ( x ( u )) dx du du. The important term to remember is the dx/du term, and this term has an analog for multiple integrals. The function x = x ( u ) given above is actually a bijection from [ c, d ] to [ a, b ]: this means that for every point t [ a, b ], there is a unique number u [ c, d ] such that x ( u ) = t . That x ( u ) is a bijection is essential in order to make the change of integration, and it means that as a function, x ( u ) has an inverse, denoted x - 1 ( u ). The inverse is itself a bijection from [ a, b ] to [ c, d ]. For example, the function x ( u ) = u 3 is clearly a bijection from R to R , since every real number has a unique real cube-root. The inverse is given by x - 1 ( u ) = u 1 / 3 x ( x - 1 ( u )) = u in this case. The function x ( u ) = u 2 is not a bijection, since x (1) = x ( - 1) = 1. To say that a function x ( u ) : R R is a bijection is equivalent to saying that the curve representing x ( u ) in the ux -plane passes both the horizontal and vertical line tests. For change of variables in multiple integrals, we have a function f ( x ) : R n R which we are integrating over a region D , and we wish to replace x with a new vector of variables u R n via the change of variables x = x ( u ). We suppose that E is the domain of x ( u ) and that x ( u ) as a function from E to D is a bijection. The inverse function theorem states that if we want to see whether x ( u ) is a bijection, then we check that det x ( u ) 6 = 0 for all u E . If that is the case, then E = { u : x ( u ) D } is the new region over which u is integrated. If we think of a function
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