11.6: Green’s Theorem
( \newcommand{\kernel}{\mathrm{null}\,}\)
Change of variables
Note: FIXME4 lectures
In one variable, we have the familiar change of variables ∫baf(g(x))g′(x)dx=∫g(b)g(a)f(x)dx. It may be surprising that the analogue in higher dimensions is quite a bit more complicated. The first complication is orientation. If we use the definition of integral from this chapter, then we do not have the notion of ∫ba versus ∫ab. We are simply integrating over an interval [a,b]. With this notation then the change of variables becomes ∫[a,b]f(g(x))|g′(x)|dx=∫g([a,b])f(x)dx. In this section we will try to obtain an analogue in this form.
First we wish to see what plays the role of |g′(x)|. If we think about it, the g′(x) is a scaling of dx. The integral measures volumes, so in one dimension it measures length. If our g was linear, that is, g(x)=Lx, then g′(x)=L. Then the length of the interval g([a,b]) is simply |L|(b−a). That is because g([a,b]) is either [La,Lb] or [Lb,La]. This property holds in higher dimension with |L| replaced by absolute value of the determinant.
[prop:volrectdet] Suppose that R⊂Rn is a rectangle and T:Rn→Rn is linear. Then T(R) is Jordan measurable and V(T(R))=|detT|V(R).
It is enough to prove for elementary matrices. The proof is left as an exercise.
We next notice that this result still holds if g is not necessarily linear, by integrating the absolute value of the Jacobian. That is, we have the following lemma
Suppose S⊂Rn is a closed bounded Jordan measurable set, and S⊂U for an open set U. If g:U→Rn is a one-to-one continuously differentiable mapping such that Jg is never zero on S. Then V(g(S))=∫S|Jg(x)|dx.
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The left hand side is ∫R′χg(S), where the integral is taken over a large enough rectangle R′ that contains g(S). The right hand side is ∫R|Jg| for a large enough rectangle R that contains S. Let ϵ>0 be given. Divide R into subrectangles, denote by R1,R2,…,RK those subrectangles which intersect S. Suppose that the partition is fine enough such that ϵ+∫S|Jg(x)|dx≥N∑j=1(supx∈S∩Rj|Jg(x)|)V(Rj) ... N∑j=1(supx∈S∩Rj|Jg(x)|)V(Rj)≥N∑j=1|Jg(xj)|V(Rj)=N∑j=1V(Dg(xj)Rj) ... FIXME ... must pick xj correctly?
Let
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So |Jg(x)| is the replacement of |g′(x)| for multiple dimensions. Note that the following theorem holds in more generality, but this statement is sufficient for many uses.
Suppose that S⊂Rn is an open bounded Jordan measurable set, and g:S→Rn is a one-to-one continuously differentiable mapping such that g(S) is Jordan measurable and Jg is never zero on S.
Suppose that f:g(S)→R is Riemann integrable, then f∘g is Riemann integrable on S and ∫g(S)f(x)dx=∫Sf(g(x))|Jg(x)|dx.
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FIXME: change of variables for functions with compact support
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Exercises
Prove .
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- If you want a funky vector space over a different field, R is an infinite dimensional vector space over the rational numbers.↩