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Mathematics LibreTexts

11.6: Green’s Theorem

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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|(ba). 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 RRn is a rectangle and T:RnRn 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 SRn is a closed bounded Jordan measurable set, and SU for an open set U. If g:URn is a one-to-one continuously differentiable mapping such that Jg is never zero on S. Then V(g(S))=S|Jg(x)|dx.

FIXME

The left hand side is Rχg(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)|dxNj=1(supxSRj|Jg(x)|)V(Rj) ... Nj=1(supxSRj|Jg(x)|)V(Rj)Nj=1|Jg(xj)|V(Rj)=Nj=1V(Dg(xj)Rj) ... FIXME ... must pick xj correctly?

Let

FIXME

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 SRn is an open bounded Jordan measurable set, and g:SRn 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 fg is Riemann integrable on S and g(S)f(x)dx=Sf(g(x))|Jg(x)|dx.

FIXME

FIXME: change of variables for functions with compact support

FIXME4

Exercises

Prove .

FIXME


  1. If you want a funky vector space over a different field, R is an infinite dimensional vector space over the rational numbers.

This page titled 11.6: Green’s Theorem is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jiří Lebl via source content that was edited to the style and standards of the LibreTexts platform.

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