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Mathematics LibreTexts

10.6: Green’s Theorem


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Change of variables

Note: FIXME4 lectures

In one variable, we have the familiar change of variables \[\int_a^b f\bigl(g(x)\bigr) g'(x)\, dx = \int_{g(a)}^{g(b)} f(x) \, dx .\] 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 \(\int_a^b\) versus \(\int_b^a\). We are simply integrating over an interval \([a,b]\). With this notation then the change of variables becomes \[\int_{[a,b]} f\bigl(g(x)\bigr) \left\lvert {g'(x)} \right\rvert\, dx = \int_{g([a,b])} f(x) \, dx .\] In this section we will try to obtain an analogue in this form.

First we wish to see what plays the role of \(\left\lvert {g'(x)} \right\rvert\). 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 \(\left\lvert {L} \right\rvert(b-a)\). That is because \(g([a,b])\) is either \([La,Lb]\) or \([Lb,La]\). This property holds in higher dimension with \(\left\lvert {L} \right\rvert\) replaced by absolute value of the determinant.

[prop:volrectdet] Suppose that \(R \subset {\mathbb{R}}^n\) is a rectangle and \(T \colon {\mathbb{R}}^n \to {\mathbb{R}}^n\) is linear. Then \(T(R)\) is Jordan measurable and \(V\bigl(T(R)\bigr) = \left\lvert {\det T} \right\rvert V(R)\).

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 \(S \subset {\mathbb{R}}^n\) is a closed bounded Jordan measurable set, and \(S \subset U\) for an open set \(U\). If \(g \colon U \to {\mathbb{R}}^n\) is a one-to-one continuously differentiable mapping such that \(J_g\) is never zero on \(S\). Then \[V\bigl(g(S)\bigr) = \int_S \left\lvert {J_g(x)} \right\rvert \, dx .\]


The left hand side is \(\int_{R'} \chi_{g(S)}\), where the integral is taken over a large enough rectangle \(R'\) that contains \(g(S)\). The right hand side is \(\int_{R} \left\lvert {J_g} \right\rvert\) for a large enough rectangle \(R\) that contains \(S\). Let \(\epsilon > 0\) be given. Divide \(R\) into subrectangles, denote by \(R_1,R_2,\ldots,R_K\) those subrectangles which intersect \(S\). Suppose that the partition is fine enough such that \[\epsilon + \int_S \left\lvert {J_g(x)} \right\rvert \, dx \geq \sum_{j=1}^N \Bigl(\sup_{x \in S \cap R_j} \left\lvert {J_g(x)} \right\rvert \Bigr) V(R_j)\] ... \[\sum_{j=1}^N \Bigl(\sup_{x \in S \cap R_j} \left\lvert {J_g(x)} \right\rvert \Bigr) V(R_j) \geq \sum_{j=1}^N \left\lvert {J_g(x_j)} \right\rvert V(R_j) = \sum_{j=1}^N V\bigl(Dg(x_j) R_j\bigr)\] ... FIXME ... must pick \(x_j\) correctly?



So \(\left\lvert {J_g(x)} \right\rvert\) is the replacement of \(\left\lvert {g'(x)} \right\rvert\) for multiple dimensions. Note that the following theorem holds in more generality, but this statement is sufficient for many uses.

Suppose that \(S \subset {\mathbb{R}}^n\) is an open bounded Jordan measurable set, and \(g \colon S \to {\mathbb{R}}^n\) is a one-to-one continuously differentiable mapping such that \(g(S)\) is Jordan measurable and \(J_g\) is never zero on \(S\).

Suppose that \(f \colon g(S) \to {\mathbb{R}}\) is Riemann integrable, then \(f \circ g\) is Riemann integrable on \(S\) and \[\int_{g(S)} f(x) \, dx = \int_S f\bigl(g(x)\bigr) \left\lvert {J_g(x)} \right\rvert \, dx .\]


FIXME: change of variables for functions with compact support



Prove .


  1. If you want a funky vector space over a different field, \({\mathbb{R}}\) is an infinite dimensional vector space over the rational numbers.