11.6: Green’s Theorem

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 $$V\bigl(g(S)\bigr) = \int_S \left\lvert {J_g(x)} \right\rvert \, dx .$$

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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?

Let

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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 $$\int_{g(S)} f(x) \, dx = \int_S f\bigl(g(x)\bigr) \left\lvert {J_g(x)} \right\rvert \, dx .$$

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FIXME: change of variables for functions with compact support

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Exercises

Prove .

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