
# 10.6: Green’s Theorem

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## Change of variables

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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.