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The second volume of Apostol's Calculus seems rather circumspect in its discussion of the change of variables formula for double integrals. Section 11.29 offers a proof under the following very limited circumstances:

Let $R$ be a rectangle, $R^*$ its image under a one-to-one mapping $u = U(x,y)$, $v = V(x,y)$, with the inverse mapping given by $x = X(u,v)$, $y = Y(u,v)$. Assume that both $X$ and $Y$ have continuous second-order partial derivatives, and that the Jacobian determinant $J(u,v)$ is never $0$ in $R^*$. Then

\begin{align} \iint\limits_R dx\, dy & = \iint\limits_{R^*}|J(u,v)| \,dx\,dy \end{align}

The proof Apostol gives is a straightforward application of Green's Theorem, except for one small issue: let $C$ be the boundary of $R$, $C^*$ be the boundary of $R^*$, and parameterize $C^*$ by \begin{align} \mathbf{\alpha}(t) = W(t)\mathbf{i} + Z(t)\mathbf{j} \end{align} with $t$ on some interval $[a,b]$. Apostol then asserts that \begin{align} \mathbf{\beta}(t) = X[W(t),Z(t)]\mathbf{i} + Y[W(t),Z(t)]\mathbf{j} \end{align} represents a parameterization of $C$. I'm afraid I don't understand the basis for this assertion. The continuity of $X$ and $Y$ along with the one-to-oneness of the map $x = X(u,v), y = Y(u,v)$ would seem to guarantee that $\mathbf{\beta}(t)$ is a piecewise smooth closed path within $R$, but it is unclear to me that we have boundaries mapping to boundaries. Is there some way to show that this is the case?

(Perhaps not so coincidentally, the proof that Munkres gives in his supplemental notes on MIT's OpenCourseware site is almost identical, except that in his statement of the theorem that $C$ maps to $C^*$ is an explicit hypothesis.)

solitaireartist
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A $C^1$ map $f$ with nonvanishing Jacobian determinant has a local $C^1$ inverse in a neighborhood of each point. The local inverses are open maps and hence $f$ must map boundary to boundary.

To elaborate: OK, I missed the direction of Apostol's claim. Letting $f\colon R\to R^*$, he's claiming that $f^{-1}$ maps $C^*$ to $C$. Suppose a point $P^*$ in $C^*$ mapped to a point $P$ in the interior of $R$. Because $f^{-1}$ is continuous, $f$ maps open sets to open sets, so a small open ball centered of $P$ (and contained in the interior of $R$) maps to a neighborhood of $P^*$, in particular, an open set, in $\Bbb R^2$. So $P^*$ cannot be a boundary point of $R^*$.

Ted Shifrin
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  • It's likely that I don't have the experience necessary to fully understand your answer. Let me attempt a rephrasing to see if I get it: Let $f$ be the map $f(x,y) = (U(x,y), V(x,y)$. Then by the inverse function theorem, there exists an open neighborhood of $(x,y)$ such that $f(x,y)$ is invertible on this neighborhood and such that this inverse ($g(x,y) = (X(u,v),Y(u,v))$ in Apostol's notation) is continuously differentiable. That much I follow -- indeed, it looks like Apostol asserts as much without needing to. What I don't see is why this implies that boundaries must map to boundaries. – solitaireartist Jan 29 '15 at 02:32
  • Is the point, then, that an open ball centered on a boundary point of $R^$ cannot be contained in $R^$? Assuming this is the case, I believe I understand your answer. Thank you. – solitaireartist Jan 29 '15 at 03:14
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    Yes, exactly. Only Interior points have that property. – Ted Shifrin Jan 29 '15 at 03:24