Clarification of "Magnification" of Homeomorphisms Note: This question is somewhat conceptual: I am seeking guidance on how to analyze a physical phenomenon topologically; I'd be happy to expand on that phenomenon if it helps clarify the question. I have a basic grasp of topology, but not a detailed-enough grasp to answer the question below (which I suspect is already well-answered in the literature).
Let $f(x,y) = (u,v)$ be a homeomorphism that maps the domain $\mathbf{D} = \{(x,y) \in \mathbb{R}\;|\; x^2 + y^2 \lt 1\}$ to some topologically isomorphic range $\mathbf{R} = \{(u,v)\in\mathbb{R}\}$. For the sake of simplicity, we can assume that $\mathbf{R} = \{(u,v)\in\mathbb{R} \;|\; u^2 + v^2 \lt 1\}$ as well (but $(x,y)$ is not generally equal to $f(x,y)$).
Let us define the "areal magnification" of $\mathbf{D}$ by $f$ at a point $(x,y)$ informally as:
$$m(x,y) = \lim_{\epsilon \rightarrow 0} \frac{\mbox{"area of disk}((x,y),\epsilon)\mbox{"}}{\mbox{"area of }f\left(\mbox{disk}((x,y),\epsilon)\right)\mbox{"}} $$
where disk$((x,y),\epsilon)$ is the disk centered at $(x,y)$ with radius $\epsilon$, and $f($disk$((x,y),\epsilon))$ is the image of that disk in $\mathbf{R}$: $\{(u,v) \;\vert\; f^{-1}(u,v) \in \mbox{disk}((x,y),\epsilon)\}$.
Alternately, we can informally define the "linear magnification" of $\mathbf{D}$ by $f$ along a continuous curve $q$ at point $(x,y) = q(t_1)$; suppose $q(t) \in \mathbf{D} \mbox{ where } 0 < t < 1$:
$$\begin{eqnarray*}
& &\mbox{Let } q(t) \mbox{ s.t. } q(t_1) = (x,y) \mbox{ be a continuous curve in } \mathbf{D} \mbox{ for } 0 < t < 1. \\
& & m_q(x,y) = \lim_{\epsilon \rightarrow 0}\frac{\mbox{"length of curve }(q(t_1-\epsilon), q(t_1+\epsilon))\mbox{"}}{\mbox{"length of curve } \left((f\circledast q)(t_1-\epsilon), (f\circledast q)(t_1-\epsilon) \right)\mbox{"}}
\end{eqnarray*}$$
Here, $(f\circledast q)(t)$ is the composition of $f(q(t))$ (the image of $q$ in $\mathbf{R}$). Note that values greater than 1 indicate some amount of stretching of the curve and values less than 1 indicate some amount of compression.
We can also define the "global" areal or linear magnification to be $\frac{\mbox{"area of }\mathbf{D}\mbox{"}}{\mbox{"area of }\mathbf{R}\mbox{"}}$ and $\frac{\mbox{"total length of }q(t)\mbox{"}}{\mbox{"total length of }(f\circledast q)(t)\mbox{"}}$, respectively.
If we imagine that $f(x,y) = (x,y)$ then it is clear that all magnification values are 1, but let us imagine a slightly more complicated scenario:
$$
\begin{eqnarray*}
f(x,y) &=& g\left(\sqrt{x^2+y^2}, \arg(x+iy)\right); \\
g(r,\theta) &=& \left(\hat{r}\cos(\theta), \hat{r}\sin(\theta)\right) \mbox{ where } \hat{r} = r + \frac{r}{4}\sin(\pi r)\sin(8\theta)
\end{eqnarray*}
$$
This version of $f$ adds a bit of wiggle to any circle in $\mathbf{D}$ according to the radius of that circle; this plot, with curves from $\mathbf{D}$ drawn on the left and the equivalent curves in $\mathbf{R}$ drawn on the right, should make the transformation clear:

The Question
It is quite unintuitive that the following are all true in the above transformation:


*

*The total area of each of $\mathbf{D}$ and $\mathbf{R}$ is 1 (in fact, $\mathbf{D} = \mathbf{R}$), so no global areal magnification has occurred.

*The curves that form radial line-segments (red lines above) are not altered in length globally either; the global linear magnification along any radial line is 1.

*But, every circular curve (blue lines above) has a global linear magnification greater than 1 (excepting the outer curve $r=1$ which is not technically part of $\mathbf{D}$ or $\mathbf{R}$).


So essentially, of the two polar dimensions ($r$, $\theta$) used to address points in $\mathbf{D}$, one of them is locally and globally magnified everywhere, one of them is locally stretched and compressed at different locations but never globally magnified, and the overall region is not magnified globally either. This feels a bit like stretching one side of a square but concluding that the elongated rectangle is the same area.
Of course, I doubt any of the above violates any basic principles, but it feels to me as if I am missing something. Or is a description of the linear magnification of orthogonal axes of a region insufficient to make any conclusions about the global magnification? Is there a third metric perhaps that explains the relationship between the linear and the areal magnification? I'd be happy if anyone could point me to somewhere in the literature that explains this phenomenon more completely, or if anyone could provide an explanation of how to think about this more clearly.
Thanks in advance.
 A: This is not really a topology problem, instead it is a multivariable calculus problem. 
What you need is the change of area formula from multivariable calculus. The key point is that your quantity $m(x,y)$ may be computed as the "Jacobian determinant" which means the determinant of the "Jacobian matrix" of partial derivatives 
$$m(x,y) = \text{det}\begin{pmatrix} \frac{\partial f_1}{\partial x}(x,y) & \frac{\partial f_1}{\partial y}(x,y) \\ \frac{\partial f_2}{\partial x}(x,y) & \frac{\partial f_2}{\partial y}(x,y) \end{pmatrix}
$$
The change of area formula is then simply the equation
$$\text{Area}(R) = \int\!\!\!\!\int_S m(x,y) \, dx \, dy
$$
Computing the linear magnification using only horizontal and vertical amounts to computation of the upper left and lower right entries of this matrix: if $q$ is the horizontal coordinate line passing through $(x,y)$ then your quantity $m_q(x,y)$ is simply the upper left entry of the matrix $\frac{\partial f_1}{\partial x}(x,y)$, whereas if $q$ is the vertical coordinate line then $m_q(x,y)$ is the lower right entry $\frac{\partial f_2}{\partial y}(x,y)$.
If the images under the mapping $f$ of the horizontal and vertical lines stay orthogonal then this is sufficient. One could do something similar with polar coordinates, using rays and circles (which are orthogonal). 
However, orthogonal coordinate systems are pesky beasts, they may not stay orthogonal under the mapping $f$. 
Intuitively, what is happening (on the infinitesmal level) is that instead of stretching a rectangle horizontally and vertically, in general the map $f$ distorts a rectangle into a parallelogram. The computation of (infinitesmal) parallelogram area is exactly what one obtains using the Jacobian determinant. 
Formally, if you take the columns of the above matrix as vectors, the quantity $m(x,y)$, which is the determinant of that matrix, is equal to the area of the parallelogram spanned by those two vectors.
If you want more details on this, any good calculus book should explain the Jacobian matrix and the Jacobian determinant and their roles in the change of variables formula. See also this.
