# Comparing areas between conformal metrics in $\mathbb{R}^2$

I would like to ask for a reference on the following subject:

Let $$f:\mathbb{R}^2\longrightarrow\mathbb{R}$$ be a radial positive function, i.e., $$f(x,y)=\lambda(x^2+y^2)$$ for some $$C^\infty$$ positive function $$\lambda:[0,\infty)\longrightarrow\mathbb{R}$$. So $$g_{(x,y)}(\cdot,\cdot)=\lambda(x^2+y^2)\langle\cdot,\cdot\rangle$$ defines a metric in each point $$(x,y)\in\mathbb{R}^2$$, conformal to the euclidean metric. If $$S_{\text{euc}}(A)$$ and $$S_g(A)$$ are the areas of $$A$$ for the euclidean and $$g$$ metric respectively, what is $$S_{\text{euc}}(A)$$ in terms of $$S_g(A)$$ or vice-versa?

Additionally, are there any special features about the quantity $$S_g(A)$$ if $$f$$ is harmonic/subharmonic?

Any references, books etc will be trully appreciated!

• Are you expecting a simple relation between $\int _A dxdy$ and $\int_ A \lambda(r^2)dxdy$ for arbitrary $A$? Commented Jul 31 at 21:45
• I believe if I just could get some expression for $S_g(A)$, even if it were not nice, that would help. Maybe some formula using change of variables on $S_g$... Commented Jul 31 at 21:48
• In fact, I am searching in books what would be that formula, but I couldn't find any reference. I even tried chatgpt, it gave me something like $S_g(A)=\iint_A(1/f)dxdy$. Commented Jul 31 at 21:52

In general if $$M$$ is a smooth $$n$$-dimensional manifold and $$g_1,g_2$$ are two Riemannian metrics which are conformal, say $$g_2=fg_1$$ for some smooth positive function $$f:M\to\Bbb{R}$$, then the induced volume measures are related as \begin{align} dV_{g_2}&=f^{n/2}\cdot dV_{g_1},\tag{1} \end{align} meaning that for all (Lebesgue-measurable) sets $$A\subset M$$, we have \begin{align} \text{vol}_{g_2}(A)&=\int_Af^{n/2}\,dV_{g_1}.\tag{2} \end{align} Now, unless $$f$$ is constant, you can’t simplify this any further to get something like $$\text{vol}_{g_2}(A)=f^{n/2}\cdot\text{vol}_{g_1}(A)$$.
In order to prove this result, you simply need to recall that the volume measure $$dV_g$$ is essentially defined such that in local coordinates $$(U,x)$$, we have for all (Lebesgue-measurable) $$A\subset U$$, $$\text{vol}_g(A):=\int_{x[A]}\sqrt{|\det g_{(x), ij}|}\,dx$$. Note that if you don’t know the language of measure theory, that’s perfectly fine; you can simply assume $$M$$ is oriented, and focus on the Riemannian volume form, and you’ll still get the same result.
Now if you have two conformally related metrics, then you simply note that the determinant is simplified as follows: \begin{align} \sqrt{|\det (g_2)_{(x),ij}|}=\sqrt{|\det f\cdot (g_1)_{(x), ij}|}=\sqrt{f^n\cdot |\det (g_1)_{(x),ij}|}=f^{n/2}\sqrt{|\det (g_1)_{(x),ij}|}, \end{align} where I simply used that if a matrix is multiplied by $$f$$, then by multi-linearity of the determinant in each of the $$n$$ columns, the determinant is multiplied by $$f^n$$. So that’s where the $$f^{n/2}$$ comes from in $$(1)$$.
Finally, in your special case, we have $$n=2$$ so $$f^{n/2}=f$$, hence \begin{align} S_g(A)&=\int_Af(x,y)\,dx\,dy. \end{align} Besides this, there isn’t a simpler way of expressing it. Even if you assumed the conformal factor is radial, or harmonic or whatever. You still need to evaluate the integral, and you need to know how the set $$A$$ looks.
But, one special case I can give you is if $$f(x,y)=\lambda(x^2+y^2)$$ is radial, and $$A$$ is a Euclidean annulus centered at the origin with radii $$0\leq r_1, then \begin{align} S_g(A)&=\int_Af(x,y)\,dx\,dy=2\pi\int_{r_1}^{r_2}\lambda(r^2)\cdot r\,dr, \end{align} which thus leaves you with a single variable integral to deal with (which could still be very complicated). Really, I would prefer you normalize the radial function as $$f(x,y)=\phi(\sqrt{x^2+y^2})$$, so the formula reads $$S_g(A)=2\pi\int_{r_1}^{r_2}\phi(r)\cdot r\,dr$$.