I am trying to show that for any region $D$ in the hyperbolic plane, $A(D) = A(\phi(D)),$ where $$A = \int\int_D \frac{1}{y^2}dxdy$$ and $\phi$ is an isometry.
I have shown that the area remains constant under horizontal translations and homotheties. I am now trying to show that the area remains constant under inversion. If I am successful, then I have proved that the area remains constant under all isometries, as every isometry is a composition of horizontal translations, homotheties, and inversions.
To demonstrate what I did for homoethies: suppose $\phi(z) = \lambda z$ for some $\lambda > 0.$ Then, $$A(\phi(D)) = \int_{\lambda c}^{\lambda d}\int_{\lambda a}^{\lambda b}\frac{1}{y^2}dxdy,$$
and, given $u = \frac{1}{\lambda x}$ and $v = \frac{1}{\lambda y},$
$$A(\phi(D)) = \int_{\lambda c}^{\lambda d}\int_{\lambda a}^{\lambda b}\frac{1}{y^2}dxdy = \int_{c}^{d}\int_{a}^{b}\frac{1}{\lambda^2 v^2}\lambda du \lambda dv = \int_{c}^{d}\int_{a}^{b}\frac{1}{v^2} dudv$$
To attempt the demonstration of the constancy under inversion, I converted the integral into polar coordinates. I have not done calculus in a while, and I am pretty rusty with practical integration. How do you express the inversion of the function in the following integration (as I did with the homothety): $$A = \int\int_D \frac{1}{r\sin(\theta)}drd\theta?$$
Thank you!
Your strategy is surely correct but a bit redundant. By the following claim, you can use the same strategy for any generic isometry of $\Bbb H^2$.
Claim: The group of orientation-preserving isometries of $\Bbb H^2$ is $\text{PSL}_2\Bbb R$. The full group of isometries is $\text{PGL}_2\Bbb R$.
Following the claim above, an isometry is an element of the form $\displaystyle\phi(z)=\frac{az+b}{cz+d}$ such that $ad-bc=1$ if orientation preserving or $ad-bc=-1$ if orientation reversing.
Set $z=x+iy$ and set $w=\phi(z)=u+iv$. Using the Cauchy-Riemann equations we can calculate the Jacobian
$$ \frac{\partial(u,v)}{\partial(x,y)}=\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}=\Big(\frac{\partial u}{\partial x}\Big)^2+\Big(\frac{\partial v}{\partial y}\Big)^2=\Big|\frac{d\phi}{dz}\Big|^2=\frac{1}{|cz+d|^4}$$
Therefore
$$A\big(\phi(D)\big)=\int_{\phi(D)}\frac{dudv}{v^2}=\int_D \frac{\partial(u,v)}{\partial(x,y)}\frac{dxdy}{v^2}=\int_D \frac{1}{|cz+d|^4}\cdot\frac{|cz+d|^4}{y^2}dxdy=A\big(D\big)$$