area form of the Poincare half plane

Question

For the upper half plane $\{(u,v)|v>0\}$, its area form is $du\wedge dv/v^2$.

How to compute the area between the u axis and the curve $\alpha(t)=(r\cos t, r\sin t)$, $0< t < \pi$?

Is this area infinite?

Answer 1

Let me write instead $\alpha(t) = (R\cos t,R\sin t)$.

On the one hand, in the polar coordinates $u = r\cos\theta$ and $v = r\sin\theta$, $du = \cos\theta dr - r\sin\theta d\theta$ and $dv = \sin\theta dr + \cos\theta d\theta$, so that $$ v^{-2} du \wedge dv = (r\sin\theta)^{-2}(\cos\theta dr - r\sin\theta d\theta) \wedge (\sin\theta dr + r\cos\theta d\theta)\\ = (r\sin\theta)^{-2} (r\cos^2\theta dr \wedge d\theta - rsin^2\theta d\theta \wedge dr)\\ = \frac{1}{r \sin^2\theta} dr \wedge d\theta \\ = - d(\log r) \wedge d(\cot \theta) $$ whilst on the other hand, the area in question $D$ is the image under polar coordinates of $(0,R) \times (0,\pi)$. But now, if we write $\theta = \operatorname{arccot}x$ and $r = e^y$ for new coordinates $x$ and $y$, so that $$ u = \frac{x}{\sqrt{1+x^2}}e^y, \quad v = \frac{1}{\sqrt{1+x^2}}e^y, $$ then on the one hand, $$ v^{-2}du \wedge dv = -d(\log r) \wedge d(\cot \theta) = -dy \wedge dx = dx \wedge dy, $$ whilst on the other, $D$ is the image of $(-\infty,\infty) \times (-\infty, \log R)$. Hence, the area of $D$ with respect to $v^{-2}du \wedge dv$ is just the Lebesgue measure (viz, usual area) of $(-\infty,\infty) \times (-\infty,\log R)$, which is infinite.