Area of the event horizon of a rotating black hole

[Afonso Campos]

The Kerr metric for a black hole of mass $M$ and angular momentum $J = aM$ is

$$ds^{2} = - \frac{\Delta(r)}{\rho^{2}}(dt-a\sin^{2}\theta d\phi)^{2} + \frac{\rho^{2}}{\Delta(r)}dr^{2} + \rho^{2} d\theta^{2} + \frac{1}{\rho^{2}}\sin^{2}\theta (adt - (r^{2}+a^{2}) d\phi)^{2},$$

where $\Delta(r) = r^{2} + a^{2} - 2Mr$, $\rho^{2} = r^{2} + a^{2} \cos^{2}\theta$ and $- M < a < M$.

The event horizon is at $r_{+} = M + \sqrt{M^{2} - a^{2}}$. This is obtained by solving the equation $\Delta(r) = 0$.How do you use this to compute the area of the horizon?

My idea is to simplify the metric to obtain

$$ds^{2} = - \left( \frac{r^{2} + a^{2} - 2Mr - a^{2} \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dt^{2} + \left( \frac{r^{2} + a^{2} \cos^{2}\theta}{r^{2} + a^{2} - 2Mr} \right) dr^{2} - \left( \frac{4aMr \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dtd\phi + \left( r^{2} + a^{2} \cos^{2}\theta \right) d\theta^{2} + \sin^{2}\theta \left( \frac{(a^{2} + r^{2})^{2} - a^{2} \sin^{2}\theta (a^{2}-2Mr+r^{2}) }{r^{2} + a^{2} \cos^{2}\theta} \right) d\phi^{2}.$$

Then, I think that the area of the horizon is given by

$$A = \int d\theta\ d\phi\ g_{\phi\phi}g_{\theta\theta}|_{r=r_{+}}.$$

Am I wrong?

 

[PeterDonis]

My idea is to simplify the metric

You don't need to go through that step. Just evaluate the original metric for the case $dt = dr = 0$ and $r = r_+$, and integrate over $\theta$ and $\phi$.