Area of Kerr-Newman event horizon I want to calculate the area of event horizon for a Kerr-Newman black hole by using boyer's coordinates. 
I searched a lot from web, but I could not find any information about calculating event horizon radius for kerr newman. Can anyone help me to find any information, or help me how to calculate ?
 A: In geometrical-Gaussian units with $G$, $c$, and $\frac{1}{4\pi\epsilon_0}$ equal to 1, the Kerr-Newman metric for a black hole of mass $M$, angular momentum $J=aM$, and charge $Q$ is
$$\begin{align}
ds^2=
&-\left(1-\frac{2Mr-Q^2}{r^2+a^2\cos^2{\theta}}\right)dt^2
+\frac{r^2+a^2\cos^2{\theta}}{r^2-2Mr+a^2+Q^2}dr^2\\
&+(r^2+a^2\cos^2{\theta)}\,d\theta^2
+\left(r^2+a^2+\frac{a^2(2Mr-Q^2)\sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right)\sin^2{\theta}\,d\phi^2\\
&-\frac{2a(2Mr-Q^2)\sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\,dt\,d\phi
\end{align}$$
in Boyer-Lindquist coordinates $(t,r,\theta,\phi)$. (When $Q$ is zero, this reduces to Wikipedia’s form for the Kerr metric. Wikipedia’s form for the Kerr-Newman metric is equivalent to the above, but seems less straightforward.)
The $g_{rr}$ component of the metric tensor is infinite when the denominator
$r^2-2Mr+a^2+Q^2$ is zero. This happens at two radial coordinates,
$$r_\pm=m\pm\sqrt{m^2-a^2-Q^2}.$$
The event horizon is at $r_+$. We want to find the area of this surface. The 2D metric on the surface $t=$ constant and $r=r_+$ is
$$ds_+^2=
(r_+^2+a^2\cos^2{\theta)}\,d\theta^2
+\left(r_+^2+a^2+\frac{a^2(2Mr_+-Q^2)\sin^2{\theta}}{r_+^2+a^2\cos^2{\theta}}\right)\sin^2{\theta}\,d\phi^2$$
and the area element on this surface is
$$\begin{align}
dA_+&=\sqrt{\det{g_+}}\,d\theta\,d\phi\\
&=\sqrt{(r_+^2+a^2\cos^2{\theta})\left(r_+^2+a^2+\frac{a^2(2Mr_+-Q^2)\sin^2{\theta}}{r_+^2+a^2\cos^2{\theta}}\right)}\sin{\theta}\,d\theta\,d\phi\\
&=\sqrt{(r_+^2+a^2\cos^2{\theta})(r_+^2+a^2)+a^2(2Mr_+-Q^2)(1-\cos^2{\theta})}\sin{\theta}\,d\theta\,d\phi\\
&=\sqrt{(r_+^4+a^2r_+^2+2Ma^2r_+-a^2Q^2)+a^2(r_+^2-2Mr_++a^2+Q^2)\cos^2{\theta}}\sin{\theta}\,d\theta\,d\phi.
\end{align}$$
Conveniently, the coefficient of $\cos^2{\theta}$ in the square root vanishes by the definition of $r_+$, 
$$r_+^2-2Mr_++a^2+Q^2=0,$$
and, using this equation to eliminate $M$ in the first term in the square root, what's left under the square root becomes the perfect square $(r_+^2+a^2)^2$. Thus the area element simplifies to the trivial-to-integrate
$$dA_+=(r_+^2+a^2)\sin{\theta}\,d\theta\,d\phi.$$
Integrating over $\theta$ from $0$ to $\pi$ and over $\phi$ from $0$ to $2\pi$ gives the area of the event horizon,
$$A_+=4\pi(r_+^2+a^2)=4\pi\left(2M^2-Q^2+2M\sqrt{M^2-a^2-Q^2}\right).$$
