What is the simplest way to calculate Gibbons-Hawking-York boundary term for Schwarzschild metric?
\begin{align} \int Kd\Sigma&=-32\pi^2m\left(1-2Mr^{-1}\right)^{1/2}\times\frac{d}{dr}\left[ir^2\left(1-2Mr^{-1}\right)^{1/2}\right]\\ &=-32\pi^2iM\left(2r-3M\right). \end{align}
The Gibbons-Hawking boundary term for a spacetime manifold is explicitly,
$$S_{GH}=\frac{1}{8\pi G}\int_{\partial M} d^3x \, \sqrt{|h|} \, K$$
where $\partial M$ is the boundary of $M$, $K$ the extrinsic curvature, and $h$ the determinant of the metric on the boundary. Let us Wick rotate the Schwarzschild metric to,
$$ds^2 = \left( 1-\frac{2GM}{r}\right)d\tau^2 + \left( 1-\frac{2GM}{r}\right)^{-1} dr^2 + r^2 d\Omega^2$$
We must impose a radial cutoff $R > GM$. The normal vector on the boundary is given by,
$$n=-\sqrt{1-\frac{2GM}{r}}\frac{\partial}{\partial r}$$
with a minus sign since we require the outward pointing normal, which points into the bulk. The metric on the boundary is then given by,
$$ds^2=\left( 1-\frac{2GM}{R}\right)d\tau^2 + R^2d\Omega^2$$
The extrinsic curvature is simply the divergence of the normal:
$$K=\nabla_a n^a = \frac{1}{r^2}\partial_r (r^2 n^r) \biggr\rvert_{r=R}= -\frac{2}{R}\sqrt{1-\frac{2GM}{R}} - \frac{GM}{R^2} \frac{1}{\sqrt{1-\frac{2GM}{R}}}$$ Can you take the calculation from here?
