# 3-cylinder surface element (Poisson's “A Relativist's Toolkit”)

From Poisson's "A Relativist's Toolkit": he introduces the non-dynamical term $$S_0=\frac{1}{8\pi}\int_{\partial\Omega}\epsilon K\sqrt{\lvert h\rvert}d^3x$$ in the GR action, where $h$ is the determinant of the induced metric on the boundary, $\epsilon=\pm1$, $K$ is the trace of the second fundamental form on $\partial\Omega$ and $\Omega$ is of course the space-time region of integration. Let $g_{ab}$ be the solution of vacuum Einstein's eqns and consider a $3$-cylinder embedded in flat space-time with its axis along the time direction with bases $t=t_1,t=t_2$; the trace of the second fundamental form on the constant time hyper-surfaces is zero while on the lateral surface is $K=2/R$, where $R$ is the radius of the $3$- cylinder and $\sqrt{\lvert h\rvert}=R^2\sin\theta$. I need to compute the integral $$\int_{\partial\Omega}\epsilon \frac{2}{R}R^2\sin\theta\,d^3x=8\pi R(t_2-t_1)$$ but I can't understand which is the surface element on the $3$-cylinder. The induced metric is $$ds^2=-dt^2+R^2(d\theta^2+\sin^2\theta d\phi^2)$$ so the $dx^3$ should be of the form $dt\,f(\theta,\phi)d\theta\,d\phi$ but how can I obtain $f(\theta,\phi)$?

The volume form is $dt f(\theta,\phi)d\theta d\phi=(dt)(Rd\theta )(R\sin\theta d\phi).$
However, you already have that factor from $\sqrt{|h|}$ so your original equation might have meant to have the coordinate differential, not the 3-volume differential, in which case you should just use $d^3x=dtd\theta d\phi$ instead of the volume element $R^2\sin\theta dtd\theta d\phi.$
And indeed if $K=0$ on the ends and $K=2/R$ elsewhere then $\int_{\partial\Omega}KR^2\sin\theta dtd\theta d\phi=8R\pi(t_2-t_1).$