Put $\mu=r_+(r_+^2+a^2)$. At large $r$, the one-form dual to $K$ has
$$
K^\flat=g_{tt}\,dt+g_{t\phi}\,d\phi,
\qquad
g_{tt}=-1+\frac{\mu}{r\Sigma}
=-1+\frac{\mu}{r^3}+O(r^{-5}).
$$
Only the $dr\wedge dt$ term contributes to the constant-$t,r$ <Komar integral>, and
$$
dK^\flat=-\frac{3\mu}{r^4},dr\wedge dt+O(r^{-5}).
$$
With the stated <orientation>, the asymptotic <Hodge star operator> gives
$$
\star(dr\wedge dt)=r^4\sin\theta\cos^2\theta\sin\lambda\,
d\theta\wedge d\phi\wedge d\lambda\wedge d\psi.
$$
The angular integral is the area of the unit four-sphere,
$$
\int\sin\theta\cos^2\theta\sin\lambda\,
d\theta,d\phi,d\lambda,d\psi=\frac{8\pi^2}{3}.
$$
Consequently
$$
\int\star dK^\flat=-8\pi^2\mu,
\qquad
\boxed{M=-\frac1{12\pi}\int\star dK^\flat
=\frac{2\pi}{3}r_+(r_+^2+a^2)}.
$$
This is the <Komar mass> of the <Singly rotating six-dimensional Myers-Perry black hole> in the units of the question.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 / 2 / b / Solution

{c}

For the singly rotating six-dimensional Myers-Perry solution with $G=1$ and $\mu=r_+(r_+^2+a^2)$,
$$
M=\frac{2\pi}{3}\mu,
\qquad
\Omega_H=\frac{a}{r_+^2+a^2},
\qquad
\kappa=\frac{3r_+^2+a^2}{2r_+(r_+^2+a^2)},
$$
and the horizon area is
$$
A_H=\frac{8\pi^2}{3}r_+^2(r_+^2+a^2).
$$


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