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A five-dimensional Myers-Perry black hole with equal rotational parameters has a metric depending on a single radial coordinate. In a fibered coordinate system, its radial functions can be written $f=1-r_0^2/r^2+{\alpha }^2{r}_0^2/r^4$, $h=1+{\alpha }^2{r}_0^2/r^4$, and $\Omega =\alpha r_0^2/(r^{4}h)$. For $0<\alpha <r_0/2$ there are outer and inner Killing horizons; at equality they form a degenerate Killing horizon. With the standard sphere identifications and $G_5=1$, its Komar mass is $3\pi r_0^2/8$.

For the magnetically charged dilaton black hole, the metric-dual one-form of $k={\partial }_t$ is $k^{♭}=-Vdt$. With orientation $dt\wedge dr\wedge d\theta \wedge d\varphi$, its Hodge star operator gives $\star d{k}^{♭}=-V^{\prime }{R}^2\sin \theta d\theta \wedge d\varphi$. Therefore the Komar mass is $R^2{V}^{\prime }/2=r_{+}/2-P^2/r$. Its radial dependence is allowed because the Ricci tensor is nonzero; the divergence of $d{k}^{♭}$ is sourced by $R_{ab}{k}^b$. At the horizon $M_K(r_{+})=(r_{+}-r_{-})/2=\kappa A_H/(4\pi )$.

Use geometrized units $G=c=1$. For a stationary asymptotically flat spacetime, normalize the timelike Killing vector field $\xi$ to have squared norm $-1$ at infinity. The Komar mass isHere the oriented binormal is chosen so that an ordinary positive-mass static solution gives a positive integral. Equivalently, on a static slice with lapse $N=\sqrt{-{\xi }^2}$ and outward spatial unit normal $s^i$,This second expression fixes the orientation unambiguously and is convenient for the Majumdar–Papapetrou solution.

The spatial metric is ${\gamma }_{ij}=H^2{\delta }_{ij}$ and $N=H^{-1}$. On a large coordinate sphere, $s^i=H^{-1}{\hat{r}}^i$ and $dA=H^2{r}^{2}d\Omega$. ThusThe original PDF has the Euclidean norm in each source denominator, which is missing in the TeX aid. With that norm, the harmonic function has asymptoticsThereforeIf $M_i$ are retained as length parameters while $G$ is restored, the physical mass is ${\sum }_i{M}_i/G$. This is the charge at infinity. A Komar integral is not generally surface-independent here, since Einstein-Maxwell theory has the electromagnetic stress-energy tensor between the surfaces; in particular it should not be confused with a sum of vacuum horizon Komar charges.

Work in geometrized units $G=c=1$, with the stationary Killing vector field $K={\partial }_t$ normalized by $K^2\to -1$ at infinity. Put $q=2m{\sinh }^2\alpha$, $f=F/\sqrt{H}$, and $R^2=r^2\sqrt{H}$; $R$ is the areal radius. The static spherical Komar mass in reciprocal radial gauge givesIndeed the future and outward unit normals to a round sphere are $n=f^{-1/2}{\partial }_t$ and $s=f^{1/2}{\partial }_r$, and $s^a{n}^b{\nabla }_a{K}_b=-f^{\prime }/2$. The Komar mass integral $-(4\pi {)}^{-1}\int s^a{n}^b{\nabla }_a{K}_{b}dA$ is therefore $R^2{f}^{\prime }/2$. Since $F^{\prime }=2m/r^2$ and $H^{\prime }=-q/r^2$,Taking the asymptotic limit yieldsThis agrees with $g_{tt}=-1+(2m+q/2)/r+O(r^{-2})$. The finite-radius Komar integral need not equal its asymptotic value, because this solution is not vacuum. Restoring $G$ divides the mass expression by $G$ when $m$ has dimensions of length.

For the Komar integral of a five-dimensional Reissner--Nordstrom metric, use the one-form metrically dual to the stationary Killing vector field: $k^{♭}=-f(r)dt$. Its exterior derivative is $d{k}^{♭}=f^{\prime }(r)dt\wedge dr$. The stated orientation gives volume formBecause the squared norm of $dt\wedge dr$ is $g^{tt}{g}^{rr}=-1$, the defining wedge identity for the Hodge star operator givesThus the Komar integral over the positively oriented angular three-sphere issince the unit three-sphere has volume $2{\pi }^2$. Write $s=r_{+}^2+r_{-}^2$, $q=r_{+}^2{r}_{-}^2$. Then $f=1-s/r^2+q/r^4$ and $r^3{f}^{\prime }=2s-4q/r^2$. With the normalization specified for the Komar mass,The orientation makes this positive. The problem's coefficient $1/(16\pi )$ must be retained: it is not the usual dimension-adjusted five-dimensional ADM-normalized Komar coefficient. With that conventional normalization the ADM mass would instead be $3\pi s/8$.

Put $q=r_0^2$ and $\eta =d\psi +\frac{1}{2}\cos \theta d\varphi$. Expanding the metric tensor in the stationary direction gives the useful exact identitiesTaking the exterior derivative givesThe fibered form of the metric tensor has determinant $\det g=-r^6{\sin }^2\theta /16$. Its inverse components needed for the flux areThereforesince $h(1-\alpha \Omega )=1$. This calculation includes the rotational mixed term; dropping it prematurely would miss the exact cancellation.

With the stipulated negative coordinate orientation, ${ϵ}_{tr\psi \theta \varphi }=-r^3\sin \theta /4$. The pullback of the Hodge star operator to the constant-$t,r$ three-dimensional spacelike submanifold is consequentlyThe coordinate ranges and standard sphere identifications giveThus the five-dimensional Komar mass, in the normalization given and $G_5=1$, isThe flux is already independent of $r$ in the vacuum region, so its limit at infinity has the same value. The negative orientation in the question is crucial for the positive sign; reversing that orientation reverses the flux.

Use the intended positive-mass family: write $b=r_{+}>0$ and $a=r_{-}=2P^2/b\ge 0$, so $R^2=r(r-a)$, $V=1-b/r$ and $ab=2P^2$. Lowering the index of the Killing vector field gives $k^{♭}=-Vdt$, henceFor the stipulated orientation and Lorentzian signature, $\sqrt{-g}=R^2\sin \theta$ and the Hodge star operator obeysThe minus sign follows from $g^{tt}{g}^{rr}=-1$. Thus the Komar mass of the magnetic dilaton black hole isIts asymptotic mass is $M=r_{+}/2$.

For any Killing vector field, its antisymmetric derivative satisfies ${\nabla }_a(d{k}^{♭}{)}^{ab}=-2R^b{}_c{k}^c$. Applying Stokes theorem between two spheres relates the difference of their Komar mass to this Ricci source. It vanishes in vacuum but need not vanish for the nonzero Ricci tensor of Einstein-Maxwell-dilaton theory. Here, explicitly, $d{M}_{\mathrm{Komar}}/dr=P^2/r^2$. The potential $A=P\cos \theta d\phi$ is only a local gauge potential near the polar axes; the mass computation uses its globally defined field strength and the metric, so does not depend on this gauge-patch issue.

Put $\mu =r_{+}(r_{+}^2+a^2)$. At large $r$, the one-form dual to $K$ hasOnly the $dr\wedge dt$ term contributes to the constant-$t,r$ Komar integral, andWith the stated orientation, the asymptotic Hodge star operator givesThe angular integral is the area of the unit four-sphere,ConsequentlyThis is the Komar mass of the Singly rotating six-dimensional Myers-Perry black hole in the units of the question.

For $d{s}^2=-f(r)d{t}^2+d{r}^2/f(r)+R(r{)}^{2}d{\Omega }^2$ and $K={\partial }_t$ normalized at infinity, the Komar mass on a round sphere is $R^2{f}^{\prime }/(2G)$. The future unit normal to a static slice is $f^{-1/2}K$, and the outward unit radial normal is $f^{1/2}{\partial }_r$; their contraction with ${\nabla }_a{K}_b$ gives $f^{\prime }/2$, whose integral over the sphere yields the formula. The asymptotic mass is its limit at infinity. In the presence of matter, the finite-radius integral need not be constant.