ed}
\nonumber
K= W = 48{m}^{2} \left( \alpha r\cos \left( \theta \right) -1 \right) ^{6} \left( \left({a}^{4}\alpha+{a}^{3} \right) \cos^{3}\theta
+ 3{a}^{2}r \left( a \alpha-1 \right) \cos^{2}\theta - 3a {r}^{2} \left( a\alpha+1 \right) \cos \theta -{r}^{3}  \left( a\alpha-1 \right)  \right)  \\
\times \dfrac{ \left(  \left( {a}^{4}\alpha - {a}^{3} \right) \cos^{3}\theta - 3{a}^{2}r \left( a \alpha+1 \right)  \cos^{2}\theta
- 3a {r}^{2}\left(a\alpha-1 \right) \cos \theta + {r}^{3}  \left( a\alpha+1 \right)  \right) } { \left( {r}^{2} + {a}^{2} \cos^{2}\theta \right) ^{6} }.\end{aligned}$$ Therefore $ P^{2}=\dfrac{W}{K}=1 $, i.e. $ P=+1 $. Hence the total gravitational entropy in this case is given by $$\label{s_grav_rot}
S_{grav}=k_{s}\int_{\sigma}\mathbf{\Psi}.\mathbf{d\sigma}=k_{s}\int_{\sigma}d\sigma=k_{s}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi C}\sqrt{g_{\theta\theta}g_{\phi\phi}}d\theta d\phi.$$ The entropy evaluated at $r_{\pm}$ is obtained as $$\label{s_grav_pm}
S_{grav_{\pm}}=k_{s}\dfrac{4\pi C(r^{2}_{\pm}+a^{2})}{(1-\alpha^{2}r_{\pm}^{2})}=k_{s}\dfrac{4\pi (r^{2}_{\pm}+a^{2})}{(1-\alpha^{2}r_{\pm}^{2})(1+2\alpha m+\alpha^2a^2)}.$$ If we substitute $ a=0 $ in (\[s\_grav\_pm\]), then we get back the expression (\[s\_grav\_nonrot\]) for the entropy of the non-rotating accelerating black holes. We see that as the acceleration parameter vanishes, i.e., $ \alpha\rightarrow 0 $, the equation (\[s\_grav\_pm\]) reduces to the expression of gravitational entropy for Kerr black holes derived in [@entropy2]. However for this axisymmetric metric, it is not possible to evaluate the spatial metric using equation (\[sm\]) because the object is rotating, and so there is a nonzero contribution from the component of $ g_{t\phi} $, which changes the spatial positions of events in course of time. Therefore the entropy density is calculated by using the full four-dimensional metric determinant $ g $ in the expression involving the covariant derivative [@entropy2], and we get $$\label{enden1}
s=k_{s}|\mathbf{\nabla}.\mathbf{