general relativity - Schiff Equation in a polar orbit

I am aware that the formula for Schiff Equation in used to determine frame-dragging is: $$\boldsymbol{\Omega} = \frac{GI}{c^2r^3}\left( \frac{3(\boldsymbol{\omega}\cdot \boldsymbol{r})\boldsymbol{r}}{r^2}-\boldsymbol{\omega} \right).$$

But why does this reduce to: $$\boldsymbol{\Omega} = \frac{GI\boldsymbol{\omega}}{c^2r^3}\frac{\int_0^{2\pi}(3\cos^2\theta - 1)\text{d}\theta}{\int_0^{2\pi}\text{d}\theta} = \frac{GI\boldsymbol{\omega}}{2c^2r^3}$$ in a polar orbit?

Answer

The angular velocity of the Earth $\boldsymbol{\omega}$ is oriented in the $z$-direction, so $$ \boldsymbol{\omega} =(0,0,\omega). $$ For a circular polar orbit we can chose the $x$-axis such that the satellite moves in the $xz$-plane. Therefore, the position of the satellite is $$ \boldsymbol{r} = (r\sin\theta,0,r\cos\theta), $$ and the frame-dragging vector can be written as $$ \boldsymbol{\Omega}(\theta) = \frac{GI}{c^2r^3}\left(3\omega\sin\theta\cos\theta,\ \ 0,\ \ 3\omega\cos^2\theta - \omega\right) $$ The average frame-dragging over one revolution is thus $$ \boldsymbol{\Omega}_\text{av} = \frac{\int_0^{2\pi}\boldsymbol{\Omega}(\theta)\,\text{d}\theta}{\int_0^{2\pi}\,\text{d}\theta}. $$ We get $$ \Omega_{\text{av},x} = \frac{GI\omega}{c^2r^3}\frac{\int_0^{2\pi} 3\sin\theta\cos\theta\,\text{d}\theta}{\int_0^{2\pi}\,\text{d}\theta}=0, $$ and $$ \Omega_{\text{av},z} = \frac{GI\omega}{c^2r^3}\frac{\int_0^{2\pi} (3\cos^2\theta -1)\,\text{d}\theta}{\int_0^{2\pi}\,\text{d}\theta} = \frac{GI\omega}{c^2r^3}\frac{\pi}{2\pi}. $$