Let's say I have an assembly of randomly aligned cylinders. The orientation is presented with orientation density function (ODF) $f(\theta, \phi)$, where $\theta$ is the polar angle (w.r.t. $z$-axis) and $\phi$ is the in-plane angle (w.r.t. $x$-axis).
And the ODF satisfies the following condition.
$$ \int_{\phi=0}^{\pi}\int_{\theta = 0}^{\pi/2} f(\theta, \phi) \sin{\theta} d\theta d\phi = 1 $$
I assume that the alignment of cylinders is in a plane i.e. $x-y$ plane, that is $f(\theta=\pi/2, \phi)$, which can be written as a function of $\phi$ only as given below,
$$ \int_{\phi=0}^{\pi}f(\phi)d\phi = 1 $$
Thus for a uniformly random planar assembly, we can write $f(\phi) = \frac{1}{\pi}$
Further, the average projection of the cylinders on $x$-axis calculated as,
$$ \int_{0}^{\pi}f(\phi)|\cos{\phi}|d\phi = \frac{1}{\pi} \int_{0}^{\pi}|\cos{\phi}|d\phi = \frac{2}{\pi} $$
Now, I rotate the same network and place it in $y-z$ plane. This means that the inplane angles $\phi$ [$0-\pi$] becomes polar angle $\theta$ [$0-\pi/2$], and all inplane angles becomes contant i.e. $\phi = 90^o$ or $\pi/2$.
Consequently, the orietation density function, $f(\theta, \phi)$ can be written as $f(\theta)$ with following normalization condition,
$$ \int_{\theta=0}^{\pi/2}f(\theta)\sin{\theta}d\theta = 1 $$
Thus the value of ODF becomes $f(\theta) = 1$.
Now we can calculate the projection on $z-axis$ as,
$$ \int_{\theta=0}^{\pi/2}f(\theta)\sin{\theta} \cos{\theta} d\theta = 1\int_{\theta=0}^{\pi/2}\sin{\theta} \cos{\theta} d\theta $$ $$ \Rightarrow \frac{1}{2}\int_{\theta=0}^{\pi/2}\sin{2\theta}d\theta = \frac{1}{2} \frac{(-cos2\theta)}{2}\bigg\rvert_{0}^{\pi/2} = \frac{1}{4}(-\cos{\pi} + \cos{0}) = \frac{1}{2} $$
I was expecting that the answer should be the same for both cases since I am just rotating the assembly from $x$-axis to $z$-axis and now looking into the $z$-axis.
I do not know what point I am missing in understanding or interpreting the ODF or the projections. Any help would be highly appreciated.