= Solution
The new angular interval has width $\pi/2$, so its <probability density function> is $2/\pi$ there. Its endpoints stay away from the <poles> of the <tangent function>, making $Y$ bounded and its <expected value> finite. Integrate over the <continuous uniform distribution> of angles:
$$
E[Y]=\frac{2x_0}{\pi}\int_{-\pi/6}^{\pi/3}\tan\theta\,d\theta
=\frac{2x_0}{\pi}[-\ln(\cos\theta)]_{-\pi/6}^{\pi/3}.
$$
The two <cosines> are $\sqrt3/2$ and $1/2$. Hence
$$
\boxed{E[Y]=\frac{2x_0}{\pi}\ln\sqrt3=\frac{x_0\ln3}{\pi}.}
$$
It has units of <length> and is positive because the permitted positive angles extend farther than the negative ones.
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