Average Momentum

Take an area-weighted average of the momenta over the integration region:

$\rho_{av} = \frac{\int \int \rho \, dA}{A}$

Parametrizations:

$m = r \, cos\theta \\ v = r \, sin\theta \\ \rho = m v = r^2 \, sin\theta \, cos\theta \\ dA = r \, dr \, d\theta$

Integral:

$\int \int \rho \, dA = \int_0^{\frac{\pi}{2}} \int_0^1 (r^2 \, sin\theta \, cos\theta) r \, dr \, d\theta \\ = \int_0^1 r^3 \, dr \int_0^{\frac{\pi}{2}} sin\theta \, cos\theta \, d\theta = \frac{1}{4} \frac{1}{2}$

Final result:

$\rho_{av} = \frac{\int \int \rho \, dA}{A} = \frac{\frac{1}{4} \frac{1}{2}}{\frac{\pi}{4}} = \frac{1}{2 \pi} \approx \boxed{0.15915}$