thermodynamics - Maxwell-Boltzmann distribution, average speed in one direction

Consider an ideal gas obeying the Maxwell-Boltzmann distribution i.e

$$f(v) = \bigg(\frac{m}{2 \pi k_{B} T}\bigg)^{3/2} \exp \left(-\frac{m v^{2}}{2 k_{B} T} \right) \, .$$

The probability distribution in 3D velocity space ($v^{2} = v_{x}^2 + v_{y}^2 + v_{z}^2$). How might you determine the average speed the particles are moving at, $\langle |v_{z}| \rangle$, in one direction?

Additionally if my ideal gas is now confined to a hemisphere in velocity space i.e we have the conditions $- \infty \leq v_{x}, v_{y} \leq \infty$ and $ 0 \leq v_{z} \leq \infty$ but it still has a Maxwell Boltzmann velocity distribution (except I think the normalization factor on $f(v)$ should change) then what is the average speed, or velocity, in the z direction $\langle v_{z} \rangle$, will this be the same as $|\langle v_{z}| \rangle$ from the previous answer?