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?
 A: The average speed of particles in a particular direction will always be smaller than the average speed of particles.  
Imagine that you have three particles with components of their velocity $(1,1,1)\, \rm ms^{-1}$.
Their average speed is $<v>= \sqrt 3\, \rm ms^{-1}$ whilst their average speed in the x-direction is $<v_{\rm x}>=1\, \rm ms^{-1}$
So what you need to do is use the distribution of velocities in the x-direction
$$f(v_\rm{x}) = \left(\frac{m}{2 \pi k_{B} T}\right)^\frac{1}{2} \exp\left ({-\frac{m v_{\rm x}^{2}}{2 k_{B} T}}\right )$$
and do the following integration
$\displaystyle \int_0^\infty v_\rm{x}\, f(v_\rm{x}) \,dv_{\rm x}$  which will give you an answer of $\dfrac {<v>}{4}$ where $<v>$ is the avergae speed of the particles.
You may find these notes of use?
A: The probability density function (PDF) for a single component (let's take $z$) is
$$f(v_z)=\sqrt{\frac{m}{2\pi k T}} \exp \left(-\frac{m v_z^2}{2kT}\right)$$
Case #1
What you want to calculate is
$$\langle |v_z| \rangle = \int_{-\infty}^{\infty} |v_z| \ f(v_z) \ d v_z = 2 \int_0^\infty v_z \ f(v_z) \ d v_z$$
Using 
$$\int_0^{\infty} u e^{-au^2} du = \frac 1 {2a} \tag{1}$$
you get
$$\langle | v_z| \rangle = 2 \cdot \sqrt{\frac{m}{2\pi k T}} \cdot \frac 1 2 \cdot \frac{2kT}{m} = 2 \sqrt{\frac{kT}{2 \pi m}} =  \sqrt{\frac{2kT}{\pi m}} $$
Notice that since
$$\langle |v| \rangle = \sqrt{\frac{8kT}{\pi m}}$$
where $|v|= \sqrt{v_x^2+v_y^2+v_z^2}$, you have 
$$\langle | v_z| \rangle = \frac 1 2 \langle |v| \rangle $$
Case #2
As you said, in this case the normalization factor must change, since we must require that
$$\int_0^\infty \tilde f(v_z) d v_z = 1$$
The correct PDF is in this case
$$\tilde f(v_z)= \frac 1 2 \sqrt{\frac{m}{2\pi k T}} \exp \left(-\frac{m v_z^2}{2kT}\right)$$
Using again (1), it is straight-forward to obtain from this distribution
$$\langle v_z \rangle_{\tilde f} = \frac 1 2 \sqrt{\frac {kT}{2 \pi m}} = \frac 1 4 \langle |v_z| \rangle = \frac 1 8 \langle |v| \rangle $$
