Can I derive the 1 dimensional Maxwell-Boltzmann Distribution using this method? I am aware that the 1D MB Distribution is usually derived by counting the momentum states in various directions. However, I would like to know if the following method using the 3 dimensional speed distribution should also work for the derivation.
This link states that the 3 dimensional MB Distribution in terms of speed $P(v)$ is.
$$P(v)=\left(\frac{m}{2\pi k_B T}\right)^{\frac{3}{2}} \cdot \exp\left(-\frac{mv^2}{2k_BT}\right) \cdot 4\pi v^2 \cdot dv$$
This would mean that the probability for a speed $v$ in 1 unit of volume in speed space would be:
$$\frac{P(v)}{4\pi v^2\cdot dv}=\left(\frac{m}{2\pi k_B T}\right)^{\frac{3}{2}} \cdot \exp\left(-\frac{mv^2}{2k_BT}\right)$$
If I want to compute the probability of speed $v$ but with a certain fixed value of a dimensional speed component, say $v_z$, I would predict that this would give the 1 dimensional speed distribution $v_z$ that belongs to $v$. I would deduce that this can be derived by calculating the following circular shell volume in speed space, shown in blue lines:

This circular shell volume $V$ basically contains a fixed value of $v_z$ while varying $v_y$ and $v_x$. I would therefore reason that if I formulate this volume and multiply it with the probability distribution for the speed per 1 unit of volume (previous formula), I'd get the 1 dimensional MB distribution for $v_z$. Since this circular shell volume can also be drawn at the below side of the sphere, a factor of $2$ must be added. 
However, I have trouble formulating this volume in the first place. It would have a width of $v\cdot d\theta$, a thickness of $dv$ and a length of $\sin(\theta) \cdot v \cdot \int^{2\pi}_0 d\phi$ (the circumference at height $v_z$) which means that the volume is equal to:
$$V = v^2 \cdot d\theta \cdot dv \cdot \sin(\theta) \cdot \int^{2\pi}_0 d\phi$$
From this link, I can see that $dv$ can be rewritten in terms of the 3 speed components and $\theta$ and $d\phi$:
$$dv = \frac{dv_xdv_ydv_z}{v^2 \cdot \sin(\theta) \cdot d\theta \cdot d\phi}$$
Substituting $dv$ with this formula would give the following formula for the volume:
$$V = \frac{dv_xdv_ydv_z}{d\phi}\cdot \int^{2\pi}_0 d\phi$$
I'm not sure how to rewrite this to be able to continue with the derivation.
 A: The derivation in the previous answer gives you $P(v_z)$, but my understanding is that you want $P(v_z |v)$, that is, the probability of finding $v_z$ at a fixed $v$.
You should integrate about the z-axis a ring of radius $R=v \sin\theta$ with fixed $v$ and $v_z$. The element of volume will be:
$dV=dR dv_z ds=v\sin\theta d\phi dv_z \sin\theta dv=v\sin^2\theta dv_z dv d\phi$ 
You can eliminate $\sin\theta$ using $v_z=v\cos\theta$ and get $\sin^2\theta= 1-(\frac{v_z}{v})^2$ and get:
$P(v,v_z)dv dv_z=2 \pi\left(\frac{m}{2\pi k_B T}\right)^{\frac{3}{2}} v(1-(\frac{v_z}{v})^2) \exp\left(-\frac{mv^2}{2k_BT}\right)dv dv_z$
A: I don't fully uderstand your reasoning, but here I show a simple derivation for $P(v_i)$. First you have
\begin{gather}
P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z=P(v)P(\theta,\phi) ~\mathrm{d}v\mathrm{d}\theta\mathrm{d}\phi
\end{gather}
And since the distribution is isotropic we have
\begin{gather}
P(\theta,\phi) ~\mathrm{d}\theta\mathrm{d}\phi = \frac{\mathrm{d}\Omega}{4\pi}= \frac{\sin \theta ~ \mathrm{d}\theta\mathrm{d}\phi}{4\pi}
\end{gather}
Hence
\begin{align}
P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z&=\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2} \exp\left(-\frac{mv^2}{2m k_BT}\right) 4\pi v^2 \mathrm{d}v\times \frac{\sin \theta ~\mathrm{d}\theta\mathrm{d}\phi}{4\pi}\\
P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z&=\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2} \exp\left(-\frac{mv^2}{2m k_BT}\right) v^2 \sin \theta ~\mathrm{d}v\mathrm{d}\theta\mathrm{d}\phi\\
P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z&=\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2} \exp\left(-\frac{m(v_x^2+v_y^2+v_z^2)}{2m k_BT}\right)~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z
\end{align}
Finally you get
\begin{gather}
P(v_i)=\left(\frac{m}{2\pi k_BT}\right)^\frac{1}{2} \exp\left(-\frac{mv_i^2}{2m k_BT}\right)
\end{gather}
