Is this equation for the rotational weight of a half disk correct? I need to know what rotational force the weight of this disk applies to the axle.

I'm not sure what the units are, whether it's torque or something else I'm looking for. The weight of the disk is $z\frac{\pi r^2}{2}$ where $z$ is the thickness however I'm only looking for the rotational force thus ignoring the weight that is supported by the axle.
After a lot of thinking I came up with the integral:
$$z\int_{0}^{r} \int_{0}^{\pi} r^2 \sin(\theta)  d\theta dr=z\frac{2r^3}{3}$$
Later I found the equation for the center of mass of a half disk: $\frac{4r}{3\pi}$ and when multiplied by the mass of the disk it equates to the same answer: $z\frac{\pi r^2}{2}\frac{4r}{3\pi}=z\frac{2r^3}{3}$
I'm not perfectly sure what I have here but my intuition tells me that a scale at the tip of a rod of length $1$ would measure $z\frac{2r^3}{3}$
 A: The volume of the disk is $z\frac{\pi r^2}{2}$ not the weight. Otherwise you have done everything correctly! :) 
Here's how I solved your problem, it's very similar to your solution but keeps track of the directions involved (and doesn't require any integration). Apologies if this does not clarify things.
The rotational force or torque ($\tau$) is equivalent to the weight ($\vec{F}$) acting at the center of mass/gravity multiplied by the distance to the center of mass ($\vec{d}$).
$$
\vec{\tau} = \vec{d} \times \vec{F} 
$$
The force is from the weight of the disk and in the downwards direction
$$
\vec{F} = -mg \ \hat y = - \rho V g \ \hat y = - \frac{1}{2}\rho z\pi r^2 g \ \hat y
$$
The distance from the axis to the center of mass is, like you said, given by $\vec{d}=\frac{4r}{3\pi} \ \hat{x}$. Therefore the torque
$$
\vec{\tau} = (\frac{4r}{3\pi}) \ \hat{x} \times (- \frac{1}{2}\rho z\pi r^2 g) \ \hat y \\
= -\frac{2r^3 \rho z g}{3} \hat z
$$
The magnitude of the torque is like how strong the turning force is, and the direction of the torque is perpendicular to both $F$ and $r$. If this concept is new to you, I suggest reading up on cross products and torque.
