Doubt about the volume of the air gap in an electrical device
The volume you are trying to calculate is an area between arcs of the same inner angle, but of different radius. It's called an annular sector:

  

Now, technically, you will need a bit of mathematics to calculate this without any reference but, in the end, the volume is equal to the area of this small region, times the height (which is h in your case). 

Practically, according to the shape above, and the picture you posted, you have an area that runs from r to R or, from your shape, it runs from \$r_1\$ to \$r_1 + g\$.

You have two coordinates (these are polar coordinates here), the radius \$r\$ and the angle \$θ\$ (a parameter), which runs from \$0\$ to \$θ\$ (the actual value) in your case. Remember from differential calculus that the area element in polar coordinates is equal to \$rdrdθ\$ (where \$θ\$ is in radians). While you 're at it, check this depiction as well, in which \$Δr\$ represents an infinitesimally small length and \$Δθ\$ an infinitesimally small angle. Because the elements are infinitesimal, this can be considered approximately equal to a rectangle, so its area (in terms of differential calculus) is the product of its sides:


So the area you want to calculate is given by a double integral of the area element over the ranges spanned by the two coordinates in the given area:

\$\int_{0}^{θ}(\int_{r_1}^{r_1+g}rdr)dθ\$

Because the function in the inner integral does not contain any \$θ\$, the two integrals can be separated and calculated as multiples:

\$\int_{0}^{θ}dθ(\int_{r_1}^{r_1+g}rdr)=\left[θ\right]_{0}^{θ} \left[\frac{r^2}{2}\right]_{r_1}^{r_1+g} =\frac{1}{2}θ \left[r^2\right]_{r_1}^{r_1+g} = \frac{1}{2}θ \left[ \left(r_1+g \right )^2 - \left(r_1\right)^2 \right] = \frac{θ}{2} \left[ r_1^2 + 2r_1g + g^2 - r_1^2 \right] = \frac{θ}{2} \left[ 2r_1g + g^2 \right] = θg \left(r_1 + \frac{g}{2} \right)  \$

This is only one side of the rotor. Twice that is equal to the area of both sides:

\$ A = 2g \left(r_1 + \frac{g}{2} \right)θ \$

This is the area. If you multiply with the height \$h\$ (vertical dimension) you get the volume you are looking for:

\$ V = 2gh\left(r_1 + 0.5g\right)θ\$