Volume of a truncated conical frustum If you have a conical frustum with a volume
$$V_f=\frac{\pi h}{3}\left( r^2+rR+R^2 \right)$$
with $h$ being the distance between the bases, $r$ the radius of the smaller circle, and $R$ the radius of the larger. 
 Then you truncate the frustum by a plane that intersects the larger base at one point at an angle $\theta$. What would the volume of the resulting figure that does not contain the large base?

The following image shows the cone with the red the volume I wish calculated.

$$m = \tan\theta$$
The formula for the red section would be the following with $V_b$ the volume of the blue section.
$$V=\frac{\pi R^{2} H}{3}\left(\frac{H - mR}{H + mR}\right)^{3/2}-\frac{\pi}{3}r(H-h) + V_b $$
I need to know what the volume of the blue section would be or the purple.
 A: Here's a sketched calculation of the volume of the "wedge", too long for a comment, but not yielding an algebraic formula (though the integral is elementary, i.e., can be evaluated in closed form in principle).

Put the origin at the vertex, the $x$-axis horizontal, and the $z$-axis vertical. Each slice of the wedge by a horizontal plane (heavy line) is a chord of a disk (shaded, top). By Cavalieri's principle, the volume of the wedge is the integral of the cross-sectional areas.


*

*If $|x_{0}| \leq r$, the chord $\{-r \leq x \leq x_{0}\}$ inside the disk of radius $r$ has area
$$
A(r, x_{0}) = r^{2} \arccos(x_{0}/r) - x_{0} \sqrt{r^{2} - x_{0}^{2}}.
$$

*Putting $m = \tan\theta$, the cutting plane has equation
$$
z - H = m(x - R),\qquad\text{or}\quad
x = \frac{z + mR - H}{m}.
$$

*The cone has profile $z = (H/R)|x|$ (hence $r = (R/H)z$ in 1.), so the wedge (shaded triangle) lies in the region
$$
z_{0} := H\, \frac{H - mR}{H + mR} \leq z \leq H - h.
$$
1. Gives the area of the cross sections; 2. gives the value of $x_{0}$ in terms of the height $z$; 3. gives the limits of integration. The wedge has volume
$$
V = \int_{z_{0}}^{H - h} \left[\frac{R^{2} z^{2}}{H^{2}} \arccos\left(\frac{(z + mR - H)H}{mRz}\right) - \frac{z + mR - H}{m} \sqrt{\frac{R^{2} z^{2}}{H^{2}} - \frac{(z + mR - H)^{2}}{m^{2}}}\right] dz.
$$
