# 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.

• It is not very clear to me how do you plan to "slice" such object, but in order to find its volume you just need to compute the area of a generic section given by a plane parallel to the bases. – Jack D'Aurizio Dec 27 '16 at 23:41
• I think this can be useful math.stackexchange.com/questions/2072846/… – cgiovanardi Dec 28 '16 at 12:08
• For something this general, you might consider employing some calculus. How comfortable are you with it? – J. M. ain't a mathematician Dec 28 '16 at 19:53
• @J.M.isn'tamathematician I only know differentiation so far but I feel fairly comfortable with that. – Joe Dec 30 '16 at 1:18

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.

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

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

3. 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 $$V = \int_{z_{0}}^{H - h} \left[\frac{R^{2} z^{2}}{H^{2}} \arccos\left(\frac{(z + mR - H)H}{mRz}\right) - \frac{z + mR - H}{m} \sqrt{\frac{R^{2} z^{2}}{H^{2}} - \frac{(z + mR - H)^{2}}{m^{2}}}\right] dz.$$

• and how would you take the derivative $\frac{dV}{d\theta}$ – Joe Jan 28 '17 at 5:57
• That's a bit of a mess; structurally, we have$$V = \int_{g(m)}^{c} f(z, m)\, dz,\qquad m = \tan\theta.$$The fundamental theorem of calculus and differentiation under the integral give$$\frac{dV}{dm} = \int_{g(m)}^{c} \frac{\partial f}{\partial m}(z, m)\, dz - f(g(m), m)\, g'(m),$$while $\dfrac{dV}{d\theta} = \dfrac{dV}{dm}\, \dfrac{dm}{d\theta} = \dfrac{dV}{dm} \sec^{2}\theta$ by the chain rule. – Andrew D. Hwang Jan 28 '17 at 14:30