How to find volume of a truncated cone by slant height If i know base radius and slant height then how i can find the volume of a truncated cone.I do some research but not able to find out how exactly these two can be used to find volume.http://www.vitutor.com/geometry/solid/truncated_cone.html This link is somehow useful but not able to find exact answer
 A: If you know the radii of the truncated cone to be $R_{1}$ and $R_{2},$ and the slant height to be $L,$ we can find the height of the frustum and the volume will follow.
Using Pythagoras on the right triangle with hypotenuse on the slant surface and base on the bottom base, we see that the height is $\sqrt{L^{2} - (R_{2} - R_{1})^{2}}.$
Now we use the formula for the volume of a frustum using our findings. We see that $\boxed{V = \frac{\pi \sqrt{L^{2} - (R_{2} - R_{1})^{2}}}{3} \cdot (R_{1}^{2} + R_{1}R_{2} + R_{2}^{2})}.$
A: That is not enough information.  Imagine two truncated cones with the same radius $r$ and a slant height of $h$ which is just a little less than $r$.  The truncated cone can be very thin with a small upper base, or the walls can be almost vertical with the upper base having a radius just less than $r$.  The two truncated cones agree on $r,h$ but not on volume.  Imagine rotating the two trapezoids below to make truncated cones.  They have the same $r$ and $h$.
A: Volume V depends on how much variable $\theta $ determines the height and radial difference by rotation.
Given base-radius is R.
It can be minimum $V=0$ when $\theta=0$, maximum $V=  \pi R^2 L$ when $\theta= \frac{\pi}{2}.$
when $ r_2= R,\, r_1= R- L\cos\theta,\, h =L \sin \theta $
$$ V= \frac{\pi h}{3} (r_1^2+ r_2^2+ r_1 r_2) $$
or
$$ V = \pi R^2 L\; \sin \theta\cdot\big(1-\dfrac{L\cos\theta}{R}+\dfrac{(L\cos\theta)^2}{3R^2}\big) $$
which can be expressed in terms of $R,h$ as:
$$ V = \pi R^2 h\cdot\big(1-\dfrac{\sqrt{L^2-h^2}}{R}+\dfrac{{L^2-h^2}}{3R^2}\big) $$

