# How to derive the volume and surface area formula for truncated cylinder without calculus?

several hours ago, I saw the problem involving geometry. The problem is tell about truncated cylinder.

I want to know how to derive the formula of volume and its surface area without calculus, but still don't get it.

I've just can found the definition and the volume without proof from:

http://encyclopedia2.thefreedictionary.com/Truncated+Cylinder

Truncated cylinder is the geometric solid produced when a cylinder is cut by a plane that is not parallel to the base.

The volume of a truncated circular cylinder is $V = \frac{\pi r^2(h1 + h2)}{2}$, where $h1$ and $h2$ are the lengths of the longest and shortest elements of the cylinder and $r$ is the radius of the base.

But, there are no proof here, how can I prove this formula? And also is it possible to find the surface area for this kind of the cylinder?

Edit: I've just found the formula of its surface area(without proof too). Here is the link:

http://www.math24.net/cylinder.html

But, how can I got the Area of the bases of a truncated cylinder? In this formula, the root form is appears. How can I got the root form?

Thanks

I would recommend the following justification. Take whatever side is cut, and call that the "top". This does not lose generality, it simply gives an orientation. Since the plane must cut through the entire cylinder, We can't solve for a version with just a corner "chopped off". Therefore, the cylinder, when viewed edge-on, looks like a trapezoid: Therefore, the extra volume on the high-side should equal the missing volume on the low side, since the top of the cylinder is a triangle $A=\frac{b\Delta h}{2}$. In this case, the $\Delta h=|h_1-h_2|$ (assuming that $h_1>h_2$, again without loss of generality).

You could cut the high-side of the cylinder off where the red line leaves the cylinder, and flip the wedge over to the other side, leaving a flat-top with average height $\bar h=\frac{h_1+h_2}{2}$.

At this point, you can use the basic form of the cylinder formula: $$V=\pi r^2*\bar h=\frac{\pi r^2(h_1+h_2)}{2}$$

Let $C(A, B)$ denote a truncated cylinder whose left edge has length $A$ and whose right edge has length $B$. (I'm assuming that it's placed so that its longest and shortest edges are to the left and right).

1. The volume of $C(A, 0)$ is $\frac{1}{2} \pi r^2 A$. Reason: Stack up $C(A, 0)$ atop $C(0, A)$ to get a cylinder of height $A$. The volume of $C(A, 0)$ is clearly half the volume of this cylinder.

2. Now for a cylinder C(A, B), where $A > B$, consider the stack

C(0, Q)
C(A, B)
C(B, A)
C(Q, 0)


where $Q = A - B$. THat's a cylinder of height $A + B + (A-B) = 2A$. If we subtract from it the volumes of the top and bottom pieces (which we computed in part 1), we get $$\pi r^2 (2A) - \pi r^2 (Q) = \pi r^2 (2A - (A-B)) = \pi r^2 (A + B)$$ But the volume we've just computed is twice the volume of the sliced cylinder $C(A,B)$ that we're interested in. So the volume of $C(A,B)$ isjust $$\pi r^2 \frac{A + B}{2}$$

A rather similar proof works for surface area.

Note that the proof given above works for a cylinder cut by ANY two planes, as long as the longest and shortest "line" of the truncated cylinder are diametrically opposite each other.

If lengths of high side & low side are $l_1$ & $l_2$ then by gluing up two such identical frustums at their elliptical sections can form a cylinder of length $(l_1+l_2)$ & radius $r$ then the volume & surface area of cylindrical frustum are calculated as follows $$V=\frac{1}{2}(\text{volume of complete cylinder})=\frac{1}{2}\left(\pi r^2(l_1+l_2)\right)=\frac{1}{2}\pi r^2(l_1+l_2)$$ $$\text{Surface area}=\frac{1}{2}(\text{surface area of complete cylinder} )=\frac{1}{2}\left(2\pi r(l_1+l_2)\right)=\pi r(l_1+l_2)$$