Surface and Volume of a cone without calculus

Bipolarity
How can one prove the formula for the surface of a cone as well as the volume of a cone without using calculus? Most of the online proofs use calculus.

I ask this because these formulas are used in proving the formula for the volume of a solid of revolution and the surface area of a surface of revolution. So in some sense using calculus to deduce the volume/SA of a cone would be circular logic right?

So is it possible to deduce the volume/SA of a cone without using calculus, and yet the reasoning would be considered analytically rigorous?

Thanks!

BiP

You can't really do it because curves are involved as opposed to solid objects that can be divided up into triangles or tetrahedrons which are the simplest objects in two and three dimensions respectively.

Possibly the only other way I can think of is to deform the object to something that can be divided up into a finite number of triangles or tetrahedrons and then you can just use summation of areas and volumes to prove it.

Whatever way you do this (if you can) you need to deform your object to something where you can divide that transformed object into triangles and tetrahedrons (a finite number of course) and if you can, then there is your answer.

If you need to use an infinite number then this is equivalent to calculus.

Studiot
Let the frustrum be P.

Denote two lines from P to the circular basal perimeter, PQ and PR

In ΔPQR, the bisector from P to QR meets QR at right angles and is nearly the slant height of the cone.

The area of triangle PQR equal 1/2 base times height = 1/2QR (slant height)

As the triangle PQR gets smaller it more nearly matches the surface of the cone.

If we plate the cone with triangles PQR and make them smaller and more numerous the the sum of their basal lengths approaches the perimeter and their bisector lengths approach the slant height

So curved surface area becomes 1/2 (perimeter of base) x (slant height)

You can also develop this to find the volume.

You can also develop formulae using physical reasoning by filling the cone with fluid and equating the pressure to the work of expansion or the sum of the segmental forces to the segmental weight.

Homework Helper
limits are not the same as calculus, they are more fundamental. euclid used limits but not calculus. calculus is a trick for computing certain limits using antiderivatives, but clever people like euclid and archimedes were able to compute some important limits without calculus.

Of course this is confusing because in many books they discuss "differential calculus" and "integral calculus" and then finally relate the two. In my opinion however "the calculus" IS the relation between the two different types of limits. We are not really disagreeing on how to do this problem, rather only on what words to use to describe the method of limits which is needed.

Actually the surface area of a cone is much simpler and is a special case of the area of a surface of revolution. If I am not wrong the point is Pappus' theorem that the area of a surface of revolution depends only on the centroid of the curve revolved, plus its length. hence the area of a cone obtained by revolving a segment around an axis, is the same as the area of the cylinder obtained by revolving that same segment after rotating it about its center until it is parallel to the axis of revolution.

For symmetrical shapes, pappus theorem also does not need "calculus" in my sense, only limits, and for revolving straight line segments it seems to use only geometry.

for volume of a cone, one does first the case of a pyramid and then takes limits to get the case of a cone. to do a p-yramid one starts from the fact that a cube can be decomposed into three congruent pyramids, so for those particular pyramids the volume formula holds.

then one uses limits again, but not calculus, to show that the volume formula for a pyramid stays the same if you change the slopes of the vertical slanted edges while maintaining the same base and height. combined with scaling, this does it for all pyramids, hence also cones.

for surface area of a cone i guess one way, at least for a right circular cone, is to view the surface area as a partial wedge of a disc, rolled up.

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If the cone has height h and base radius r, the distance from the apex to the base, along such a straight line, has length $\sqrt{h^2+ r^2}$. That becomes the radius of the disk we are flattening the cone onto- and so the disk has area $\pi(h^2+ r^2)$. But the base has radius r and so circumference $2\pi r$ rather than the $2\pi\sqrt{r^2+ h^2}$ or the full disk.
That is, because the flattened cone has circumference $2\pi r$ rather than the full $2\pi\sqrt{r^2+h^2}$ of the disk, it only occupies $\frac{2\pi r}{2\pi\sqrt{r^2+ h^2}}= \frac{r}{\sqrt{r^2+h^2}}$ of the disk and so the area of the flattened cone, and so the "slant area" of the cone, is that fraction of the area of the disk:
$$\frac{r}{\sqrt{r^2+h^2}(\pi (r^2+h^2}= \pi r\sqrt{r^2+ h^2}$$