Are limits necassary to prove the volume of a cone?

Yes and no. Calculus (including the theory of limits) is actually necessary to rigorously *define* the volume of any solid that is not a polytope (i.e. anything enclosed by curved surfaces). So in particular, you can’t rigorously define the volume of a cone, a cylinder, a sphere, etc. without limits.

Of course, the ancient Greeks knew the volumes of these objects and they certainly didn’t have calculus, so how did they do it? One can *assume* that there is a concept of volume that has certain “obvious” properties (which are often called “axioms”). One obvious property of volume is that if one solid is contained in another, then it has smaller volume. This is what the ancient Greeks used to compute volume, together with something called the “method of exhaustion.” If you can put your solid inside a polytope, that gives you an upper bound for the volume, and by choosing better polytopes, you can get more accurate upper bounds. You can obtain lower bounds similarly. If those upper and lower bounds can get arbitrarily close together, you have an exact computation of volume. If you’re familiar with limits, you can see that this is just a limit argument, and from a modern perspective, the method of exhaustion is just a precursor to the idea of taking a limit. So in that sense, the answer to your question is “yes” but the answer is “no” in the sense that the method of exhaustion pre-dates Newton’s theory of limits by over a thousand years.

Also worth noting that even something as basic as the area enclosed by a circle also needs the method of exhaustion, or something similar, if you want to compute it. Btw, the entire concept of the definite integral is an extremely natural outgrowth of this idea.
Would you consider an integral as a limit?
No, you can do it using integrals to revolve a line with non-zero slope about the x or y axis

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