How can you forebode that the answer shall be independent of the number of sides of the base of the polygon ?
Stewart, Clegg, Watson. Calculus Early Transcendentals, 2021 9th edition. page 370. Problem 24.
- Given a sphere with radius r, find the height of a pyramid of minimum volume whose base is a square and whose base and triangular faces are all tangent to the sphere. What if the base of the pyramid is a regular n-gon? (A regular n-gon is a polygon with n equal sides and angles.) (Use the fact that the volume of a pyramid is $\dfrac{1}3Ah$, where A is the area of the base.)
When I attempted this question, I thought I flubbed up, because my answer lacked the number of sides of the base of pyramid! Even after seeing the solution, I still can’t intuit why the answer is independent of the number of sides of the base of the pyramid. Before attempting any paperwork, how can a student forefeel that the answer shall be independent of the number of sides of the base of the polygon?
1 answer
Cavalieri's principle will tell you that the shape of the base doesn't matter. The area of the base is the only information about the base that is needed.
Cavalieri's principle (which I suspect was enunciated by Archimedes of Syracuse a couple of millennia before the man named Cavalieri was born) says that if every horizontal cross-section of one three-dimensional figure has the same area as the same horizontal cross-section of another three dimensional figure, then the two figures have the same volume.
Suppose you have two pyramids whose bases have equal areas (but do not necessarily both have the same shape), and that have equal heights. Now suppose, for example, that you look at a horizontal plane ("horizontal" means parallel to the base) that is $1/3$ of the way from the apex down to the base. The intersection of the pyramid with that plane has the same shape as the base, but every distance between two points in that intersection is $1/3$ times the distance between the corresponding pair of points in the base. Consequently the area of that intersection is $1/9$ the area of the base. That applies to both pyramids: the areas of the intersections are the same in both. Thus if one of them is one-third-base-times-height, then so is the other.
Thus in order to prove, without using antiderivatives, that the volume is $Ah/3,$ it is enough to prove it with just one pyramid, with just one shape of the base. That can be done as follows: the cube $0\le x\le1,\,\,0\le y\le1, \,\,0\le z\le1$ can be partitioned into three congruent pyramids, as follows: The base of one of them is the unit square in the $(x,y)$-plane and the apex is $(x,y,z)=(1,1,1).$ The base of another is the unit square in the $(x,z)$-plane and the apex is that same point, $(x,y,z)=(1,1,1).$ The base of the third is the unit square in the $(y,z)$-plane and the apex is again that same point. Thus each has one-third of the volume of the whole cube.
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