Wednesday, August 26, 2009

Volume of a cone: simple calculus

I've been brushing up on my calculus skills, though it can be challenging for an old fart like me. (In one ear and out the other). I saw a book online (Strang's Calculus), which I like because it is succinct, and also the author has a very interesting, idiosyncratic style. I liked it so much I got a hard copy. Now that I've studied a bit, I realize that the book is more than succinct, it's dense, filled with math, and contains a lot more than just calculus. It's a great book.

In Chapter 8, Applications of the Integral, we encounter Example 11, to find the volume of a cone. I've redrawn the diagram from the book, below. The idea is to move vertically from the top to the bottom, letting x be equal to the radius at each point. At each value for x, we draw a shell (the outside surface of a cylinder) as shown in red. As we move from top to bottom, we accumulate a collection of these cylinders whose volumes are summed to get the volume of the cone.

The key is to recognize that the distance from the top of the cone to the top of each cylinder has the same relationship to x as b does to r (e.g. when x = r, then this distance = b). The smallest and largest rectangles in the figure are similar.

The height of the cylinder plus this distance equals b.

So the height of each cylinder is

h = b-bx/r

From this point it's easy. The area of the cross-section of each shell is its diameter (2*x), times &pi times the small width dx; the volume is this area times the height. I'm going to leave out the multiplication symbol I usually use (*) for clarity:

  = 2πxhdx
= 2πx(b-bx/r)dx

We sum (integrate) all these cylinders:

  = ∫2πx(b-bx/r) dx
= ∫2πxb dx - ∫2πxb(x/r) dx
= πx2b - (2/3)πx3b/r

evaluated between x = 0 and x = r:

  = πr2b - (2/3)πr2b = (1/3)πr2b

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