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
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:
We sum (integrate) all these cylinders:
evaluated between x = 0 and x = r: