Deriving the Volume of a Cone with Integration

We start by drawing an arbitrary line in the coordinate plane with y-intercept of some radius r and x-intercept of some height h. These will be dimensions of the cone we will use later on.

We will create an equation of the line using r and h as shown above

Notice that the region closed by the and the x and y axes is a right triangle . If we revolve this region around the x-axis, then a cone is formed. Just try to picture it in your head

Notice that the cross sections of a cone are circles. We could break the cone up into a bunch of circular disks and add the volumes of the disks together.This will only be an approximation of the volume, but as the number of disks approaches infinity, we will have an exact value for the volume of the cone.

We can create a bunch of rectangles in the region bounded by the line and the axes. Revolving the rectangles will create disks whose radii vary as a function of the line. That is, the radius of each disk is equal to the y value of the line at its specific location on the graph.

The volume of each disk represents a small change in the volume of the cone as shown by the expression below.

As delta x approaches zero, the number of disks that can fit in the region bounded by the lines approaches infinity. We can now write the equation as follows.

We will create a summation of these disks by integrating this equation from 0 to h. We must also rewrite y in terms of x. If you work it all out correctly, you should come up with the right formula for the volume of a cone.

Using u-substitution makes the integral A LOT less messy.