Start with an area --- a planar region --- which you can imagine as a
piece of cardboard. The cardboard is attached by one edge to a stick
(the * axis of revolution*). As you spin the
stick, the area revolves and sweeps out a region in space.

The problem is to find the * volume of
revolution* --- the volume of the region in space which is swept
out by the area.

In the simplest case --- the one I've described above --- you can find the volume by cutting the region into circular slices perpendicular to the axis.

If the radius of a slice is r, then its volume is

In the simplest case, the radius is given by a nonegative function , and the volume is generated by revolving the area under the graph of f from to around the x-axis. As usual, I divide the interval up into pieces, with the k-th piece having width . I pick an x-value in the k-th piece, say . Then the volume of the k-th circular slice will be

The total volume is approximated by adding up the volumes of the slices, as you can see in the picture above. So if I have n slices, then

To get the exact volume, I shrink the slices, letting the thickness of a typical slice go to 0:

The expression on the right is a Riemann sum for . So

In setting up problems, I'll use a shortcut rather than writing down the Riemann sum. I'll simply write

In the examples I do, the axis of revolution will always be parallel to the x-axis or the y-axis. If the axis is parallel to the x-axis, the thickness is ; if the axis is parallel to the y-axis, the thickness is .

The radius r is the distance from the axis of revolution to the edge of the slice. It will usually be given by a function specified in the problem which determines the region which is being revolved. You'll see how this works in the examples below.

* Example.* Here is the region under from to :

The region is revolved about the x-axis. It sweeps out a * volume of revolution*:

To find the volume, cut the solid into circular slices --- like sausage slices --- perpendicular to the axis:

The radius of a typical slice is the height of the curve: . The thickness is . Thus, the volume of a typical slice is --- (circle area) times (thickness). So the total volume is

* Example.* The area bounded by and the x-axis is revolved about the x-axis. Find the
volume of the solid generated.

The region is the area under the parabola from to .

In the picture above, I've superimposed a typical slice over the picture of the area being revolved. You can see that the radius of a typical slice is the height of the curve: . The thickness of a typical slice is .

Thus, the total volume is

In the situations above, the axis of revolution lay along one edge of the area. If it does not, the volume of revolution may have a "hole" in the middle:

If I cut such a volume into slices, I get circular rings or washers:

The area of such a washer is

Thus, the volume of a typical washer is

As before, I integrate to find the total volume:

* Example.* The region bounded by
and is revolved about the x-axis. Find the volume of the
solid that is generated.

For a typical washer, the inner radius is and the outer radius is . You can see this in the picture above, where I've superimposed a typical washer on top of the picture of the region.

The region extends from to . The volume is

* Example.* The region bounded by
and is revolved about the y-axis. Find the volume
generated.

The curves intersect at and at . The inner radius is and the outer radius is . The volume of the solid is

Copyright 2005 by Bruce Ikenaga