Volumes

If you take a reasonable region in space, and slice it by planes perpendicular to an axis (we'll think about the x - axis) with a machete, you might be lucky and get something whose area is easy to describe. Now, if you think about cutting the region with a chain saw, instead of a machete, so that you actually cut out some bit dV of its volume, how much do you cut out? It's the area A_x of the slice at the point x along the axis, times the thickness dx of the saw. So, if you were to saw and saw until there was no region left, you would have cut up

n å i = 1  Axi dx

volume of stuff. Well, approximately. Of course, you have to think about the slope on the edges and all, but not for long. In the limit, as the thickness of the saw blade goes to 0, the sum approaches a limit of

ó õ b a  Axdx,

where x = a is the first slice you need, and x = b is the last, to cut through the whole region.

Try to find integral formulas to find the volume of:

1. A sphere of radius a .
   Solution: If the sphere has center at the origin, and you slice the sphere by planes perpendicular to the x -axis, the slice will be a circle, but not one of radius a , the radius will be
   

   Ö

   a2-x2

   
   if the slice crosses the x -axis at x . So, the area of the slice will be A_x = p(a²-x²) , so that dV = p(a²-x²)dx . The sphere would be built from slices beginning at x = -a and continuing until x = a . So, the volume would be:
   

   V = ó õ a -a  p(a2-x2)dx = p æ ç è a2x- x3 3 ö ÷ ø ê ê ê a -a  = 4pa3 3 .

2. A cone, of base radius 2 and height 3.
   Solution: Here the slices would again be circles. If the cone is laid out with its axis along the x -axis, point at the origin, then the radius of the circle cut out by a plane cutting through perpendicularly, meeting the axis at x would be 2x/3 , because the ratio of radius to height (distance from the apex to the slice) is 2/3 for the full height, and would remain that ratio for the small cone that remains after being sliced. Since the ``height'' is x , the radius would be
   

   2 3

   x

   
   . Thus, the volume is
   

   V = ó õ 3 0  p æ ç è 2 3 x ö ÷ ø 2   dx = 4p 27 x3 ê ê ê 3 0  = 4p.

3. A pyramid, with a square base of 200 cubits×200 cubits and a height of 150 cubits .
   Solution: Slice this one horizontally, a distance x cubits down from the top. Again the ratios of corresponding sides is preserved, since the little pyramid at the top is similar to the whole pyramid, so the ratio height¸base = 150¸200 = 3/4 . That means that the square base cut through at height x cubits down from the top has side 4x/3 , and the volume would be
   

   V = ó õ 150 0  æ ç è 4x 3 ö ÷ ø 2   dx = 16 27 x3 ê ê ê 150 0  = 2,000,000 cubic cubits.

6.3.1  Volumes of revolution; disks and washers

Many examples of this techniques compute volumes of regions which are rotationally symmetric about some axis. These are called ``volumes of revolution'' for obvious reasons. A volume of revolution can be described by ``sweeping out'' the region by revolving a plane area about the axis of symmetry - so a sphere is swept out by spinning a disk about a diameter, or a football by spinning an ellipse about the major axis. If you describe the region that way, you can use that planar description to find the areas of the slices A_x . In this case, if the region being spun about the x -axis is the region between the curve y = f(x) and the x -axis, for x Î [a,b] , then A_x = pr² = p(f(x))² .

If the axis of rotation is not the x -axis, then the radius function has to be thought about. For example, find the volume of the region generated when the semi-circle

y = Ö   1-x2   ,

for x Î [-1,1] , is spun about the axis y = -1 . Also, you can find the volumes of more complicated regions the same way. If a region in the plane between two curves is revolved around an axis, then the slice is not a disk, but an annulus, or ``washer''. The area is simply the difference between the outer area and the inner, pR²-pr² , so the volume really should be thought of as:

Volume = ó õ b a  (pR2-pr2)dx,

where we assume that the axis of rotation is the x -axis.

Here are a couple of examples:

1. Find the volume of the region obtained when the area bounded between the curves y = x and y = x² is revolved about the x -axis. Then, what is the volume when the same region is revolved about the y -axis (switch variables in the formula).

2. Find the volumes of the football-shaped region swept out when the ellipse

x2 36 + y2 9 = 1

is revolved around the x -axis.

6.3.2  Shells

The point of these volume problems is supposed to get you to see how integration applications take shape - you decompose the problem up into a bunch of little, easy problems. Then you add them back together (integrate), and the combination of the solutions to the easy problems solves the hard one.

The shell method is just another example of that. You still compute volumes, but you do it by decomposing the region into cylinders rather than slices. The direction of the thickness of the cylinder tells you the variable to use. The area of the cylinders, (instead of the slices), is: A_x = 2prh , where r is the radius of the cylinder, and h is the height. The volume, by this method, is:

ó õ b a  2prhdx,

if x gives the direction of the thickness of the cylinder. You have to figure out what r and h are for each problem.

You need to decompose the region into just enough cylinders to fill out the region once. While that's pretty obvious with slices, but not so with shells - because each one goes all the way around the axis. You don't have to go from one edge to the opposite one, but from the center out.

Example 1 Find the volume of a ``cored apple'', a sphere of radius 2 minus the stuff inside a cylinder of radius 1 that passes through the center of the sphere.

Solution

Assume that things are arranged so that the axis of the ``corer'' is the y -axis. Then, in the xy -plane, before you rotate the picture, the direction radially away from the axis is the x direction, so that is what you use for the variable. The height of a representative cylinder, expressed in terms of x , is

h = 2y = 2 Ö   4-x2   .

Why the 2? You go up and down from the x axis to get to the top and bottom of the cylinder. The radius r = x , because we rotate around the x -axis. So, the volume becomes:

Volume = ó õ b a  dV = ó õ b a  2prhdx = ó õ 2 1  2p2 Ö   4-x2   x dx.

The limits of integration come from the fact that the innermost slice you need is at r = 1 , and the outermost slice you need is at r = 2 . I'll leave the final answer - or, rather, figuring it out, to you. The answer is 4 Ö3p, as my machine told me.

Here are some more examples:

Example 2 Find the volume obtained when the region bounded by the curves y = sin(x) , y = -1 , x = 0 , and x = p is revolved about the y -axis.

Example 3 Find the volume of the solid obtained by revolving the region bounded by the curves y = 2x-x²+2 , x = 0 , x = 2 , and y = x about the y -axis.

Example 4 A torus is the doughnut-shaped region obtained by revolving a circle about a line outside the circle. find the volume of the torus obtained by revolving the circle (x-2)²+y² = 1 about the y -axis.

Copyright 2000 David L. Johnson