Calculating Simple Volumes

We can use integration to find the volumes of a variety of geometric solids. On this page, we will apply this idea to some particularly simple examples, the sphere and the hyperboloid. The nice property of these is that they are symmetric about some axis of rotation. Indeed, we can describe the surfaces that enclose these shapes by a single curve (the "profile") which is rotated about an axis. We call these shapes surfaces of revolution, and as will be evident below, computing the enclosed volume is a simple procedure involving integration.


The sphere

The simple formula for the volume of a sphere was not known in antiquity, and it was a struggle to develop it. However, what was certainly known is the volume of a cylinder with base area A and height h. In particular it was known that this volume is


\[ 
V_{cyl} = A h 
\]


For the case of a right circular cylinder (whose base is a circle of radius r, and whose sides are perpendicular to the base) the volume is


\[ 
V_{circ ~cyl} = \pi r^2 h 
\]


We will find this result useful in the "dissection" below.
Consider the sphere shown here, and the set of little disks (i.e cylinders) stacked up so as to approximate its volume. Basically, we have sliced up the sphere into slices of equal thickness, and approximated each slice as a cylinder. Naturally, the sum total of the volumes in all the disks is not exactly the same as the volume of the sphere, (there are bits due to the curved edge of the sphere that are not included in the disks) but if we make the slices very very thin, the approximation becomes better and better. Indeed, if we take a limit, as the number of slices, n, goes to infinity, we expect to obtain a definite integral, much in the same way as it was obtained in problems involving the dissection of an irregular area into a set of rectangles.
We will now focus attention on just one of the disks that make up the slices, and get a handle on its volume. Here is a picture of what this little disk looks like. Now we need to figure out what its radius and height (thickness) is, so we can calculate the volume.
Suppose that sphere shown above has a radius $ R $ . If we include a coordinate axis (the y axis) and note that the sphere extends from $ y=-R $ to $  y=R $ , (a length of 2R), we see that the width of each of the n slices (or disks) is

\[ 
\Delta y = \frac{2 R}{n} 
\]

We need to determine the area of the base of each of the disks, and to do so we must find how the radius of a disk varies according to its position.
We see from the picture that, using the Pythagorean triangle, the radius of the disk is related to the the coordinate at which the slice occurs, $ y_k $ and to the radius of the sphere, $ R $ as follows:

\[ 
r^2 = R^2 - y_k^2 
\]



\[ 
\pi~ r^2 = \pi~ (R^2~ - y_k^2~) 
\]

From this argument, we see that the volume of the k'th disk is

\[ 
v_k = A h = \pi~ (~R^2 - y_k^2~)~ \Delta y 
\]

Thus the volume contained within all n disks is:

\[ 
V_n =\sum_{k=1}^n~~ \pi~ (R^2 - y_k^2~)~ \Delta y 
\]

It should come as no surprise that when we increase the number of disks used to approximate the volume of the sphere, we get a closer and closer approximation. You can see this yourself by moving the red dot in the demo below. (The red dot controls the number of slices used in the approximation.) To see what the figure looks like in 3D, you might want to point your mouse at the 3D shape and move or rotate it a bit.

Returning to our calculation, and taking a limit as $ n \to \infty $ , we find that


\[ 
V =\int_{-R}^R \pi (R^2 - y^2)~ dy 
\]

Here $ R $ is a constant. We can compute this integral, and we find that

\[ 
V =\pi [R^2 y - \frac{y^3}{3}] |_{-R}^R 
\]

Plugging in the endpoints and simplifying leads to:

\[ 
V =\pi [R^2 (R+R) - \frac{1}{3}(R^3+R^3)] = \frac{4 \pi}{3} R^3 
\]

Volume of a Surface of Revolution
The procedure we have used for a sphere can be used more generally for any surface of revolution. All we need to specify is the function that generates the profile of the surface, and the axis about which it is revolved. For example, suppose that $ y = f(x) $ is the curve that generates the surface, and that it is rotated about the x axis to produce the surface of revolution. Then in this case, the width of a disk in a dissection of the surface would be $ \Delta x $ and the radius of a disk located at $ x $ is given by $ y = f(x) $ . Thus, carrying out the same procedure as above would lead to


\[ 
A = \int_a^b~~ [f(x)]^2~ dx 
\]


In some cases, actually computing this integral may be problematical, but in the examples chosen here, the function $  f(x) $ is of a particularly simple nature so that this integration step can be easily done.


The volume of a hyperboloid

We will use the general result above to calculate the volume of a hyperboloid. This type of surface looks like a cooling tower of a nuclear power plant. Its generating curve is

\[ 
x^2 - y^2 = k 
\]

where k is a constant. The curve is rotated about the y axis, as shown below, and we will compute the enclosed volume for $  -1<y<1 $ Note that the summation step is done along the y axis.

Here is a picture of the hyperbolid, with the same kind of dissection into disks. Experiment with the number of disks, as before, and look at the surface from different directions by rotating the 3D image with your mouse.

The radius of the little disks is determined by the x coordinate along this curve, namely,


\[ 
x^2= [f(y)]^2 =k + y^2 
\]

Thus, the volume of the hyperboloid is


\[ 
V = \int_{-1}^1 [k + y^2] dy 
\]


Computing this integral, we find that


\[ 
V =  [ky + \frac{y^3}{3}] |_{-1}^1 
\]


Which, after simplification, can be expressed as:


\[ 
V =  [k(1+1) + \frac{1}{3}(1+1)] = 2k (1 + \frac{1}{3}) = \frac{8}{3} k 
\]





For your consideration: