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### Surface Area of Solid of Revolution

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Date: 05/21/2001 at 09:32:13
From: Stan Winston
Subject: Alternate Surface Area of Solid of Revolution Formula

My teacher mentioned that we were going to be looking at surface areas
of solids of revolution shortly after we did volumes of rotation. So,
before checking it out in the book, I tried to derive it myself. I
figured that area is just height times width, so why not take the
integral of the circumference of the solid of revolution? When the
multiple circumferences are accumulated by integration, it's the same
as adding the areas of multiple cylindrical strips with infinitely
small widths, which should give me the surface area. For example, the
surface area of the cone found by rotating y = x around the x-axis
from x = 0 to x = 4 should be:

2*pi*(integral of x dx) from 0 to 4

Unfortunately this doesn't give me the answer I know to be correct, as
you have to use the length of a curve formula within the integral for
it to work. In fact, the only solid that I've found my formula to work
for is the cylinder. What's wrong with my analysis? Why doesn't it
work?
```

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Date: 05/21/2001 at 13:07:48
From: Doctor Rick
Subject: Re: Alternate Surface Area of Solid of Revolution Formula

Hi, Stan.

You've done a pretty good analysis of your own problem. All the pieces
are there for you to see why your formula doesn't work.

Consider the strips or sections of surface whose areas you want to sum
- before they are reduced to infinitesimal width. These strips are not
cylindrical in shape (as your formula assumes), but conical. (That is,
they are shaped like sections, or frustums, of cones.) This explains
your observations: in the case of a cylinder, the strips are indeed
cylindrical, and so your formula works, but for any other shape - and
most obviously for cones - the strips are conical.

The situation is analogous to the curve-length formula: following the
same approach in that case, you would just integrate the width of each
infinitesimal segment, giving the very uninteresting integral[dx] = x.
The length along the curve of the sloped segment, sqrt(dx^2 + dy^2),
does NOT approach dx as dx goes to 0; rather it approaches
sqrt(1+(dy/dx)^2)*dx. Therefore this factor must be retained in the
curve length formula.

In the same way, the surface area of a conical section does not
approach the surface area of a cylindrical section as dx goes to zero.
The difference in the radii at the two ends of the conical section is
second-order in dx, so it does go away as dx goes to zero. The length
perpendicular to the circumference, however, is not dx but the length
along the curve; thus the same factor sqrt(1+(dy/dx)^2) must be
retained in the formula. The product of the (mean) circumference of
the segment and the length along the curve gives the area of the
conical section, as dx goes to zero.

You might take a look at the answer in the Dr. Math archives
concerning the derivation of the surface area of a sphere. This answer
does not use calculus, but it shows the conical section I am talking
about, and compares it with a cylindrical section.

Volume of a Sphere
http://mathforum.org/dr.math/problems/banijamal05.28.99.html

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/21/2001 at 13:23:30
From: Doctor Rob
Subject: Re: Alternate Surface Area of Solid of Revolution Formula

Thanks for writing to Ask Dr. Math, Stan!

Don't feel upset with this. It is a common error. In fact, I fell into
the same trap when I first saw this stuff.

Let delta-x and delta-y be small, so that the curve is very well
approximated by a straight line near (x,y). Then you have the
following picture, where the diagonal line "is" the curve:

(x+delta-x,y+delta-y)
_,o
_,-'  |
delta-s _,-'      |
_,-'          |delta-y
_,-'              |
_,-' t                |
(x,y) o-----------------------o
delta-x         (x+delta-x,y)

The problem is that the ratio of the length delta-s of a small segment
of the curve being revolved to the length delta-x of its projection on
the x-axis does not approach 1 as those lengths approach zero. In
fact, it approaches the secant of the angle of inclination t of the
tangent line there. That also means that the limit of the ratio of the
lateral area of the frustum of a cone generated by revolving the
segment delta-s about the x-axis to the area of the cylinder generated
by revolving the segment delta-x about the x-axis is also not 1, but
in fact also equals sec(t). Now dy/dx is the slope of the tangent
line, which is the tangent of that same angle t.

Now recall the trigonometric identity:

tan(t)^2 + 1 = sec^2(t)

Putting this all together the limit of the ratio delta-s/delta-x is:

ds/dx = sqrt(1+[dy/dx]^2)

In general, this is not equal to 1. It is equal to 1 if the tangent
line is horizontal throughout the interval, that is, the curve is a
horizontal line and the surface is a cylinder.

Then the surface area is given by the integral

b
S = INT [2*Pi*y*ds]     (ds, not dx !!)
x=a

b
S = INT [2*Pi*y*(ds/dx)*dx]
x=a

b
S = 2*Pi*INT [y*sqrt(1+[dy/dx]^2)*dx]
a

If this is not clear, or you want to discuss the matter further,
please feel free to write again.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
College Higher-Dimensional Geometry

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