Math Is Fun Forum

Hi LuisRodg,

I sympathise with you on this one. The reason is subtle. Here is why you can't use the circumference:

Section 1

If f(x)=k, where k is constant, then the area under the curve y = f(x) from x = a to x = b is rectangular and has the value k(b-a). The idea of integration extends this simple concept by defining a value for the area when f is any continuous function. The situation is typical of many where:

(a) Corresponding to an interval a <= x <= b, we wish to define a value

(say) of some quantity Q (in the usual case Q refers to area).

(b)

depends on the values of a continuous function q, and

if q(x) = k where k is constant.

(c) For intuitive reasons we require that Q be additive, i.e. if the interval a <= x <= b is subdivided by points x1,x2,x3,......,xn,xn+1 where

a = x1 < x2 < x3 < ..... < xn < xn+1 = b

then

or writing

,

(d) For intuitive reasons we require that if

for values of x in the interval

then

where

The particular case where

is the area under the curve y = q(x) from x = a to x = b is the mose familiar one. In all such cases we define:

Section 2

In Section 1 we have described the situation in which the value

of a quantity Q is given by the integral:

In practice the required integrand q(x) is usually found by seeking an approximate value of the form:

for

For example, in the case of the area under the curve y = f(x) the most obvious approximation for the area

of the rth strip is:

(Call this "Equation 1")

and from this approximation f(x) emerges as the likely integrand.

Now if:

then writing

we have

(Call this "Equation 2")

Hence if we write

that is

then letting

we see that this agrees with equation 2 if, and only if (THIS IS THE VITAL POINT TO UNDERSTAND IN RELTAION TO WHY THE CIRCUMFERENCE METHOD IS NOT GOOD ENOUGH)

This shows how good the approximation to

must be; IT IS NOT ENOUGH that the error

shoud tend to zero as

and instead WE REQUIRE THAT

Section 3 - Surface Area

Denote by

the surface area swept over by the curve y = f(x) from x = a to x = b when it revolves through 360 degrees about the x-axis. Slicing up this surface area by planes perpendicular to the x-axis, we divide it into a number of bands. Let

be the length of the arc of the curve (say from points P to Q) so that:

The typical band of surface area is the area swept over by the arc PQ, if this area is denoted by

we have:

It follows that:

This is the formula we are all familiar with.

Now, WHY WILL THE CIRCUMFERENCE METHOD NOT WORK in the above derivation?

Because the approximation:

for

DOES NOT satisfy the criterion of Section 2.

You can check it quite simply for yourself. This is the key point.

With the circumference method will will NOT have

Which we MUST have for the integration to be valid.