Now, let us use what we know about the service area of this cone.

It is pi times r times L, but how do we get dS?

Well, of course, we apply the differentiation operator and we see that

dS is in fact pi times, through the product rule, rdL plus Ldr.

Now, dL we know that's the arch length element.

What do we do with dr? Well, consider the right triangle with

hypotenuse L and height r. Then, if we extend things a little bit,

into d l and d r Then through similar triangles we can argue that dL over L is

dr over r. Cross-multiplying we see that L times dr

is equal to r times dL, unless we can simplify the surface area element to

twice pie r dL. And this is wonderful because we know dL,

the arc length element and we know r, because r is typically given in terms of

y of x, where this curve is some function of y of x.

Therefore, using what we know about the arc length element of such a graft.

We have dS is pi times y times square root of 1 plus the of ydx squared dx,

well, let's see how this works in a context of a fun problem.

Consider a round ball of radius R divide this ball into slices of equal width.

The question is which of these slices has the most surface area.

Is it the one in the middle or maybe the one at the end?

Well, let's set things up in terms of some coordinates.

If this is a ball of radius r, then it can be considered a surface of revolution

where we take the curve, y equals square root of r squared minus x squared, and

rotate that out the x-axis. Therefore, we know that the surface area

element is 2 pi y dL, in this case, it's 2 pi times the square root of r squared

minus x squared times the square root of 1 plus the derivative squared, dx.

We've done this derivative before. It's minus x over square root of r

squared minus x squared. And so, substituting that in for the

surface area element, we see something that looks a little frightening, but it's

not so bad. If we put the arc length element above

the common denominator r squared minus x squared, then indeed we see that, that

denominator cancels with the r squared minus x squared to the left.

We're left with 2pi, square root of r squared dx.

That is 2 pi rdx. That is a simple service area element,

but it's not only simple. It is independent of x.

It does not matter as long as your thickness is dx.

That means that in the question of which slice has the most area.

None of them or rather all of them, they all have the same area.

In fact, we can take this surface area element and very easily integrate it as x

goes from negative r to r to obtain in our heads.

The familiar formula for surface area of a ball of radius r for pi r squared.

Now, that might seem like a trick or a curiosity, but if we slice horizontally

and consider a slightly different setting, then we have a basis for what is

called the Lambert cylindrical projection.