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Welcome to Calculus. I'm Professor Ghrist.

We're about to begin Lecture 36, on surface area.

In previous lessons, we've computed areas of flat, two-dimensional regions, but

what about surface area? The area of a shape that might be wavy or

curve in three-dimensional space. In this lesson, we'll see how the surface

area element is related to the arc length element.

Let's begin with the computation of surface area for a very simple surface.

In this case, a cone over a circle. Let's say of height h, where the circle

has radius r. Now, the best way to compute this is to

decompose into triangles and integrate. So let's take our surface area element to

the sum triangle. Defined by a segment of the circle with

angle, d theta. The length of the base of this triangle

may be approximated by rd theta. What is the height?

Not h, but rather L, the slant length of this cone.

Although, expressable in terms of r and h, let's just use L.

So our surface area element is 1 half the base, rd theta times the height, L.

Integrating to obtain the surface area, we get the integral from 0 to 2pi, from 1

half rLd theta. That is simply pi rL.

That part is easy enough. We're going to put this to use in

determining the surface area of a surface of revolution, obtained by revolving a

curve. In the x-y plane, about an axis, let's

say about the x-axis. What is the appropriate surface area

element? We're going to fix a value of x, and

consider a vertical slice of this surface, obtaining a circle.

We then thicken it up by dS. What is this surface area element?

Well, this is not a cylinder. It is rather the tail end of some cone, a

cone with slat length L and of radius r. Now, what we care about is not the slat

length. But rather dL, the small change in that

length. That is indeed the arc length element.

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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.

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Let's turn to a different example. This time, we're going to take the curve

1 over x to the p as x goes from 1 to infinity and rotate that curve about the

x-axis. This is going to give us some improper

integrals. The question is what is the surface area?

Well, let's begin by doing something a little different.

Let's compute the volume of this solid, obtained if we slice by orthogonal discs.

Then, what do we obtain? For the volume.

Well, at some point x we have a volume element given by pi times the radius x to

the minus p squared times dx. We can integrate this to obtain the

volume. That gives us the integral from 1 to

infinity of pi times x to the minus 2P dx.

This is a simple integral. We've done this before when we did the P

integrals. When 2P is bigger than 1, we get pi or 2P

minus 1. We could rewrite that as saying, P is

bigger than a half. Otherwise, when P is less than or equal

to 1 half, we have a divergent integral and the volume is infinite.

Now that we've done volume, let's do surface area.

Surface area element is 2 pi times the height x to the minus p times dL.

We can rewrite that dL as square root of 1 plus the derivative negative px to the

minus p minus 1 quantity squared. Now, what do we get when we integrate

that? That looks like an unpleasant integral.

It is an unpleasant integral. But, we can at least determine whether

it's convergent or divergent. How do we do that?

Recall, what matters is the leading-order term.

We could apply the binomial expansion to that square root.

Since when x is very large the P squared x to the negative 2P minus 2 is very

small. And we obtain 2 pi x to the minus P times

1 plus some higher order terms. Then we can be very specific about what

order those are. What really matters is that the leading

order term is 2 pi times x to the minus p.

And therefore, we know when the integral converges and when it diverges.

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It converges when p is strictly bigger than 1 and diverges otherwise.

Now, let's compare that surface area to what we saw in the case of volume.

There are different constraints on P for surface area.

What we have is something that is finite, if P is strictly bigger than 1.

But for volume, we have something that is finite, when P is strictly bigger than 1

half, and this leads us to the very curious result.

That if your value of P is somewhere between 1 half and 1, then you can have

an object with finite volume, but infinite surface area.

That seems a bit counterintuitive at first, but it's very cool.

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Let's try an example, that's a bit more finite.

Let's consider, simply, a parable y equals x squared, but let's rotate it

about the y-axis. As x goes from 0 to 1, that's going to

give us some bowl shaped region. In this case, we're going to want to

slice horizontally and use the appropriate surface area element to pi

rdL. What is r?

Well, in this case it's equal to x. That is the square root of y.

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And in this case, what is dL? Well, it's 1 plus the derivative squared.

Now, that derivative is what? We need to differentiate square root of

y. That gives us 1 over 2 root y.

Simplifying, we get a surface area element that is 2 pi times square root of

y plus 1 4th dy. And now, to obtain the surface area, we

integrate this surface area element as y goes from 0 to 1.

This is not too hard of an integral. We get 2 pi times quantity y plus 1

quarter to the 3 halves times 2 3rds. Evaluating that from 0 to 1, I'll let you

check that that yields pi divided by 6 times 5 square root of 5 minus 1.

And it's worth noting briefly that this problem can also be solved by integrating

with respect to x instead of y. In this case, r is equal to x.

And dL is square root of one plus dy, dx squared dx.

That derivative simply 2 times x. I'll leave it to you to set up the

integral and show that with a simple u substitution, it's possible to compute

the exact same answer with just about the same amount of work.

Sometimes, it's better to integrate the other way.

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Let's end with one last example, a catenoid.

This is what happens when you rotate an catenary about the axis.

Recall the catenary, the shape obtained by hanging chain or rope.

Now, we know the equation for that. It is y is some constant 1 over kappa

times the hyperbolic cosine of kappa x plus sum initial height y naught.

In this case, what is the surface area element?

2 pi y square root of 1 plus dy, dx squared dx.

It's easy to differentiate a hyperbolic cosine and so we obtain a surface area

element. Well, doesn't look so nice we can

simplify that square root, but in the end, we've got a COSH squared term and a

COSH term. Now, this is not impossible to integrate.

It is doable, but we're not going to do it here.

What I do want to point out is that this is a surface that you can see, it's a

wonderful example of something called a minimal.

Surface something that minimizes the surface area for a fixed boundary.

If you took two wire loops and dipped them in a soap film solution and held

them apart, if you do it just right, then the surface that you get will be a

catenoid. And that's kind of fun to play with.

And now, you know how to compute its surface area.

With this, we complete our applications of definite integrals to problems in

geometry. But of course, there are many other

applications we could explore. In our next lesson, we'll consider a more

physical application related to work.