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Welcome to Calculus. I'm Professor Ghrist.

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

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

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

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

Â area element is related to the arc length element.

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

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

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

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

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

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

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

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

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

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

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

Â half rLd theta. That is simply pi rL.

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

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

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

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

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

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

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

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

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

Â length. But rather dL, the small change in that

Â 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â 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

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

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

Â integrals. The question is what is the surface area?

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

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

Â Then, what do we obtain? For the volume.

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

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

Â volume. That gives us the integral from 1 to

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

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

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

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

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

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

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

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

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

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

Â that? That looks like an unpleasant integral.

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

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

Â Recall, what matters is the leading-order term.

Â We could apply the binomial expansion to that square root.

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

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

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

Â order those are. What really matters is that the leading

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

Â 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.

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

Â There are different constraints on P for surface area.

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

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

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

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

Â an object with finite volume, but infinite surface area.

Â 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.

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

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

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

Â slice horizontally and use the appropriate surface area element to pi

Â rdL. What is r?

Â 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.

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

Â y. That gives us 1 over 2 root y.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â In this case, what is the surface area element?

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

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

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

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

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

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

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

Â wonderful example of something called a minimal.

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

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

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

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

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

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

Â geometry. But of course, there are many other

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

Â physical application related to work.

Â