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 a 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 be some 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 expressible in terms of r and h, Let's just use L. So our surface area element is one-half the base, r d theta, times the height, L. Integrating to obtain the surface area, we get the integral from 0 to 2 pi of one-half r L d theta. That is simply pi r L. 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 xy plane about an axis, let's say about the x-axis. What is the appropriate surface area element? We are going to fix a value of x and consider a vertical slice of this surface obtaining a circle. We then thicken it up by dx. What is this surface area element? Well this is not a cylinder. It is rather the tail end of some cone, a cone with slant length L and of radius r. Now what we care about is not the slant length, L, but rather dL the small change in that length. That is indeed the arc length element. Now, let us use what we know about the surface 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 + Ldr. Now dL we know. That's the arc 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 dL and dr, then through similar triangles, we can argue that dL over L is dr over r. Cross-multiplying, we see that L times dr = r times dL. And thus, we can simply the surface area element to twice pi 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(x) where this curve is some function of y(x). Therefore, using what we know about the arc length element for such a graph, we have dS is 2 pi times y, times square root of 1 + the derivative, dy, dx squared, dx. Well, let's see how this works in the 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 = the square root of r squared- x squared and rotate that about the x-axis. Therefore, we know that the surface are element is 2 pi y dL. In this case it's 2 pi times square root of r squared- x squared times the square root of 1 + the derivative squared dx. We've done this derivative before. It's -x over square root of r squared- 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 over the common denominator, r squared- x squared, then indeed we see that that denominator cancels with the r squared- x squared to the left. And we're left with 2 pi square root of r squared dx. That is, 2 pi r dx. That is a simple surface 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 -r to r to obtain in our heads the familiar formula for surface area of a ball of radius r, 4 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. In cartography this is a solution to the problem of how to make a map of the world where everything has the same area as on the surface of the ball, even though the map is flat. It distorts length but not area. 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. And 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 -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 -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 over 2p -1. We could rewrite that as saying p is bigger than a half. Otherwise, when p is less than or equal to one-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 -p, times cL. We can rewrite that cL as square root of 1 + the derivative -px to the -p-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 can apply the binomial expansion to that square root, since when x is very large, the p squared x to the -2p-2 is very small. And we obtain 2 pi x to the -p times 1 + plus some higher order terms. And 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 -p. And therefore, we know when the integral converges and when it diverges. It converges when p os 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 one-half. And this leads us to the very curious result that if your value of p is somewhere between one-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. Well, let's try an example that's a bit more finite. Let's consider, simply a parabola, y = 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, 2 pi rdL. What is r? Well, in this case it's equal to x, and that is the square root of y. And in this case what is dL? Well, it's 1 + 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 + one-fourth 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 + one-quarter to the three-halves times two-thirds. Evaluating that from 0 to 1, I'll let you check that that yields pi over 6 times 5 square root of 5- 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 1 + dy dx squared dx. That derivative is 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. Let's end with one last example, a catenoid. This is what happens when you rotate a 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 + some initial heights y naught. In this case what is the surface area element? 2 pi y square root of 1 + dy dx squared, dx. It's easy to differentiate a hyperbolic cosine, and so we obtain a surface area element that, 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 gonna 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 kinda 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.