 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Ange Misseldine. This video represents the first one in lecture 21 for which we're going to talk about the concept of area of a surface of revolution. It turns out that this topic is going to be in many ways familiar because many of the principles we talk about will be very similar to volume problems we did in the past, particularly volume of solids of revolution. But this time, instead of looking at the volume of the solid, we're going to be looking at the area of the surface that surrounds it. Also, this idea that this lecture here is going to combine with some topics of arc length that we talked about in lecture 20. So it's going to be a fun sort of like reunion show for some topics that we've seen previously. So a surface of revolution, unlike a solid of revolution is formed when a curve is rotated about an axis of revolution. So solids of revolution, we took a region, we rotated. This time we just have to rotate a curve. And that's why arc length is going to play a critical role right here. So in particular, the surface of revolution is, it's really just the boundary of a solid of revolution. So that's why there's some similarities we're going to see. And it's our purpose in this lecture here to calculate the area of surfaces of revolution. That is, find the surface area of a solid of revolution, like the solids we talked about, we did the volumes before. Now, we'll start off with a couple of examples before we get to the general formula here. You can see in this diagram, which this diagram was taken courtesy from James Stewart's Calico's textbook. We want to calculate the surface area of a cylinder, which the cylinder has a radius r and it has a height of h, like so. Now, if you were to, if you take the circle on the top, you really can't see white right now. If you take the circle on the top, this isn't so bad, you know the radius r. And so the area on top is just going to be a pi r squared. And there's also going to be a pi r squared on the bottom. So we have a two pi r squared if we're looking for the area, the surface area of this cylinder. But there's more to it than that, right? There's also this lateral side, this side that kind of goes around. If you think of the cylinder as like a soup label, right? If we rip off the soup label, how big would that be? Well, if we slice along the lateral side along the line perpendicular to the circle on top and the circle on bottom, and we lay that lateral side flat, we're going to get a rectangle that we see right here, right? And as it's a rectangle, the area of that thing is going to be length times width. Now, what are the length and widths of this rectangle? Well, the cut gives us the length and that cut will coincide with the height of the cylinder. So backing up the height and the length are going to be identical here. And then the next thing to consider is what is going to be the width of this rectangle. The width is actually going to be the circumference of the associated circle right here, which would be two pi r. And so getting that in there, we're going to get a two pi r times h. And so we see that the surface area of the cylinder is going to be two pi r squared plus two pi rh. And in particular, we're going to be interested in this lateral side. The surface area of the lateral here is this two pi rh, okay? This lateral side is going to be significant to us as we try to derive a general formula for surface area. All right, so let's look at a slightly harder problem. Let us consider the surface area of a cone where the cone has a base radius of r. And then we're going to say that the length of the slant here is this number L, okay? Now, kind of like with the cylinder, right? The area, you have this circle on the bottom and that circle on the bottom will have an area of pi r squared. It's the lateral side that I actually care about. So we're only going to focus on that for this one because, again, just take the circle area plus this other thing here. So if we remove the circle, this kind of looks like a snow cone or like an ice cream cone type of thing. We don't care about the circle. There's actually a place for ice cream to put into it. How much surface area does the rest of this thing have? Well, like the cylinder, if we cut along some line here, so we want to pick a line that's going to be perpendicular to the base circle. And it connects to the apex of the cone right there. If we cut along this side and then flatten that object, we're actually going to get something that looks like the sector of a circle. So it's not a whole circle, but it's a really big pizza slice. And if you're not sure about this yourself, what I want you to do is at home, you know, pause your video right now, go get a piece of paper, cut out a sector of a circle, pick whatever angle you want, pick whatever radius you want. I don't care. But pick a sector of the circle and then like tape or glue together the two radii of that circle. And you'll actually make a cone and then you'll be tempted to eat a snow cone. Now that you have a paper cone in front of you. So yeah, this will be even more dangerous for students who are studying this during the summer, for which case we're eating snow cones at all moments of our lives. So we get this sector of a circle, which is actually good news for us because we can calculate the area of a sector of a circle. So the typical formula is you're going to take one half. Let's see, you're going to take one half. Well, maybe I'll get to that in just a second. Yeah. So if we'll actually kind of explain to you where it comes from, right? If you take the area of a circle, pi R square is typically we're going to do here. Although for this circle, the radius is actually L. So if you take the whole area of the circle, this is going to be a pi L squared right here. Now the important thing to know here is that when you measure an angle using radians, this is supposed to be a proportionality here. So if you take the whole area of the circle, this is going to be pi L squared. So if you take that whole area and you multiply it by the ratio theta over 2 pi, this is meant to be a proportionality. What percentage of the circle do we actually have here? If you have the whole circle, then the theta would be 2 pi. 1 times the whole area, you get the whole area. But this is just a proportionality thing here. The whole area of the circle was just pi L squared like we saw before. You get this theta over 2 pi. You will notice that the pi's cancel here and you end up with the formula 1 half theta L squared. Now this is the formula I was trying to write on the screen just a little bit ago, but I want to again include a derivation. Or sometimes people write 1 half L squared theta. It doesn't matter the order you write them in. Now this, do be aware that in order for this to work, the angle theta does have to be in radians. Otherwise this proportionality argument would crumble to pieces. So we get the area of this sector. So this right here is good news right here. 1 half L squared theta, that gives us the area of this lateral side of the cone. But there is someone of a problem in this description is that the area of formula does involve L which we can measure from the original cone. But what about theta? Theta is sort of like this new variable we introduced. So let's try to get rid of theta here. And actually the way we're going to do that is similar to how we got this area of formula in the first place. If we take the whole circumference, the whole circumference of the circle, and we times it again by this proportionality argument here, theta over 2 pi, right? Well the circumference of the whole of the whole circle here would be 2 pi L, right? Pi times the diameter 2 L would be the diameter there. Times it by theta over 2 pi. Notice here what happens is that again the pi's cancel. In fact the 2's cancel this time. And this gives us a theta times L. But the good news here is that we can actually measure, so this right here would give us the circumference of the sector, so this portion right here. But the thing is we already know what this is. This distance would be the circumference of the base circle of the original cone. And that's going to be a 2 pi R. And so this equals 2 pi R. And so the reason this is useful is you can then solve for theta here. Theta will equal 2 pi R over L. We then make that substitution in up here like so. And so then the area of the side of this cone would equal one half. Well I'll come back to that. We're going to do it right here on the right hand side. We have one half L squared. We're going to get 2 pi R over L. And so some simplifications do happen. The 2 cancels with the one half. One of the L's cancels with that. And so what we're left with is going to be a pi R times L. So we get pi RL. Kind of sounds like a girl's name there. Pi RL come over here please. Anyways the area of the side of a cone is going to be this pi RL. Pretty simple formula. It takes a little bit to derive where it comes from. But a pretty simple formula to calculate that. Alright so this is getting one stepping stone onto where we're trying to go to. Alright so the next step is now we want to consider what's called a frustum. A frustum is basically a cone with the top of the cone cut off. So if a cone had some bad dealings with the queen of hearts we get a frustum right here. Now again all of these images here are courtesy of James Stewart's Calico's textbook. So to calculate the frustum we kind of have to envision what was the cone that got cut off right here. Because if we know what that cone is we could find the surface area of that. Because again we're ignoring the base circle. We know that circle would just be a pi R squared or pi R2 squared right here. So let's think of what would be the area, the surface area of the whole cone. Well what we saw before, if you take the whole cone, the whole cone's area, this is going to be the pi RL we saw before. So we're going to get pi, the R, the radius is going to be the radius at the bottom which is R2. And then the length, let's say that the frustum has a length inside the length of L. And let's say that the small cone has a length of L1. So putting that together you're going to get L plus L1. That would be the whole cone's area. If we focus on just the small cone, the small cone that's missing, the small cone's area, that's going to equal pi, its radius which its radius actually is going to be R1. And then its slant, which is just going to be L1, we get the following. So the whole cone's area, just the side area there, is going to be pi R2L plus L1. And then the small cone's area is going to be pi R1L1, like so. Now the reason we introduced all of these is that if we want to look for the area of the frustum, we're going to take the difference of these two things. Take the whole cone, pi R2L plus L1, and then subtract from it pi R1L1, like so. Distribute the pi R2 across this sum. We're going to get pi R2L plus pi R2L1 minus pi R1. That pi looks pretty bad. Pi, I guess you call that a burnt pi or something like that. But you get a pi R1L1, like so. Now taking these last two terms, the pi R2L1 and the pi R1L1, I'm going to factor these things because they both have a L1 and a pi together. So this looks like pi R2L plus pi parentheses R2 minus R1L1, like so. So we're going to come back to this guy in just a second, but just be aware of what we've done so far. What I want to do next is actually, because we have a lot of symbols here. We have this L1, we have this R2, R1, and L. Now for the frustum, we can see R1 by measuring it. R2, we can also get an L. But L1 is measuring this fictitious part of the cone that doesn't actually exist. So we want to remove L1 from the consideration, and we can make a proportionality argument here. Because we have some similar triangles here. There's a similar triangle here. So there's one triangle here that has R1 and L1. Let's kind of redraw it to the side. You're going to get R1 and L1. And then there's a larger triangle right here for the whole cone, and then the small cone. That one, you get R2 at the bottom, and then you get L1 plus L. So you have these similar triangles, and therefore a proportionality argument follows from that. We see that L1 over R1 is equal to L1 plus L over R2. Which if you cross multiply here, this is going to give you, this should be an R2. If you cross multiply here, you're going to get that R2 L1 is equal to R1 times L1 plus L, which distribute that if you need to. R1 L1 plus R1 L1. Now if you move things around a little bit, we're going to move, we're going to keep the R1, the R1 L1 by itself on the right hand side. We're going to subtract the R1 L1 to the other side. Notice we're going to get that R, we're going to get R1 L1. This is equal to R2 L1 minus R1 L1. Which when you factor that you get R2 minus R1 with an L1 out in front. So now you can see kind of the reasons why we did that argument there. Because this R2 minus R1 L1 is exactly what we found right here. We can make a substitution. Making the substitution, we'll just keep the pi R2 L from before. We're now going to get pi R1 L1 when we make that substitution there. And so now we have the following pi R2 L plus pi R1 L. Did I have the wrong symbol somewhere? Yes, I did, my bad. There should be, let me fix this real quick. This should be an R1 L. Because where did this R1 L come from? It came from this thing right here. You had an R1 L. I'll switch the order of these. So we had an L plus L1. And it looks like, it looks like I wrote L1 twice. It should just be an R1 L. So this is, this one right here is an R1 L. And therefore this should be, oops, that's not an eraser. This should be a pi R1 L. Sorry about that, everyone. So noticing now here, we have a pi R2 L plus a pi R1 L. Factoring out the common pi in L, we see that the area of the lateral side of this frustum is a pi times R1 plus R2 L, like so. And so the last step we're going to take right here is that notice that with this frustum, we have R1 on the top. We have R2 on the bottom. What we're going to do is we're going to place R1 with R2 with the average, the average of these things. We're just going to call it R, where R is one half R1 plus R2. So if we look at the average radius, notice we have, we have an R1 plus R2 right here. If we replace this with R1 plus R2 over two, and we'll slap a two in front, we can rewrite this area formula in the format that we want. A equals two pi RL. Oh boy, that took a long time to get to, but this is the formula we want to because with this formula in mind, we're now ready to talk about the surface area of the area of the area of a surface of revolution here. So, like we saw before, with our function, so we have some function whose derivative is continuous, so f is f's derivative is continuous on the domain a to b right here. Then the area of the surface of revolution formed by rotating the function y equals f of x so that there's a curve associated to that from a to b. We're going to rotate this around some line L, which in this graphic right here, which is courtesy of James Stewart's calculus textbook, we're just rotating on the x-axis for simplicity. Then the surface area capital S is going to equal the integral of two pi RDS, right, where R is the radius of this function as you go from, as you go from the function to the axis. Then DS is the arc length of the associated function. So where did the arc length come from right here? Well, you'll notice that in this picture, what does this cross-section look like? As we take our axis and we subdivide it into pieces like this, we're going to just take a small piece of the function and we're going to rotate this around. We rotate this around and we form a so-called band. A band is just the surface area of a frustum. And so we see right here in this picture what something looks like the surface area of our frustum, a so-called band. So surface area is constructed by taking bands. We approximate the surface using bands. And so each band will have a surface area. So the ith surface area will equal the formula we saw in the previous slide. Remember this two pi RL. So we get this two pi Ri Li. And so the surface area will be an approximation of this by taking the sum as i goes from one to n. We take the sum of this as well. i goes from one to n. And so then the thing to recognize here is that what is Li? Li was the slant distance of the band. And so that's where the ds comes into play here. Because for our approximation purposes, this is a delta s. And as we take the limit delta s, so here notice Li is equal to delta si. And so as we take the limit as n goes to infinity, we're going to see that this Li is going to approach a ds. And so we get this formula right here. The surface area is equal to two pi R times ds. Where ds has its meaning from before, the square root of dx squared plus dy squared. And so that is a bit of a long derivation on where this arc length formula comes from. But with this formula, we can very naturally start to calculate the area of surfaces of revolution. And one thing I want to warn you about is despite the difficulty that arc length formulas had, the integrals had with calculation, it turns out that surface area is typically much easier because the two pi R often helps us either do a u-substitution or an integration by parts that wasn't necessarily available for arc length calculations. So in the next video, I'll actually show you how you can compute surface area and it turns out it's not as complicated as it might seem.