 Welcome back, lecture 9, Math 241. We left a problem on the table yesterday, so let's pick back up with that, and we'll be able to go into our next application of integration, which is volume. So the thing that's going to carry over in just about every application in chapter 6 is that we're going to break what it is that we're trying to find area under a curve, area between two curves, volume, which we'll start on today. We're going to break it into pieces, we'll try to analyze a description of one of those pieces, one of those little skinny little rectangles, one of those little slices of volume, whether it's a desk or a washer or whatever it is, cylindrical shell. We'll talk about the kind of generic description of that piece, and then we'll let integration add together an infinite number of those pieces for us. So remember what that funny looking integration s is related to, it's related to that other s, that sigma s, which is a summation of an infinite amount of things. So that's kind of how we're going to handle just about everything in chapter 6 as far as an application of integration. So we had a couple of parametric equations, and we wanted the area under that cycloid, one of the little loops of the cycloid, and we finished class yesterday with this equation. This is related to f of x dx from a to b, which we use in finding area under a curve. And f of x is really the same as y, we've treated it that way to this point, and y in the parametric equation was defined as a function of t, excuse me, a function of theta. This was what y was defined to be in terms of theta. f of x dx, well dx, this is the derivative of what x was defined to be in terms of theta. And then we're going to be integrating with respect to theta, so these values are kind of starting and stopping values as far as theta is concerned. So it's the same format, it looks a little different, it looks significantly different, but it still is kind of f of x dx in the integrand, and our limits are in terms of theta now, so therefore we'll be integrating with respect to theta. So we left with that, anything that we can simplify by pulling out in front of that. What things have to stay to the right of the integral sign, and what things can we bring out front? Bring out the r times the r, so we can bring out an r squared. And once we do that, we have one minus cosine theta, we have two of those, so that thing times itself d theta. So let's square that, two of the three terms that we'll generate will be kind of easily integrable, the third we're going to have to mess with a little bit more. So we'll square one minus cosine theta, we'll get one, the middle term ought to be twice their product, and the last term, which is a little bit of a stubborn term is that last term squared, which would be cosine squared. So we've got r squared times the integral of one d theta, gosh I can do that one, minus two integral of cosine theta d theta, I can do that one too. The last one is a little stubborn, but we've done this before. What did we do when we encountered either a sine squared or a cosine squared or a sine to the fourth, cosine to the sixth, we're stuck with an even power of sine or cosine. Okay, so we had that double angle or half angle identity formula, so for cosine squared of theta, what are we going to replace it with? Two theta, does that work? Now let's suppose you forget that, how could we come up with that in 45 seconds or less? I'll be honest with you, at this very moment I don't know that that's right. I mean I trust the two people that gave me that, but if I had to come up with that on my own right now, I don't know that I could do that. So do you have a backup if the memory fails? Yes. Cosine of two theta equals cosine squared theta minus sine squared theta. Okay, I would start with that. Cosine of two theta, that's a double angle identity, is that right? And if you have to go a step back from that, it would actually be cosine of theta plus theta, right? Cosine theta, cosine theta, minus sine theta, sine theta, so anyway you could get to that point. Well we want the cosine squared, so I guess we want to keep this around, right? And we want to, we're stuck with this, but we can get rid of sine squared and replace it with what? One minus cosine squared? Yes. I guess I'd better hurry because I said 45 seconds or less, right? What's cosine squared minus a negative cosine squared? Two cosine squared minus one, so what's left to do? Kind of validate that we're using the right double angle identity here, okay? So we'd need to add one to both sides and take half of each side. Does that work? I don't know if that's 45 seconds or less, but it's not too much to validate that we in fact have the correct substitution for cosine squared. Questions or issues with that? What if we wanted sine squared and you forget it? Can we still use this? And replace, keep that and replace that, right? With one minus sine squared. How does it differ from this one? One minus, right? Still got a half out in front, one minus cosine of two theta. So if we replace that in there, let's go ahead and integrate the things that we can integrate. What's the integral of one d theta? Would be theta? What's the integral of cosine theta d theta? So you're really asking yourself what has cosine for its derivative, right? Sine. Would be sine. And now for cosine squared we're going to replace it with this double angle identity. I'll bring the one half out in front. Okay, all this is the same. Let's get to the new part. We're going to have a one half. And then we're going to be integrating once again one d theta. What's that? Theta. Theta again. And we're going to be integrating cosine two theta d theta. Let u equal two theta. We need a what? Two in the integrand. Correct for it with a half out front? One half. We've already got the other one half. Sine. Sine two theta. So we've got everything at least either integrated. Well, everything's integrated now. This is done. All we have to do is evaluate. It's done. Evaluate from zero to two pi. And now we've integrated each piece of this so we could also evaluate from zero to two pi. So, and that's kind of typical I guess with a x in terms of theta and y in terms of theta that you're going to get stuck with something that requires this kind of substitution. So we're kind of stuck with a cosine squared. But we've handled that before. We just kind of have to dig back into what we substituted for that previously. Anybody question kind of how to get to an answer from this point? Everybody feel confident? You can plug in zero and two pi and subtract and get a solution. And that would be the area under one of the loops of the cycloid. I usually don't talk on my cell phone when I come from my office to here. And like yesterday it was really pretty and all the snows on the ground kind of contemplate different things. I talk enough. I don't need to talk additionally on the cell phone. Sometimes I get a call. I urge you to do that by the way is to kind of have some contemplation time rather than get out of class and you know start dialing everybody and where are you going? What's for lunch? You know, constant chatter. That's kind of what the world is like right now a little bit but just deny that sometimes. My cell phone's off right now. Sometimes I just leave it off and it's just wonderful. So when I walk places I get to look at nature and snow and you know it's just beautiful. I urge you to do that to turn the cell phone off. But I was doing that contemplation thing. You've already heard that before. Hemachinicole and Chandler and Kelly and Kate. Sorry you had to hear it again. But I was contemplating the fact that I didn't think I did a great job of the diagram yesterday of the cycloid. So let's address this once again and see if I can kind of fill in. So we've got this circle, circular something, that's here we're putting a dot on it and we're going to roll it along the desktop, this line, the x-axis and as we roll this circle along this thing tracks out what amounts to a cycloid and when we've rolled it along so that the circle is right back where we started, the dot on the circle is right back here. This is what I felt like I didn't do an adequate job of. Now we said to simplify things that the radius would be one, that helps but it doesn't have to be one. If we've rolled this thing along and we've tracked the position of that dot and it now is right back where it started we said that was 2 pi but in reality it's only 2 pi if the radius is one but isn't that distance from here to here the circumference of that circle, is that correct? So as you roll it out and the dot is right back where it started this ought to be the circumference and if the radius is one the circumference is 2 pi probably didn't do a good job with that yesterday so let's let me try to clean that up today. If it's not one what is the circumference? 2 pi times the radius. So if the radius is one it is 2 pi well we plotted a point yesterday and I kind of had to backtrack because I had already said that this was pi and this was 2 pi, well that's only true when the radius is one. If the radius is just open-ended and it's r then we would say this would be 2 pi r and this would be pi r, right? So I think we ended up with a point that we were trying to plot that was in fact pi r and I said let's make r equal to one just to make it fit the diagram but if r is not one this thing still works you won't be integrating from 0 to 2 pi you'd be integrating from 0 to 2 pi r and if you want to do it in terms of r that's fine you could have a final answer in terms of r we did the problem where r was one so this was pi and this was 2 pi but it can be done, let r just be r some open-ended number see what weird things I contemplate when I walk from here to Harrelson but I knew that I had not done justice to that yesterday let's go into 6.2 so remember the skinny little rectangle concept we want an infinite number of those added together we do that not by this summation but remember as we have an infinite number of these skinny little things that we're adding together that s, that Greek sigma becomes this s, the integration so we are able to add an infinite number of these things together as long as we describe them right here so if we describe one of them and we tell it where to start and where to end then we're in business we've added an infinite number of those things together so in 6.2 we're going to take I guess three distinctly different looks at volume and we're going to slice it up these are volumes of solids of revolution so we're going to have a figure in the plane we're going to rotate it around some axis either the x-axis or the y-axis or a line that's parallel to the x or y-axis and we're going to generate some three dimensional figure and we're going to use this concept of getting skinny little pieces or slices of it and add them all together using integral calculus so it is an application of integration this is probably way overly simplified but let's say we have some f of x in the plane and we start and stop the region bounded by f of x with some vertical line x equals a and x equals b so we start and stop it just like we did with the area and we want that region that's bounded by this vertical line, this vertical line the curve itself and the x-axis and we want to revolve that around the x-axis so here is our axis of revolution so I'm not going to be good at this but I have some pictures that are better than mine and you have shaded pictures in your book in section 6.2 much better than what I'm going to come up with here but if you can envision taking that region holding this fixed right here as an axis of revolution spinning that around the x-axis you'll come up with a three dimensional solid and it looks something like that and I'll show a picture that's better than this so you can envision this solid 3D figure our job is to first of all create the pieces that we can describe one of them it represents all of them and then we integrate it that adds together an infinite number of these skinny little pieces or slices together from where we tell it to start to where we tell it to stop so one of these slices much thicker than we actually want it to be we really want it to be as thin as we could possibly make them stack them all side to side and we have all the volume that we want from x equals a to x equals b but there's one of the slices now how do we form the slices so we're going to slice it up perpendicular to the axis of revolution so our axis of revolution is the x-axis so we form our slices this way chopping it up perpendicular to the axis of revolution we need to describe how much volume there is in this slice and hoping that all the slices and you may actually want to draw several of them to see that if we describe one of them in that fashion are they all really described by that same description so here's our slice actually let's position it a little differently that slice on its edge looks like this all of them do so what's the volume of something that looks like this what is it first of all that's a cylinder right little short squatty little cylinder what's the volume of a cylinder pi r squared h now if you know that then you can do any of these volume problems if you know that the volume of a cylinder is pi r squared h then you're going to be able to do this that'll kind of start you know you can memorize other formulas if you want to but this is the one that will kind of go all the way back to the volume of a cylinder that will get you started so we need to be able to describe the radius of each slice that we generate which is a little disc, a solid disc in this case everybody agree that those are solid if we take that region spin it around the x-axis we don't have anything hollow in the middle it's a solid disc here's a better picture this one at the bottom is what we're dealing with I'm trying some different colors I really have a problem with light blue so I'm trying some different colors even though we kind of have a stack of light blue paper up here not my favorite color so we've got this function we take the region they've divided theirs up into skinny little rectangles that's fine but when we revolve that around the x-axis we slice it up perpendicular to the axis of revolution and we get a bunch of discs, solid discs so that's what we're dealing with now how skinny do we want them to be a whole lot skinnier than this picture paper thin, in fact, even skinnier than paper thin we want our thickness to approach zero so what is the thickness what would we say that is? that's a delta x on this picture we want the thickness, which is the height really isn't it of each little short squatty cylinder we want that to diminish to practically zero when we allow that to happen then we have zero error we get the exact volume of this figure now that's if we're going around the x-axis where the x-axis is the axis of revolution we might take a curve and say the y-axis is the axis of revolution so we take this region, bounded y equals some initial value y equals some terminal value spin that around the y-axis we get a figure that looks like this since we're going around the y-axis we chop it up perpendicular to the axis of revolution and that creates solid discs in this fashion so these are really the same if we can describe one of the discs in this case a solid disc we'll use integral calculus to add them all together from beginning y value to ending y value so up here again we want the thickness to approach zero well what is the thickness up here it's a delta y so we're going to have dy's in our integrand and we're going to be integrating from y something y0 to y1 down here we're going to have a dx in the integrand and we're going to be integrating from x0 to xn but it's the same process now that's if we use one of the two axes as the axis of revolution and we might use a line that's parallel to them and we'll hopefully have enough time to look at an example I don't know, not looking good but at some point in time, maybe not today we'll get to that example as well so each piece is a solid disc and we want to find the volume alright, first example pictures help, pictures aren't absolutely necessary but pictures I think are going to help you to decide if do we have a solid disc well that's all we have at this point so that's what the first example is going to be but the next example is not going to be a solid disc we'll have to analyze how we find its volume so let's take, let's make it pretty simple and I've got a better picture of this so we're not just going to depend on my picture let's take the region bounded by y equals x squared x equals negative 1 x equals 1 and the x axis and we're going to take that region and we're going to spin it around the x axis so we'll have this three-dimensional solid so y equals x squared plus 1 looks like this negative 1 to 1 so there's the region that we're going to spin around the x axis so I think it helps to get that symmetric image again whether that helps give it a three-dimensional feel I don't know but it helps me kind of decide what I'm dealing with so I'll try to give this a little three-dimensional feel and I want to chop it up, slice it up perpendicular to the axis of revolution so there's a representative piece or slice so if we take that bounded region spin it around the x axis can you visualize the three-dimensional solid that we have? does that slice or that piece seem to represent any other slice or piece that I could draw? so if we can describe what the volume of this one is all of them are pretty much the same do they all have as we start at the axis of revolution and go out to generate the radius that's the radius would all of them have that same line segment as their radius? not that same one but could we describe it in such a way that it represents the radius of every disc? I think we can so from the x axis up to a point on the curve that's going to be the radius so we're going to I said that that would get you started it can get you started that's the volume of one of these little discs one of these little cylinders but we're going to need to know what the radius is how would you describe coming from the x axis up to a point on the curve what is that typically? that's f of x that's how we plot the point we put it into the function we go over that many units x units but we go up f of x units so the radius seems to be y or f of x what about the height or thickness what's the thickness of this particular slice or disc? it's going to be an increment of x some little small distance in terms of x that's created by how we chopped it up so it's going to be a delta x or I guess a little further down the road it's going to be a dx in the integrand we've got a better picture of this because mine is not all that good so here's a better picture of what we're dealing with here's our curve y equal x squared plus one stopping at one starting at negative one it's a solid three-dimensional solid and one of the slices or discs is pictured here so we need the radius now they have this described as w sub i squared plus one wherever we are w sub i we would put that into the function that's just telling you that that distance is the f of whatever the x value we're plugging in so if it's w sub i we'd plug it into this function and we'd get w sub i squared plus one so better figure but we don't have to be quite so subscripted that we put in all of the subscripted values we'll describe one of them we know what integral calculus does it adds them all up for us so pi this number we'll just bring it out front where are we starting these little pieces or slices negative one and that is an x value and aren't we going to be integrating with respect to x we are because we're going to have a dx in the integrand and we're going to stop it at one that's where to start and stop the skinny little discs pi is already taken care of we want the radius squared what is the radius it's the y value and what is the y value for this particular function x squared plus one so that's our radius squared and our height of each disc is an increment of x a little delta x which in the integrand for us becomes a dx so there's the integral calculus problem that will get us the volume of this three-dimensional region questions about getting to that point so that'll get you started pi r squared h volume of a cylinder if you want to memorize more beyond that then that's completely up to you but that should work so we've got a square this x squared plus one well maybe if we don't have to that's not squared if we don't have to use a u substitution can we let u equal x squared plus one we could try I don't think it's going to work why is it not going to work because we don't have another x in the integrand we need another x for du right we can't manufacture variable quantities inside the integrand we could change any problem we wanted to if we could do that so we are forced to square it first term squared twice their product last term squared it's not going to be bad so we're going to integrate x to the fourth dx x to the fifth over five integrate two x squared with respect to x two-thirds x cubed correct me if I'm wrong now and integrate one with respect to x I think I can handle that one negative one to one so we're going to have one-fifth plus two-thirds plus one and what all the negatives stay negative right everything's odd so negative one-fifth minus two-thirds minus one everybody feel confident you can put all that together right we're going to have two-fifths plus four-thirds plus two add that together multiply it by pi that's the number of cubic units of volume right that's the volume of a three-dimensional region questions about that integration is not going to be the hard part I mean occasionally we'll have to do some testy little double-angle identity or something like that that we've already done before but when you get to that point you're three-fourths of the way there questions before that one gets moved what would change if we took a similar shaped region and wrapped it around the y-axis instead of the x-axis our thickness or height would be delta y we'd be going from a y-value some initial y-value to some terminal y-value basically nothing changes right you just reference all your disks on how they're related how the radius is related to the y-axis instead of the x-axis okay let's try this picture and see how we think the region is going to be different from what we've looked at so far in terms of a solid disk we want to take the region bounded by these curves this is kind of a bizarre way to give you the equation but I'm giving it to you the way it was written I don't think this is a problem in this book I think I got it out of another source so we've got a parabola and a line two vertical lines and we want to take this region we're going to do two problems with this diagram we'll draw the diagram the first time let's go around the x-axis so it kind of looks like it's going to be the same type problem it's actually going to be very different so if we solve this for y this would be y equals x squared plus 2 very similar to what we had now we've got another line if we solve that for 2y 2y would be x plus 2 and therefore y would be one-half x plus 1 is that right? I think that's right so y equals x squared plus 2 and I'm going to start at x equals 0 so I don't need the other part of that so I want to start at x equals 0 stop at x equals 1 and the other one is this line y equals one-half x plus 1 so the y-intercept is 1 and the slope is one-half so there's our region that's bounded by the parabola of the line and the two vertical lines x equals 0 and x equals 1 so we want to go around the x-axis a little different isn't it? what's the result that's going to be different? why is this a different problem? it's not bounded by the x-axis okay we don't have the region all the way down to the x-axis like we did before or in the reverse situation we don't have it all the way over to the y-axis so when we take this and we spin it around the x-axis I'm not going to clutter it too much but this is going to probably clutter it what about that disk once we take that region and spin it around the x-axis it's missing some volume out of the middle so the disk is not a solid disk depending on the part of the country you're from it's either a washer or a washer I grew up where this was a washer but I've adapted and now I call it a washer but it's one of those two it's something that looks like this a donut or a bagel we'll go with that too but all those conjure up the idea that there's something missing right in the middle so this is not a solid disk it's a washer so we're not going to be able to start with pi r squared h because that's a solid disk so we need to come up with what the volume of a washer would be so a washer is really the entire solid disk minus the part that's missing out of the middle so if we want to come up with something like this we'll start out with the entire solid disk and we'll subtract out the volume that's missing so we're going to have two radii here so let's call one capital R and the other one lowercase r so how much volume is there in the larger solid disk pi capital R squared h does that work and the other one has a lowercase r pi R squared h so we'll kind of approach this from what we have done which is very Ossobellian if you want to know what that is look up David Ossobel is a psychologist and that was his method of cognition is the way we're doing this right here I'm an Ossobellian learner myself look it up you might be as well so if we want to come up with a volume of this can't we take this guy and subtract this from it what do they have in common that we could factor out front and h or let me put the h at the end because the h is going to be a dx or a dy right so we usually have that at the end of the integrand capital R we can call the outer radius and this lowercase r we can call the inner radius there's the volume of a washer does that work now don't take outer radius minus inner radius and then square the result that's not what we have it's the outer radius squared minus the inner radius squared so that's our volume that's our description of a washer regardless of where that washer is located how it's positioned in the plane if we slice up our three dimensional region and each slice is not a solid disc but rather a washer then we're going to come back to this thing right here here's a better picture I said I would come up with a better picture so if we have some region we've revolved it around the x-axis and we have something like this and the dotted lines show you the part of the volume that's actually missing in there and you slice it up perpendicular to the axis of revolution you're going to get one of these does that work and we just need to make sure we don't count the volume that's in here so we do that by taking the outer radius subtracting the inner radius from it and in this case our thickness or height is delta x and this would be from it looks like on this diagram from A to B so that's what one of our slices is going to be we describe one of the slices what does the little elongated S do it adds together an infinite number of these skinny little slices for us Chandler you had a question yeah I guess I set it up differently and it's I think I just did it wrong, never mind is that right? can you see that slice? my figure I didn't kind of draw the three-dimensional version because it's going to get cluttered in a hurry and you're going to see that you're going to lose the fact that this is missing some volume out of the middle so you might just want to kind of visualize what that looks like and then picture one of your slices so we had these curves y equal x squared plus 2 and y equal one-half x plus 1 so I'm going to refer back to this diagram sorry it's going to get kind of cluttered we're going to need the outer radius well the outer radius comes from the axis of revolution which is right here so from here all the way out to this curve describe to me what that distance is isn't that the y value of every point on this curve right? so the outer radius is really y on the parabola and what is y on the parabola? x squared plus 2 so our outer radius is the y value on the I'll just put a sub p y value on the parabola which is x squared plus 2 so that's our capital R our lower case R it's going to get cluttered I'm already overlapping what's this distance from the x axis up to here which is really the inner radius that's the y value on the line that's right and the y value on the line is one-half x plus 1 right? so we'll farm the pi out to the front we're starting we're going to integrate with respect to x so we're going to start this at 0 and end at 1 capital R was x squared plus 2 that's the outer radius we're going to square that we're going to subtract from that the square of the inner radius inner radius was the y value on the line so that's capital R squared minus lower case R squared and what's the height or thickness dx delta x so again when we get to this point we're probably two-thirds to three-fourths of the way done with the problem and this should be kind of old stuff let's see what it looks like any questions before we go just a little bit further so we'll square that x to the fourth plus 4 x squared plus 4 we'll square one-half x one-fourth x squared the middle term would be twice their product what's the middle term? x so we're going to have an x to the fourth how many x squared three-and-a-half, seven-halves x squared does that work? oh sorry, 3.75 where'd I get seven-halves? so what do we have there? 15-fourths I need to check my arithmetic thank you so we got 4x squared minus a fourth x squared 15-fourths x squared we've got a what? minus x and a plus 3 everybody feel confident you can integrate each of those separate terms once you're done integrating what are you going to get when you plug in zero? everything's going to have an x in it so those will all be zero so you're going to get whatever you get when you plug in one so again the integration is not going to be the difficult part it's the setup identifying capital R identifying little r outer radius, inner radius that should be the volume of this region which has part of the volume missing out of the inside each slice is a washer alright we're out of time we're going to take this same problem and this next time go around a line it's parallel to the x-axis instead and see how that changes things I will see you tomorrow