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Calculus: The Disk Method along the y-Axis

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About this Lesson

  • Type: Video Tutorial
  • Length: 11:43
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 126 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Applications of Integral Calculus (18 lessons, $26.73)
Calculus: Disks and Washers (5 lessons, $8.91)

Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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Recent Reviews

Nopic_dkb
Superior
02/04/2011
~ greenshoes

great explanation! Thank you!

Nopic_dkb
Superior
02/04/2011
~ greenshoes

great explanation! Thank you!

Applications of Integral Calculus
Disks and Washers
The Disk Method Along the y-Axis Page [1 of 3]
So let's take a look at these solids that are made by taking an area in the lane and sweeping it through space in order to make a collective whole three-dimensional thing. Once we have a collective whole three-dimensional thing, we can ask the question, what's the volume of it? So these are knows as volumes of solids of revolution. That is to say, you take some two-dimensional thing, and then you rotate it around some axis and you pretend, for example, the thing is all inked up. And as it's inked up and you drag it through space, there's an inked residue that's left behind. And that inked residue actually describes, now, a solid object. Pretend now, I solidified. I freeze it now. So I have a solid object, and the question is what's the volume? It turns out we can use integration in an fairly systematic way to figure out what the volume of these things are.
So what I'd like to do is take a look at some examples, and here's one. Let's, first of all, take the area bounded by the following curves: y = x^3, the y-axis, and y = 1, y = 8. So I want to take that area and then I want to revolve it around the y-axis. And that's going to produce now a particular solid. And the question is, what is the volume of that solid? Now the way that I actually proceed in these things is to begin just by drawing a picture. If you draw a picture, that really will sort of capture the essence of what's going on here.
So what would this look like? Well, let's graph this thing and see what it would look like. Now the sketch doesn't have to be extremely accurate. You just want to get a sense of what's going on here. So we have y = x^3, so I know how that looks. That's a cubic function. Let me sketch it out for you. It sort of goes like this. It comes down like this and it comes off here and then it sort of comes down like this. So it looks like that. That's sort of how it looks, just like that. So that's the y = x^3 function.
Then I have the y-axis. Well, that's just the y-axis. There's not much action and excitement there. And then I have y = 1, which let's put right here. There's y = 1. That's now going to be a horizontal line. And a y = 8, so let's say 8's way up here. Now, there's a region that's bounded by this, this, this and this, and it's right in here. So let me show you that green region. So this region right here is the region that's bounded by all those curves. And what I want to do is take this area and I want to revolve that area around the y-axis. So imagine taking this thing - pretend it was on a hinge right here. So if we're hinged up, I could take it and sweep it out. And I sweep it, not only this way, but sweep it all the way underneath, and then it comes out the other side. And when I do that, I get some sort of surface, a surface of revolution, in fact. And let me try to draw that for you.
So if you imagine, now, taking this and reflecting it over to here, it would sort of take on this shape. So this piece would sort of reflect over here and would look kind of like this. And again, the accuracy is not as important as just to help us with the intuition, so I'm not going to worry too much about it. So it sweeps out here. And what does it look like? Well, it sort of looks like a solid bowl. It's all filled in inside, after it spun around. It's like a chunk of - it's like a round piece of cheese. It's like a Jarlsberg cheese. You ever seen those round cheeses, where I just sort of cut off this part and cut off this part, and I have that round cheese thing right here, left over.
Now how would you compute the volume of this thing? Well, what you have to realize is that, in fact, if I slice it in a particular direction, I get very simple shapes. Now, if I slice this way, the shapes are actually not so clearly defined, because I'm going to run against these two walls for awhile. And then when I get out to here, I start to run against this curve. So those slices seem a little bit complicated to figure out. But what if I sliced this way? Well, see, if I sliced this way, if you now visualize this thing being revolved around the y-axis, what we see is every slice will give me a circle. So let me show you some slices here. In fact, let me just draw in one particular slice - sort of an arbitrary slice. If I draw an arbitrary slice, it would look like this. So this is just a generic slice, if I were to cut this solid now - so you have to visualize that it's already been rotated around - that's what it would look like. I would just spin it around, and every time you slice, you would get a different little circle wafer thing. Here, the circles would be very, very small. Here, the circles would be quite large.
So what I really want to do is just sum up the volume of those disks. And so this is sometimes referred to as the method of computing volume using the disks, because every single slice, in fact, is a little disk.
So if we think this way, then what would the volume be? Well, I'm going to sum up the disks, and I'm going to sum up the disks from where to where? My first disk is going to be way down here, which is at, notice, y = 1. So the y is going to go from 1, and I sum all the disks up until I get to a height of 8. So I'm going to sum from 1 to 8. And now what's the volume of this little wafer-thin disk? Well, it's going to be the area of the circle multiplied by the thickness. But that thickness notice, is a vertical thickness. So it's a small change in which direction? Well, it's a small change in the up and down direction. It's a small change in y. So, in fact, this integral is going to look maybe a little peculiar to us, because, in fact, this is going to be a integral that is going to be a dy integral, because I'm summing up in this direction. I'm summing up y values. Usually, when we think about areas into curves, we sort of sum this way, in which case it's a dx. But this one going up or down is going to be a dy. And, again, the thinking is that this is a small change in the y direction.
Now, so that's the thickness. Now what about the area of the top? Well, suppose that I'm an arbitrary point here. Now what's an arbitrary point? If I'm going from 1 to 8, those are y values. So let's pick an arbitrary y value here. I'll call it y. And that little y, in my mind, I'm thinking of actually scanning from 1 and it's going to move all the way up - like a thermometer - all the way up to 8. So that y is just some generic point that's going to slide up. But at this generic point y, I've got to figure out the area associated with that circle, and what is it? Well, it's r^2. Now what's r? Well r is actually this length right here, isn't it? It's not y. y is just this height. But r is the radius. And so what is that? Well, that's just this value right here. And how would I figure out that value. Well, if I know this is y, I have to work backwards to find out what the x would have to be. So if I take this equation and figure out what x has to be, what would it have to be?
Well, if I solve this for x, I would see that x would equal the cubed root of y. So this must be the cubed root of y. Let me say that again. I'm letting my y values range from 1 to 8, because that's how I'm slicing. I'm slice, slice, slice, slice, slice, slice, slice, slice, slice up to here, so that's from 1 to 8. And if I pick an arbitrary point in between there and call it y - since I'm on the y-axis - I've got to compute the radius. And the radius is just the distance from here to here. Well, that means I've got to find that length, which is nothing more than the x value I have to plug in, in order to get to y. So if I take this thing and solve it for x, I see that x would have to equal the cubed root of y.
So that's the radius. And what's the area of a disk? It's r^2. So I have , and then I have this r - that's the radius - squared. So that's going to be y cube root squared, which is to the power. So that's the integral I have to evaluate in order to figure out the volume of this solid generated by revolution.
So, in fact, now the integration - you can already see into the future - is not going to be that difficult. The real challenging thing here is to set up the integral. Setting this up is really the challenge at hand. Once you set it up, evaluating it is no big deal. So again, the thinking is to realize that if I slice it in this particular direction - and in fact, slicing it sort of perpendicular to the y-axis, slicing from 1 all the way up to 8 - every slice looks like a little thin disc. And I can actually compute the volume of the thin disk. The height is just a tiny change in y, so that's the dy, which tips me off that this is going to be an integral in terms of y, rather than x. And then all I've got to do now is find the area of this disk. And the area of a disk is r^2. And so we have the , and now what's the r? If I'm at an arbitrary point y, I just can figure out what that r is by taking this and solving this for x.
Okay, so let's now evaluate this integral. Well, the is a multiplicative constant, so I can pull it out in front. I'm integrating from 1 to 8, . And what does that equal? That equals . And so that's going to be and you add 1, so that's going to be like . And then I've got to divide by , which is to multiply by . And then I evaluate this going from 8 to 1. So when I plug in 8, what do I see? I see , and now I've got to plug in 8 here. Let's do the arithmetic in our head. So what is ? The first thing I have to do is I'll take the cube root of 8. The cube root of 8 is 2. And then if you take 2 and raise it to the 5^th power, you get 32. So this is just 32. And then subtract off what you get when you plug in 1, which is just . And so this works out to be , which if you want to know numerically what that is, that's around 58.433 units cubed, because it's volume. So the volume of this is , and I did that by summing up a whole bunch of disks. But I summed them up with respect to y.
All right, great. This is the idea of finding volumes of revolutions using a disk method, because every slice produces a disk. I'll see you at the next lecture.
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