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About this Lesson
 Type: Video Tutorial
 Length: 12:59
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 139 MB
 Posted: 06/26/2009
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus: Parametric Equations, Polar Coordinates (18 lessons, $27.72)
Calculus: Polar Functions and Area (5 lessons, $7.92)
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 UTAustin 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, padic 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.
About this Author
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More..Recent Reviews
Parametric Equations and Polar Coordinates
Polar Functions and Area
Heading Towards the Area of a Polar Region Page [1 of 3]
Thinking about differentiation and calculus with respect to polar coordinates and polar forms and polar function raises the obvious question: How would you find the integral or areas of some of these polar regions  these polar graphs? What would they look like? So you have these beautiful rose petals, all of these [ ] and so forth, what is the area, for example, of a rose petal? How would you figure that out? What I'd like to do is just try to inspire that and really head towards the area of polar regions. And I want to do this in sort of a first step just to sort of get our feet wet because really if we just look at the formula, it will make no sense, but if we just think about it for a little teeny bit, it will make complete sense, and that way we'll be able to understand really what's going on and set up the appropriate integrals and find the areas correctly.
Let's begin by just heading towards the areas of polar regions. First of all, what would the area of a polar region even look like visually? Well, visually if you have the plane and you have some polar regions, which might be very wiggly or something, so this might be r = f(). Suppose I give you two points, and I want you to find the area of this between two points. First of all, what does that even mean between two points? It means I have to give you two points from the domain, which means I give you two values. So if I give you two values, like = a and = b, what is = a? = a, remember, is a line that emanates out of the origin  emanates out of the pole  and has an angle from here up to that line of a radians. So in fact, I would look at a little piece like this  this might be = a. Then maybe I'd go to here, and this might be = b. So the region I sweep out actually is no longer the kind of region we've seen in early calculus, but it's like this funny squiggly pieshaped type wedge. You see, it's a wedge like that, and the question is what's the area of that red wedge? So that's the kind of thing I want to figure out, what's the area of that? How would I compute that?
Well, we're faced with a hard problem, and so we know what to do: Don't do it. Instead, let's think of an easier problem. In fact, let's think of a problem that was actually already resolved. Let's think about what happens when we think about the good, old Cartesian coordinates and we're looking at areas under curves in the Cartesian plane.
So let me just bring that back to you really fast. If you've got some crazy function  suppose you have a function that looks like this  and I want to find the area from here, a, to b. That area is complicated, and who knows what the formula for that is? I don't know. So instead, we don't do that question. What we did was we actually said, "Well, let's do something easier." Let's actually instead of thinking of this exotic shape, let's actually make a whole bunch of shapes that are each easy to evaluate. So, when you're faced with a big, hard question, sometimes it's better to do a lot of little teeny questions, each of which are solvable.
One thing we could do is just break this up into little regions, and put rectangles in. In fact, here they are. So you put in these rectangles here, and you can find the area of each rectangle. That's not too bad. It's base times height. As long as you put in a lot of rectangles, in fact if you sort of take the limit and put in infinite many rectangles and they get thinner and thinner and thinner, what happens is in that limit is you actually start to fill up space and you actually go from this sort of bumpy thing to the perfectly smooth curve.
So to get the integral, what do you do? Let me just recap. So in this case  in the olden days  we would sum up from a to b  from our starting point to our ending point  these little areas. And what's the area of one of these rectangles? We're going to sum up all of those rectangles. Well, the area of the rectangle, that's actually easy  it's base times height. What's the base? Remember, I'm putting in a lot of them, so the base is just a tiny, tiny, tiny little bit, and it's a tiny change. It's a tiny change in which direction? Well, this is the xdirection. So it's a tiny change in the xdirection, which we refer to in calculus as dx. That means a small change in the xdirection. And what about the ydirection, base times height, what's the height? Well, it's the value of the function  we're after the function  and that's f(x). So we saw that the area under this Cartesian curve turns out to be just , and that's where it came from.
You may say, "Okay, so if we argue by analogy, maybe the answer to our question"  so here's a guess, and the guess would be  "let's just dabble and go to this picture, let's just do the same thing." So let's go from a to b  from the small angle to the big angle  so I'm going to go from here to here. So actually would you allow me right now to make this a and make this b? If I do that, then this could keep in consistency with this. Then we go from a to b  sweep out that  and I'll just write f()d. And the question is: Is that right? Well, that's a great guess. In fact, it's a great guess, great guess. Of course, all guesses are great. The question is is it a correct guess? Is it correct? And the answer is no, it's not, because we used this principle. You see, this was actually finding areas of rectangles  base times height  but there are no rectangles here. It's very difficult to see how to put in rectangles in a systematic way. So in fact this guess, although it's great, turns out to be incorrect. So it's sort of nice to see this thing, because everyone wants to sort of say it, and then realize that it's not right because this came from the old method where we were using rectangular coordinate. If we're going to use polar coordinates, we've got to do something else.
So what do we do? Let's adopt the same strategy, which is let's consider an easier question, but now the easier question should be little wedges. Let's break this up into little pieces that look like this and sum up those wedges. If I make those wedges small enough, instead of rectangles, what they would look like is little teeny slices of pizza or little pieces of pie. So what I'd see is the following: wedge now would look something like this. This is the analog  this is the polar analog  of a rectangle. You see, in rectangular coordinates, we get a rectangle. In polar coordinates, we get this sort of little pieceofapie thing, and I want to sum those up. How do you sum up pieces of a pie?
Let me remind you how this goes. Suppose that the angle here is , and suppose that this length here is r  that's the radius of this arc of a circle. Then, in fact, this length here, this arc length, actually is conveniently given by just r and if we have represented in a radian. In fact, that's the whole point of having radians instead of degrees. We're all familiar with degrees, but radians are much better because they allow me to tell you exactly how long that arc is. That arc is just going to be how much we sweep out and how far r is. Think about it for a second. If I keep that same angle but I make it much, much longer, well I'm sweeping out more. Do you see that? And if I keep the same r but make smaller, then I'm sweeping out less. So it depends on both r and in the following way: r times .
For example, think about the good, oldfashioned circle of radius 1. If I sweep around one, that's an angle of 2. And what is the circumference of the circle? It's 2r, which is 2. So notice that in that case I sweep around once, and I go 2, that length is 2. If I go out further  make a bigger circle  then the circumference would be larger. So, in fact, the general formula is r.
Now the question is what's the area of this? This is what I want to figure out. I want to figure out the area of this little piece. That seems a little bit too hard, but if we think about it we can just approximate it and say that roughly speaking, that's a triangle. Now this is not correct, but so what? These things are so small. Remember, these little wedges are going to be so small that you couldn't tell this from that. It's the same size roughly. This is r  a little bit of fantasy math here, folks  and, in fact, what's this? Well, that's basically r too because look, it's so tiny that you can't really tell the difference. Technically it's not, but it's roughly r. And this here, that's roughly this. As drawn, this is all wrong; but if you imagine this being really, really thin  let me show you one actual size right now  look at that, look how thin that is. They all look the same. Everything is the same  this length, this length, and that center length that you can't even see in there would be the same. So, in fact, this is not a bad picture, but the area of this we could figure out. The area is onehalf this  base, if you look at it sideways  times height  this. So it would be r^2^^^^. And, in fact, if you take a look at this  so here's actually a slice of a pie that I've worked out in advance  you can actually see this even more clearly if I cut it in half. If I cut this in half right now, then I have half as much here. So this would be r divided by two, and look what happens if I just put these together in a different way  put this here, and instead of putting this one like this, let's put it like this. Then essentially that looks like a little rectangle, doesn't it? Now, that's actually a huge one. In actual size it looks like this, so it even looks more like a rectangle. And what's the area of a rectangle? Well, it's base times height, and the base is r, and the height is half of this, which is . So notice the area of that, we're again seeing r^2^^^^. That is the area of a little teeny, teeny wedge. What if I sum up all those wedges? If I sum up all of those wedges, then what I would see is the following: I would see an integral. I'm going to sum up from = a to = b, and I'm not going to sum up rectangles, I'm now going to sum up polar rectangles, and a polar rectangle is actually this shape. What are they? Well, it's going to be the area is this: onehalf, and then I have r^2. Now what's r^2? Well, r = f(), so this is just f()^2. So I put f()^2, and then . Now what's ? Well, that's a very, very, very tiny angle. It's a tiny change in . What do you think we call that? In calculus we call that d.
So, in fact, this integral represents the area of this red wedge. So the area of a polar wedge between two angles like this is . It's a lot different than our initial guess, but hopefully it makes a little bit of sense. It's the exact same process. We try to change things into simpler objects and analyze the simpler objects, and that shows us where the comes from, where the r^2 comes from, and where the little, teeny change in comes from.
So, in fact, if you want to find areas of exotic regions that are polar, this is what we use  same idea, but now new formula. Let's see how this is going to be used in practice at the next lecture. See you then.
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