Preview
Buy lesson
Buy lesson
(only $1.98) 
You Might Also Like

Calculus: Differentiability 
Calculus: Power Series  Differentiation 
Calculus: Power Series DifferentiationIntegration 
Calculus: An Introduction to the Integral Test 
Calculus: Trig Substitution to Integrate Radicals 
Calculus: Welcome to Calculus II 
Calculus: Fibonacci Numbers 
Calculus: DomainRestricted Function & Derivative 
Calculus: Applying Implicit Differentiation 
Calculus: Introduction to Implicit Differentiation 
College Algebra: Solving for x in Log Equations 
College Algebra: Finding Log Function Values 
College Algebra: Exponential to Log Functions 
College Algebra: Using Exponent Properties 
College Algebra: Finding the Inverse of a Function 
College Algebra: Graphing Polynomial Functions 
College Algebra: Polynomial Zeros & Multiplicities 
College Algebra: PiecewiseDefined Functions 
College Algebra: Decoding the Circle Formula 
College Algebra: Rationalizing Denominators

Calculus: Introduction to Implicit Differentiation 
Calculus: Applying Implicit Differentiation 
Calculus: DomainRestricted Function & Derivative 
Calculus: Fibonacci Numbers 
Calculus: Welcome to Calculus II 
Calculus: Trig Substitution to Integrate Radicals 
Calculus: An Introduction to the Integral Test 
Calculus: Power Series DifferentiationIntegration 
Calculus: Power Series  Differentiation 
Calculus: Differentiability
About this Lesson
 Type: Video Tutorial
 Length: 8:50
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 94 MB
 Posted: 06/27/2009
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
 Thinkwell
 2174 lessons
 Joined:
11/14/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
 AAAAA+++++
 04/25/2010

He's ALWAYS GREAT!!!! MUST WATCH!! THANK YOU!!!! Hopefully people can truly appreciate how rare it is to have the ability to watch one of the best professors utilize all skills of teaching to produce such power packed lessons that stick in the brain. Every detail is thought of in the video presentation. THANKS AGAIN!
Parametric Equations and Polar Coordinates
Polar Functions and Area
Finding the Area of a Polar Region Part One Page [1 of 2]
It's one thing to head toward the idea of understanding of finding areas of polar regions, but now actually let's start finding areas of polar regions. What I want to do is take a look at the example of figuring out the area of one of the petals on one of these rose curves. So here's a rose, for example, and the petals, and this is what I want to figure out: I want to find out what is the area of that petal  a polar petal? It smells so good. I'll keep it right here so we can appreciate that.
So what do we do? Well, we know the formula, and here's the particular petal I want to look at. My particular petal is going to be the r = cos 3, which we know has three petals, and I'd like to find the area of this particular petal. Let's find the area of this particular petal right here. So that's the area I want to figure out. How do I do that? Well, what I do is we know how to compute these areas. That area is going to equal the integral from the smallest angle I sweep out to the largest angle I sweep out  I'll write that  and now we'll no longer have those rectangular features chopped up  that's what we used when we had rectangular coordinates. Now we're going to have these polar rectangles, which are these little wedges  little pieces of pie  and so the area of a piece of pie we saw was f^2 times the tiny change in . And by the way, the is just a multiplic of the constants, so in fact a lot of times you just write it on the outside. This is just a constant number. So , and then f()^2d.
So how do we set up the integral for finding the area for this? Well, I've got to find out the angles, first of all. What are the angles? Well, the angle, if you think about it, the small angle is actually this angle right here. That angle right there is the small angle because this curve just nestles and touches this tangent right at that point. So it starts here, and then I'm going to sweep up until I hit this line, which again we saw already is tangent to this curve. So those are precisely the two angles where this curve lives inside because they just sort of nestle in there and just graze it perfectly.
So I'm going to want to actually integrate from whatever this angle is up to this angle. Now this is actually going to be a little bit tricky because what is this angle? Well, there are many names for it, right? There are many names for it. I could sweep all the way around and get to it that way. I could actually sweep around two times and then end up there, or I could actually go backwards, and look at that! What is the correct angle to look at? Well, I want to think about moving smoothly from here up to here, so I want to start small and increase up to this angle. So thinking about that angle this way, actually, would not work, because that would actually now be covering all of these pieces right in here. If I just want to start here and go up to here, I should actually use an angle that's going to be smaller than this and will allow me to flow beautifully to here. So the representation of this angle that I'll use is going to be just taking this and scooting down. That would be . Let me recap that. The idea is I want to go from very continuously up. This would be zero, then I would keep going to . So it's a very smooth progression of angles. So I start with , and I progress out to here. So I can now write this in. This is . And then what's the function? Well, there's the function right there, so this would be cos^23d. So there's the integral I have to evaluate. If I evaluate that interval, that will give me exactly the area of the shaded region.
So my question is: How do you figure this out? If it was just cosine alone  if it was just that  we'd be in high heaven because all I would do there is just take the integral or do a sine something. But I've got the square there, so this is actually a major problem. So when we're faced with a major problem, what we have to do is we have to actually, of course, go to a major aside. So I want to actually show you a little technique that can be used to help untangle this thing. It's an integration technique that comes in quite handy when we see trig functions like this. I just want to show it to you really fast. It's a great little trick to actually resolve this.
So here is the beginning of the major aside. I'm actually going to remove this, and we'll come back to that in a second. Remember, the whole point is to find the area of the petal. Now we're doing this little aside  this little fact about trig stuff. Now the first trig fact I want to tell you about is the following: Suppose that I have a double angle. Suppose I'm taking the cos^2^^^^  two times some angle. What does that equal? Well, back in high school, some teacher told us the answer told us the answer and we didn't pay attention at all because we'd never use that again in our lives. Well, this is a point in your life where you're going to use it. I'm going to remind you what that formula is. That formula actually equals cos^2^^^^  sin^2^^^^. Now why is this at all interesting? Well, it's at all interesting because remember, our major problem, which is part of the major aside, was to deal with the cosine squared. So I'd like to actually take that two and somehow get rid of it, so I'd like to find a trig identity that would allow me to get rid of that square. The reason why I'm particularly fond of this fact is because this fact allows us to look at this and see a cosine squared there. Of course, I see a sine squared  it's sort of an expensive price to pay  but then I use the Pythagorean theorem identity, which says that sin^2^^^^ + cos^2^^^^ = 1. So if I solve for sin^2^^^^, I see that equals 1  cos^2^^^^, just by bringing this over to the other side. So I actually can remove that term, and I can see that, in fact, this thing in turn equals cos^2^^^^ minus  and then instead of sin^2, I'm going to put this in: 1  cos^2^^^^. So I just inserted this for that. Actually, I made a mistake. I made it on purpose, and I hope that you actually caught it immediately, and that is that if you're going to subtract all of this, then we have to make sure that we subtract everything. We need parentheses right there to make sure that that negative sign not only hits there but also hits there. So if we now distribute that negative sign, what I see is a cos^2 and a +cos^2, so I have two of them, so I see 2cos^2^^^^, and then I see 1.
So the moral of the story is that cos2 is actually equal to  this is now an identity, it's a fact  2cos^2^^^^  1. So let's now solve for this. If I solve for that, what I would see is  and watch this now, I'm going to do it to you live  I bring this over and it's going to be a +1 here. So I'd see cos2 + 1. I just brought that over. And if I divide now by two, I'll get this. So there is a really important identity that we've just proved, we just derived, and it says that if you ever see a cosine squared of something, you can always replace it by something that has no square in it. So the whole point of this major aside was to allow us to find this really neat, really, really cool substitution; namely, if you have cosine squared of an angle, that equals cosine of two times the angle, plus one, divided by two.
Now, back to our question at hand, we come back to this. You see, now I'm in perfect position to actually apply this substitution; and if you apply this substitution, we can now actually figure out this particular integral. So, in fact, what I'll do is I'll actually do this for you. I'll actually do this for you and show you how to make the substitution in the next lesson. I'll see you then.
Get it Now and Start Learning
Embed this video on your site
Copy and paste the following snippet:
Link to this page
Copy and paste the following snippet:
He's ALWAYS GREAT!!!! MUST WATCH!! THANK YOU!!!! Hopefully people can truly appreciate how rare it is to have the ability to watch one of the best professors utilize all skills of teaching to produce such power packed lessons that stick in the brain. Every detail is thought of in the video presentation. THANKS AGAIN!