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Calculus: Finding the Area of a Polar Region: Pt 2

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About this Lesson

  • Type: Video Tutorial
  • Length: 9:17
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 99 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 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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Parametric Equations and Polar Coordinates
Polar Functions and Area
Finding the Area of a Polar Region- Part Two Page [1 of 2]
When we last left polar coordinates, we were trying to figure out the area of a rose petal. The way we were going to do that was we wanted to find the area of this petal right here. How would we do that? Well, we set up an integral. So, in fact, let me now bring us back to finding the area of a rose petal, and the way we set it up was we set up an integral , and we start with - and scan all the way up to ( ), and the function itself squared, d. Now in fact, when we last left our last episode, you may notice I made a little thing here, I forgot to carry down the , so that's sort of a sad mistake. Let me put that in right now. I'll use a different color just to show you that it's happening live. Now this integral represents that area of that rose petal, and the question is: How do you evaluate that integral? Well, that involves a major aside, and, in fact, we did a little trig identity, and we found this really, really substitution. That makes sense - if you're doing polar coordinates, you should have cool substitutions.
Here you go. The cool substitution is - and it follows from the double-angle formula - that if you ever see a cosine squared of anything, it just equals the cosine of two times the anything, plus one, divided by two. So the problem with this integral is that that squared is there, and who knows how to integrate cosine squared of anything? It's a hard integral. But this conversion will allow us to remove the square, and then we have some extra stuff to deal with. But it actually will simplify the question and just change it into an equivalent but doable question.
So now I'll use this substitution right here, so I have the - let's not forget that - and now I have this integral, - to . Now in place of cosine squared of junk, I'm going to now use this identity and replace it by cosine of two times the junk - cos(2junk) - plus . So I have cos(2junk). Now the junk itself is 3, so two times that is 6, plus one, all divided by two - . So I just applied this really, really cool substitution, and I've done nothing else. So these two things are identical because these are identical. Now this is a cool substitution. And why is it really cool? It's really cool because now I have no more squares. Look, Ma, no squares! All I've got now is this. Now this maybe looks like a harder integral to tackle, but actually it's not because I can break it up into a whole bunch of pieces. I'm going to do that for you now.
For example, notice that this is just , right? This is right in front of those things, so I can actually take that two downstairs and pull it all the way out and have it join this two, and that would be a because this is just a multiple in front of all of that. So I can clean this up a little teeny bit, and write the following.
Let me just, in fact, look at the integral itself. I want to evaluate it now - [cos(6) +1] d. If I just put that in front - in fact, if you really want to see it happening live, you can sort of visualize an invisible one in front - and then you can really see the out in front multiplying, so I can just now push it out and it makes this a . So this is the integral that I want to evaluate. Notice that I did not write an equal sign here because I removed the points of integration. Let me just find an anti-derivative, and then at the end of the day, I'll plug in my points of integration and figure out what this value actually is. So right now, integrate this.
Well, I have two pieces I'm adding together, so I can integrate each individually. So I'll probably do this the a-la-carte method. The first thing I'll have is , then I have this now indefinite integral of [cos(6) + 1] d. Now I'll just write that, and there will be two little pieces I'll look at. I'll look at this piece, and then I'll add to it the other piece because the integral of a sum is the sum of the integral. Notice how I wrote the everywhere because that is multiplying everything. So I'll just write this up in two integrals, and I've got to distribute the way over to here too. Now this integral is really easy. What's the anti-derivative of one with respect to ? It's just because the derivative of is one. So, in fact, that part is actually pretty easy. Let me actually do that here - I'll write . What about this? Well, this requires a little teeny bit of substitution, or we can just think about it for a second. I've got cosine of something, what's a function whose derivative is cos(something)? Well, the answer is sin(something). So let me put the in front, and I'll put sin(something). But if you take the derivative of this, you don't get quite the right answer because the derivative of this involves a little chain rule. We see cos(6), and then we've got to multiply it by the derivative of the inside, which would be a six. So this is actually six times too much. So to actually make it the right amount, I should divide this by six. So when I take the derivative, that's six, and the six down in the denominator will cancel, and I'll just be left with the cosine. This is an example of the u to u substitution technique. So, in fact, what I need in front here is a 1/6.
If you take the derivative of that now, let's see what happens. That's a constant. The derivative of sin(blah) is cos(blah), times the derivative of the inside. The derivative of the inside is six; six divided by six is one. So, in fact, I just get this. So that, indeed, is the right answer. I'm adding these two things together, and so there we have this is an anti-derivative for this integral.
Great! So what is that? Well, that equals sin(6)+ . So there's an anti-derivative. Now remember, I've got to now plug in my end points. Let's see if I can bring this back into the picture for you. So now you see that this integral equals this, and now what I want to evaluate this at the points of integration. So let's do that. So if I evaluate this from - to , what does that equal? Well, first I plug in wherever I see a and see what I get. So let's do that. I'll carry this up right to here. So if I plug in , I see sin(6), so that's just , so sin(), and then I add , which is . Then I subtract from that - notice the parentheses - what I get when I plug in -. So I get - here, I see sin(-), and then when I plug in - here, I see a negative - so -1/4. That's what I'm looking like, so it looks like that. I've got to distribute that negative sign. Now what is sin()? sin() is zero, so, in fact, this is just zero. And sin(-) is zero, so that makes things a lot easier. All I'm left with here is , which is , so this equals . Then I have a minus, and then if you distribute, you see that I have that minus with this minus, and it makes another plus, and again we see . So what I see here is + is 2(), which is . So, in fact, this equals the number .
So, what does that represent? It's such a long time ago, you probably forgot. The whole question was to find the area of this rose petal. Now what we're seeing is the area of this rose petal can be expressed as this integral right here, and to evaluate that integral, we use this really cool substitution, and after the cool substitution, we came to here; and then to evaluate that, we just had to do a whole bunch of little integration stuff. And when we evaluate it at the very end, we see that the area of that red shaded region - that red rose - turns out to be .
Congratulations! That was sort of a neat integral that involves a whole bunch of little steps, and you can see that this very, very delicate curve - that delicate petal on this rose curve - the area of that can be computed, and actually it's sort of an interesting number, it's just . You might have thought it would be really complicated with a lot of cosines and sines and maybe square roots and things, but it turns out it's very elegant. Nature always is elegant. I'll see you at the next lecture.

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