Calculus: Area of Region Bounded by Polar Curves 1

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Parametric Equations and Polar Coordinates
Polar Functions and Area
Area of a Region Bounded by Two Polar Curves- Part One Page [1 of 2]
We can figure out the areas of the polar regions that are given by a particular curve, but what about if you wanted to find the area of a polar region that's actually down in between two polar curves? How would that work? Well, let's just think about that together and just see that the way we think about that with respect to just the good, old-fashioned Cartesian coordinates will work exactly the same.
So suppose I have two functions: One that's way out here, and that's the function r = f() - and this is a polar function - and suppose I have another one, and this one is r = g(). And I give you two different 's, so that means I give you two different little pieces like this, and I want you to find the area of the region that's in between that. Suppose I give you two angles - two 's, let's say this is = a and this is = b - and the question is: How do I find the area in between in this red shaded area? Well, really, this is not a big deal at all because all we have to do is think about it. I know how to find the area of this whole wedge under this curve, and I can figure out the area under the curve between this whole wedge under here. So all I've got to do is find the area of the entire thing and subtract the area of this little piece right here. So all I've got to do is find the area in this wedge of this curve and subtract it from the area between this wedge of this curve. That's all I have to do.
So, in fact, the formula is actually pretty straight forward, and it would look like this: So that area, if I call it area a, that area would equal of the far curve - first I find the area of the whole thing, which is the far curve, which in this case is f()^2. Of course, that with the, g() there, gives me the area of that wedge. I have to subtract off the area of this little wedge in here, which is the area with respect to the close curve, which is g()^2e. That's all there is to it.
So, in fact, if you think about this the way I do - which, I don't think you should because it's a little too sick - but if you think of this as the far and you think of this as the near, then, in fact, the formula - and this is the way I would look at the formula because it makes more sense than f and g and q and w and z and you know - but if you think of it this way, then you just are integrating one-half, and you're integrating from a to b, and you're taking the far one first - the one that's farthest from the origin - squared, minus the closest one, squared, d. And here's the formula. This is a more fancy way of saying it, and this is a more intuitive way. Just first find this big thing, and then subtract off the little ones.
So, let's actually see an example of this in action. The function that I want to look at are these two - these are polar functions, now - r = 2 and r = 1+2 cos. What I want to do is find the area bounded by these two curves. So I want to see sort of where they fit together, and then find the area that's bounded in between these two curves. Well, what do they look like? Well, in fact, you can see the really pretty graph. This is r = 2; it is just a circle. And r = 1 + 2 cos is just a limacon, and you can see that they actually do intersect. There are these two points in this intersection where they sort of share the same point. Then there is this sort of crescent moon-shaped region that's bounded by these two curves. What I want to do now is figure out what the area of that crescent moon-shaped region is. So that's the question. And you can see that it is bounded between two curves. So what I'm going to do is just use this idea that for finding the area of that, I just take one-half the integral from the low to the high , and I take the far value function - which in this case is the red one, squared - minus the near function, which is the function closest to the origin - closest to the pole - and I square that. Then I integrate, and that should be the answer.
So let's set up the integral and see what this looks like. The answer to the question to me that is not so clear right now is where are these points of intersection? I'm going to start down here, and then I'm going to integrate all the way up to this particular angle. So I have to figure out two angles. I have to figure out this angle - this pitch - and then scan all the way up to that angle - or that pitch. So how do I do that? Well, what I want to do is I want to find out those values on these two curves for which the r is the same - for which they are the same distance away from the origin. So I set the r's equal to each other and solve.
So let's actually do that. If we do that - setting the r's equal - we see the following: We see that 2 = 1 + 2cos. So when I bring this over, I see that 1 = 2cos. If I divide both sides by two, I see = cos. Now I've got to think to myself, "Okay, now what are the 's which make this a half?" If you look at the picture just one more time, we'll see that if I'm going to start way down here and then go up, this is going to be thinking of this as this negative angle. So this is some negative angle, and then we sort of migrate up to zero radians and then keep going up to this positive angle. So I actually want in some sense the first negative angle for which the cosine is one-half, and then I'm going to go all the way up to the first positive angle for which the cosine is one-half.
You can do this a lot of different ways. In fact, some of you could probably just look at it and tell what the answer is and so forth. I always look at a picture because it makes it so easy and you can never make a mistake when you look at a picture. That's the great thing about these pictures. They constantly prove.
All right, so if we just grab the cosine function a little bit, and we look for one-half, now where is the half? Well, cosine equals one-half right here - here's one-half. So where's the first negative place? The first negative place I hit is right here. So there's the negative angle. And the first positive place I hit is right here. And what is this? Well, when is cos = in sort of the small angle between zero and ? Well, that's at . So by symmetry, this must be -. So those are the angles. So this here must be -, then we go and sweep out to zero, and keep going up to . Okay, those are the points of integration...those are the points of integration. = - to .
Okay, let's set up our integral now. I'm going to be looking for far minus near. So with do I do? The integral will be , and then I take the far function - the function that's furthest away from the origin, and that's the red function - so I take 1 + 2cos, and I square that, and then I subtract the r that's close by. And the r that's closer to the origin is for the graph r = 2. So I take r = 2 and I square that, and I just see four. So I see four would be there. So that's the integral. The value of that integral represents the area of this crescent-shaped region.
Now we're in the position to evaluate that integral. I say "just," but actually that's a little bit of work right now because what do we see? Well, this whole thing needs to be squared out, so let's square that out and see what we get. This would equal , and if I square that, I've got to foil that out now; and if you foil that out, we'll see a 1^2, which is 1, and then the inside and the outside terms combine to get 4cos, and then the last times the last is +4cos^2^^^^ - and don't forget the -4 d.
Well, if you look at this, there are some sort of happy parts and some sort of unhappy parts. In particular, the one and the minus four, they combine to get negative three, and integrating cosine is not that bad, but it's this person here that's going to be sort of a problem child. So let's actually break this up into two pieces so we can sort of look at them separately. I'll write the happy piece first.
So 1 - 4 = -3 +...and then I have... 4cos. That's the integral that actually is not too bad to do. Then plus the other integral - don't forget the out in front and the points of integration, - to . Then I've got this particular person - 4cos^2^^^^ d. Now this integral is actually not that hard to do. We can integrate that almost by looking at it. In fact, now actually try this. In fact, I want you to really try this right now. So why don't you try to evaluate this integral? This one's actually not too difficult to do. Integrating a constant is not so bad, integrating a cosine is not so bad. Then what I think you really should try to do is practice doing this more exotic integral. Let me give you two little pieces of hint. The first hint is that this is just a constant, so you can pull that out in front. That's no big deal. Then you have a cos2, and there you should use this really cool fact - this really cool substitution - that you can replace a square by this kind of thing - so replace the cos^2^^^^ by cos2 + 1, all divided by 2. Then try to actually evaluate the integral. It's great practice for integration techniques in general. Then together we'll do it at the next lecture. I'll see you there.