Calculus: Area of Region Bounded by Polar Curves 2

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Parametric Equations and Polar Coordinates
Polar Functions and Area
Area of a Region Bounded by Two Polar Curves- Part Two Page [1 of 3]
Okay, well how did you make out? Let's try it together and see if we can get an answer, and hopefully it will be the right one. What I want to do is why don't we evaluate this integral, which I think is sort of the easy part of it, and then we'll do this integral, which is the more challenging one. So let's do the easy part first since it's easier. In fact, let me just do it on a clean sheet of paper, and I'll bring this back.
So, what we've got here is the integral with the out in front - - and we're looking at (-3+4cos)d, that's what's over there. So, this is pretty easy to evaluate. I've got the out in front, and then the integral of -3 is just -3+4, and the integral of cosine is just sin. So I want to evaluate this now, when ranges from - to . So what does this equal? Well, I have a out in front still. Then if I plug in the , when I put that right in here, notice the 3 x , the three's cancel, so I'm just left with -, and then plus 4sin(). Then I subtract from that , and I'm going to plug in this other endpoint. So I see now -3 x - - so the negative three's cancel - I'm just left with . And then here I'm left with +4sin(-).
So now what's going on here? Well, let's take a look at this very carefully. By the way, notice that I put the parentheses here because I have to remember to subtract everything. So here I see -, and then here I see if I distribute that , I just see a two - four divided by two is two - and then sin(). Well, what's sin()? Well, that's the . So that's that piece, now what about here? Well, here I see another -, so there's that piece. And then what do I get here? Here I see - don't forget to distribute that minus sign, so that makes this a minus - and then here I see a two - 4 divided by 2 is 2 - and what's sin(-)? Well, the sine function is odd, so, in fact, this is going to be -. The reference angle or however you think about it, make it a -. So what does this equal? Well, here I've got - and another -, so that's -2(), which is just -. And what do I have here? Here I've got a lot of cancellation. I'm just left with . Then here I get more cancellation, and a negative times a negative produces a positive, so I have another , so that's +2.
So the value of the easy part of our integral turns out to be this. Let's actually record that down here before we forget it. So this equals - + 2. That's just this little piece right here. Now I've got to write a plus, and I'll fill in this integral here.
All right, so let's do that integral. That integral is the one that requires a little more work. You think, "Gee, that was a lot of work right there, how could it be more work?" Well, watch! Even more work now because now I'm looking at the following integral: , and I'm integrating 4cos^2^^^^ d. All right, now how do we integrate this? Well, the four I'm going to pull out. This is a constant multiple, and the four divided by two just gives me a two. So this is just the 2. To integrate cos^2^^^^, I use this really, really cool substitution, but I can replace cos^2^^^^ by . So I'm actually going to use this great substitution - this identity - right now. If I do that, what I see is 2, and then I see d. Now that's actually going to be a little easier to do because, in fact, if you notice, first of all if I bring this constant - this two downstairs - down here, they cancel - the constant multiple - so in fact really I'm just left with the following: of just cos2+1. And what's the integral of cos2? Well, it's going to be some kind of sine - sin2 - because the derivative of sin2 is cos2, but when I take the derivative of the inside, I get an extra factor of two out in front. So since there's no factor of two here, I need to compensate for that right now by dividing this by two. So let's just check and see. If I take the derivative of this, I see the derivative of sin(blop) is cos(blop), and the derivative of blop is two, and that two cancels with this constant two, and so I just see this. A little application of the u-substitution technique. Then what's the integral of one? Well, that's just .
Now I evaluate this from - to . So what do I see? When I plug in in here, I see - plugging that in for - and then I've got to subtract from that the quantity I get when I plug this in. So that's going to be , so that's going to be a -.
Okay, now what does that equal? Well, let me just do that here as a sort of a little side calculation. So I'm picking up the action right here, right to here, and what does that equal? What it equals is - and what does that equal? We've got this whole thing here, let's take a look at this. First of all, what's ? Well, that remains, so I have out in front, , and then I have a + still. Then what about this? Well, again, the function's odd, so this is going to be -, and I have this over two, but don't forget I'm subtracting all of this, so I have to subtract, so minus - a minus is a plus . Then don't forget I have a minus - it's hard to see it here, but if you sort of visualize putting it right here - I have to distribute it, making that a plus, so I see a plus another . So what this equals is this is going to be , so that's just going to be two square roots of three over four, which is , plus two copies of . Whew! So that's what this integral equals. That was the hard integral.
Now if we come back and put everything together, we should be in pretty good shape because this little piece right here we just discovered equals . So, if we add all of this up, what do we see? Well, we can combine these guys together. So here I've got , and I want to subtract. So if I subtract , then I'm left with -. This is , and then I add two of them, so I just get a minus. Then here if I combine these two things, here I see a two, so to make this over two off of four - - and +, so this is going to be +. So if you want to write that piece first, you could write this as , and then -. Whew! That number represents this integral, and this integral - you might have forgotten from about a thousand lectures ago - actually represents the area of this crescent moon. The area of this crescent moon turns out to be exactly .
Well, I hope you got the same answer I did, and I hope we're both correct. I'll see you at the next lecture.