7.4 Area and Arc Length in Polar Coordinates

[STARTS AT 10:54]

Instructor: All right, next we have arc length of a curve, defined by a polar function. And this is actually based off of the arc length from a parametric perspective, that we saw previously. So there is our-- I mean, there's the equation we're going to use. We're going to have effectively two functions. r, and d r d theta. So we're going to treat it like we have two parametric functions in a sort. All right. 

So the arc length of a cardioid r equals 2 plus 2 cosine theta, that we just saw. Well first, we want to say r is, well that's sort of redundant. We have r, how about d r d theta. That is going to be 2, negative 2 sine theta. Now to find our interval, alpha to beta here. Well, if we deal with the function, r equals 2 plus 2 cosine theta, if we evaluate this at theta of 0. This is the point 0, 4. 

Now I'm not going to do this here. I suggest you do this. If you make a table, make a table. You find that this starts to return back where it started, at 2 pi. And that's really just because cosine is cyclic in that way. But if we looked at a table, we could see that it actually starts repeating itself at that point. So we have a starting point of 0 in terms of theta. And we have an ending point of 2 pi, which just makes sense when we look at a cardioid. If you look at that original graph that I showed you from the textbook, it's not a circle. But it has some of the same properties. So it just makes sense if we are a bounded between 0 to 2 pi that might be our ending and beginning points. 

So our length is the integral from 0 to 2 pi of the square root of r squared 2 plus 2 cosine theta squared plus d r d theta squared. So this would be negative 2 sine theta squared d theta. Well if we multiply that first argument here out, that would be 4 plus 8 cosine theta plus 4 cosine squared theta. Multiplying the second out, that'd be plus 4 sine squared theta. And of course, we have that all under that radical. 

Now this actually simplifies quite well, because sine squared plus cosine squared equals one. That goes back to those original equations we started with a section ago. So this part right here is, rather than being equal to 1 is equal to 4, if you factored the four out. So this is going to be the integral from 0 to 2 pi square root of 8 plus 8 cosine theta d theta. All right. Factoring a 2 out. Actually no, let's do some trigonometry on the side, here. 

We know of a way to relate square roots, or squares, with single powers. An exponent of 1, with our cosine and sine. So let's go ahead and take this formula right here. Cosine squared theta is 1/2 plus 1/2 cosine 2 theta. 2 theta. All right. Well if we rearrange this, actually let's multiply both sides by 16. Multiply both sides by 16, That would be 8 plus 8 cosine 2 theta. Not quite what I want. 

So how I'm going to write this, then, is 16 cosine squared theta over 2 equals 8 plus 8 cosine theta. And I'll just divide theta in half, both places. Now these are not-- these are equivalent forms of the same equation, but really I'm just manipulating to see if I can get something that works. OK? So these are in effect, three different equations. Well actually, the first two were perfectly fine. Those two were equal. However, this one I've manipulated that just a little bit. I'd probably need to change a variable, really, there. 

So what I can then go back over here and do, is this is equal to 0 to 2 pi the square root of 16 cosine squared theta over 2, which will be equal to 4. And this is the integral of 0 to 2 pi of cosine of theta over 2. OK. 

Now that would be sine theta over 2 times 2 because we're dividing by 1/2 there. Leave our 4 evaluate it at 0 and 2 pi. OK. I need to make a note way back here, that when I took the square root here, that really should have had absolute value bars, there. Because cosine is an even function. Because it's an even function, we can actually just double from 0 to pi. So let's go ahead and change that real quick. I meant to make note of that. 

So this is actually going to be equal to the interval from 0 to pi. And we're going to double that. Cosine's even. So then, that actually should be an 8 out there. Changing that. All right. So evaluating that at 0 and at pi, that would actually be a 0 area, because it's an even function. But we actually want to know the total length. All right, evaluating that, we should get a value of 16. And that is our arc length.

[ENDS AT 17:57]