 This video is on the calculus of functions that we express with polar coordinates, so we're going to do differentiation and integration. So there will be these two parts. And the first question of each, of course, is to derive equations for the derivative and the integral of these functions. Remember the homework set that's available on GitHub as a PDF file and also the set of notes. Just go to the link in the description down below and you'll find those GitHub files so you can try these problems on your own first. And of course there's also going to be a Wolfram language file, just a notebook, so that you can code along if you wish to do so. As mentioned, we're now dealing with the calculus of functions that are given in polar coordinates. And to understand these kind of problems first we have to just have an equation or a ways to take the derivative of a polar coordinate function. So in problem one here is then derive an equation for the first derivative of a polar coordinate function. And what we need to do to solve this little problem is remember three things. In polar coordinates we have the fact that x equals r times the cosine of the angle and y equals r times the sine of that angle with the positive x axis. And we also just have to know that if we're taking the derivative dy dx because we still have this univariate function f of x or y equals f of x, is that we can take this derivative dy with respect to d theta and if we divide that by dx d theta what we're going to have then is just dy dx. And I suppose you can see that that we have in the denominator of the numerator and in the denominator of the denominator we have d theta which we can cancel out. So let's have a look at what these things are. And let's just grab a pen name. So if we take dx d theta, let's start with dy d theta then. And what we have here is the fact, I suppose we can put that in green as well, that r is some function of theta. That r is some function of theta. So if we have this now, x equals r times the cosine of theta, what we actually have there is some function of theta cosine theta and we have some function of theta sine theta. Just get a darker gray there. So if we take the derivative of dy with respect to theta, y with respect to theta, so dy d theta. We have here the product rule that we have to apply and remember that is going to be u prime v plus u v prime. If one of the two is u and one is v, so let's make that u and let's make that v. So we're going to have f prime of theta and then very simply the sine of theta. And we add to that f of theta and now we have to take the derivative of v which is sine theta and that's cosine of theta. And if we write dx d theta, that's going to be f prime of theta, cosine of theta. And then it's going to be a negative because we have f of theta but the first derivative of the cosine of theta with respect to theta is minus the sine of theta. So that's why we have the negative there. And if we divide these two by each other, that is going to be our derivative. So let's write it out there so that we remember it. The y dx of this polar coordinate function that is going to be f prime of theta, sine of theta plus the f of theta, cosine of theta. And we divide that all by the f prime of theta, cosine of theta minus the f of theta, sine of theta. So it's actually quite an easy derivation and of course this last definition here, this function dy dx is the one that you can commit to memory. But as you can see, even if you can't remember, it's quite an easy derivation. So problem two, 1.2 in the section on the derivatives of polar coordinate functions. We also calculate the slope of the polar coordinate function r equals the sine of three times theta. And you can see here, using the Wolfram language, we've got the polar plot function. We pass the function itself, sine of three times theta. Remember you have to put a little space between the three and the theta. And I get that little theta symbol by just hitting the escape key, typing th and hitting escape again. And then comma and we want theta to go from zero to two pi. And you can see the three little leaves there, which is this polar coordinate function. So let's get the first derivative of this function. And in the end, remember, what we're expressing here is a function y equals the f of x. And we've just converted those to polar form. And that's how we get r equals sine of three times theta. And if you think about it, it's actually quite wonderful that any point on this curve will have that the radius is going to be the sine of three times the angle of that, the vector that points towards that point. So it's actually quite, quite fascinating. So let's remember, let's go for green. We have the fact that dy dx, that is going to equal f prime of theta. And remember y equals r times the sine of theta. So that'll be the sine of theta. Plus we have f of theta. And then we have the cosine of the cosine of theta. And let me just get my ruler and we divide that by the f prime of theta. And we have the cosine of theta minus the f of theta times the sine of theta. So let's get those going. So dy dx. Remembering now, let's just put that there. We have the fact that x equals, we have sine of theta. And remember x is the f, is r times the cosine of theta. So that's sine of, we should say three theta times the cosine of theta. And we have the fact that y equals then the sine of three theta times the sine of theta. So those are the two functions that we have to get the derivatives off. So let's get the first derivative of y with respect to theta. So we're going to treat this as u and we're going to treat that as v. So it's u prime. So that'll be three times the cosine of three theta. And as much as we have to use the chain rule, the sine of an angle, its first derivative is the cosine of that angle. The cosine of the angle, which is three theta. And then of the inside function, the three theta, we have to take that derivative as well, which is going to be three. And then we still have to have the v as well. So we still have the sine theta there. And then plus we have to have the f of theta, which is this r. That's the sine of three theta. And then we'd have to take the derivative of cosine of theta, because remember that is going to be u and that's going to be v. So we've left u in place and we just have to take the derivative of v. So that's going to be the cosine of theta. And we divide that by, let's do the same for x. So this is going to be our u. This is going to be our v. And remember what we're after is u prime v plus u v prime. That's what we're after. So let's get u prime. So that's the first derivative of the sine of three theta. That's going to be three times the cosine of three theta. Three times the cosine of three theta. And then we leave v. We leave that alone. That's just the cosine of theta. And we subtract from that. We have just the u that we have to take now. That's the sine of three theta. And then the first derivative of the cosine of theta. theta, remember that's the sine of theta and we have the negative out front. And that is the first derivative of this function, this polar coordinate function r equals the sine of 3 theta. Problem 1.3, find the coordinates where the slope of the polar coordinate function r equals sine 3 theta is horizontal. So we've just had that, let's have a look what it looks like using the Warframe language. There we go. There's our function, the curve of our function and where do we think that we're going to have that the slope, the slope remember, this is the first derivative. And I think we can identify three spots, I hope you can see there. I mean, as it curves around here, we're going to have a slope that's zero. If we curve there, that we can have a slope of zero. Yet the bottom where x equals zero, we're going to have this slope of zero and right down here we're going to have a slope of zero. So we should have, actually have, have these four spots where the slope is going to be zero. Let's have a look at that. As always, and this is probably not one that you, you know, you have to memorize because it was so easy to remember in problem 1.1 to work this out. But we're going to have the fact that dy dx of these functions that is going to be f prime of theta times the sine of theta plus the f of theta cosine of theta. And if I grab my ruler there, there we go. That's going to be over the f prime of theta cosine of theta minus f of theta sine of theta. So that's what we have. And if we remember our previous problem, problem 1.2, we had the fact that dy dx. And let's just think about what that was. Or we can just quickly do it. It's not a problem. So that's going to be three times the cosine of three theta sine of theta. And we're going to add to that the sine of three theta cosine of theta. So all I'm doing here is I'm just using, I'm just using the product rule. And we divide that over dx d theta. And that's going to be three times the cosine of three theta cosine of theta. We subtract from that the sine of three times theta, three times theta times the sine of theta. So we need, we need this to be zero. And the way that we do that is to have this equals zero over not zero. Let me write it down like that. We can't have the denominator being zero. We can't divide by zero. And if zero over zero, we have indeterminate form, etc. So what we need is we need for a fine values for theta for which the numerator is zero. So what we really need here is three times the cosine of three theta sine of theta plus the sine of three times the angle cosine of the angle that has got to equal zero. If you sort of got to solve that little problem, let's have a look at the Wolfram language. And we see problem 1.3 there. Let's increase the size a little bit. So it's not too difficult to see. We can see what we've done there. So what I've plotted is the numerator and denominator. And we can see that they are not zero, you know, the spots where they are zero do not coincide. So actually quite safe then. It's only this blue line, the numerator, where it is zero. We can see it's zero there. It's zero there. It's zero there. It's zero there. It's zero there. And of course, we're going to have this periodicity. And we just got to see that we don't go over two pi because then we repeat, you know, we go round and round the circle. So we kind of need those first four spots where that is zero. And I'll just quickly write that down because you'll actually need, you know, just your calculator or of course the Wolfram language just to do that. And what we're going to have, we're going to have theta, at least for those spots, is going to be about 0.659. It's going to happen at pi over two. It's going to happen at 2.48. And it's going to happen at pi. So those are the four. And then, of course, we can work out, we can work out what R is going to be because remember, R is just some function of theta. And in this instance, this problem that we have, it's the sign of three theta. So then it becomes easy to work out the three values for R. And you can just plug that in sine times the three of that sine of three pi over two, the sine of three times this bit or the sine of three pi. And you can work all of those out. And then you'll have these values R comma theta. And what you'll have to do then is just convert that to X and Y. And that's easy enough because remember X equals, I should do this in green, because these are the things we always have to remember. We have that X equals R times the cosine of theta and Y equals R times the sine of theta. And if you do that, check on the Wolfram language, see that you get the same results as I do. So we're going to get these four values. So it's 0.726 thereabouts by 0.563. That's the one. The second one we're going to get is 0 comma minus one. The third one we're going to get is negative 0.722 and 0.562. And the last one we're going to get is 0 comma zero. And that's sort of from the graph that we expect these things to be. So in this problem, the point is not to do these calculations, the simple numerical anyone can press a button on a calculator to do this. This is the crucial part here where we're concerned about how do we get this to be 0. And we have to have that the numerator equals 0 and the denominator is not. And hopefully in the exams you'll be given some sort of clue as to work out values for theta or just be able to use the Wolfram language. And at least when you do this on your own, it's easy enough to use the Wolfram language just to see where the numerator is going to be 0 and to note that at those spots the denominator is not 0 at the same time. And you can see using the Wolfram language, we've got those values for the angles which I've just saved. And I may create a little function x coordinate and y coordinate here, which will take a value and that's what it will do as far as solving those, you know, how to get to the x and y coordinates. So we can work out the x coordinate, y coordinate. And if I pass all those four angles to the x coordinate and the y coordinate, we get these, you know, the two values for each as I've shown in the results. Problem 1.4, we want to find a coordinate where the absolute value of r, the length of r of the polar coordinate function is at a maximum and the polar coordinate function we're referring to is still this r equals sine of 3 theta. And if we look at the Wolfram language here in this first plot we did for problem 1.2 and 1.3, is that the maximum radius is going to be right out here. So from the origin right out there, from the origin right out there and from the origin right out here. So you can work out any one or three of those. It's just once the problem is asked for one of them. So how would we go, you know, how would we go about that just this maximum distance that we are there? Well, there's a few things that we can remember. And so first of all, let's just have as far as the problem is concerned, we have r equals the sine of three times this angle. Now the sine function, this is what it is, the sine of some angle equals r. The sine, the maximum value that it can go, remember, is from negative one to one. So one would be the maximum. So what we're actually going to have here is that the sine of three times theta, the maximum value that it can take is one. We just need to find out what those angles are, you know, what what angles this will be. And if we if we wish to do some work on the side, just take some other function, some other angle, say the sine of alpha, the sine of alpha, alpha equals one, that means alpha equals the arc sine of one. And you know, at what angles is this, you know, is this going to happen? We can write down a couple of these angles. It's going to be at pi over two. And then it's going to have this plus two times or two times n times pi, where n is just this natural number. So we're going to have five pi over two. So they're going to happen there, it's going to happen at nine pi over two. And it's just going to carry on like that. So let's choose alpha equals pi over two. We're just going to choose one of those. But that, remember, equals three times theta. We had alpha there and we have three there. Those are equal to each other. In other words, theta is going to be pi over six. So let's just take theta or theta over six. So what we have here is the fact that we have this r comma theta. So r is going to be at the maximum value that sine can take that is going to be one comma. And we're going to have pi over six. But we remember x we need x and y coordinates that is r times the cosine of theta should do this in green. And anyway, that's going to be one times the cosine of pi over six. And y is going to be the r times the sine of theta. So that's going to be one times the sine of pi over six. Can you remember that little how to draw that little triangle? It's always easy just to draw this over and over again. So that sticks in your mind as well. So if you do that, this is the angle we're concerned about. That's the pi over six angle right there. So that's going to be one, two, square root of three. So if we think about what x comma y is going to be. So what is the cosine of pi over six that is adjacent divided by hypotenuse. So that's going to be the square root of three divided by two. And as far as a half is concerned for this angle up here, opposite divided by hypotenuse, that's going to be a half. So that is going to be one of the coordinates where we're going to find that this value r here, that is at its maximum. And if you take the second one, you can have the fact that alpha equals five pi over two. And that is just going to equal three theta. And here we're going to have theta equals five pi over six. And you're going to have r equals one for that point as well. So you can work out what x and y is going to be there. And so you would carry on for those three values. You just have one more to do. And that's going to be nine pi over six. That you just have to simplify as far as that's concerned. And you can work out the values, the third value. And from the graph, it's easy enough to see that it's going to be at x equals zero. And let's bring up that graph. So it's going to be this one down here. And that's going to be x equals zero and y equals negative one. That's going to be the last angle. So we worked out the one right on top there. Problem 1.5, find the slope of the tangent line to the polar coordinate curve. We're still stuck with this curve because there's so much you can ask about these things. r equals sine of three theta, where r is at a maximum. So we've done that now. We know where one of those points are. And we also remember from the previous problem, so please check out the previous problem, that we have the fact that r comma theta at one of these points, the one that we worked out at least, was going to be one and pi over six. So we know that. And we know what dy dx is. We worked out dy dx in a previous problem. So let's just write that out that was three times the cosine of three theta sine of theta. And we added to that the sine of three theta times the cosine of theta. And that was divided by three times the cosine of three theta, cosine of theta. Whoa, look at that. I want to make an over theta cosine of theta minus the sine of three theta sine of theta. So there we go. And now all we have to do, that's the slope remember, and we have a point where we have theta equals, so theta equals a pi over six. So we have to get dy dx at this, this theta equals pi over six. So all we have to do is plug in that. And if you plug that in and use the Wolfram language or your calculator, you're going to get negative five square root of three over two. And that makes, that makes sense. That's going to be negative slope. So I just want to show you the other two. We were also going to have the fact that we have dy dx. And that next angle we had was, remember that was five pi over six from the previous problem. And if you plug those in, that is just going to be five times three over five times the square root of three over two. And that's just from symmetry there. And then the last time we're going to have dy dx that is going to be at theta equals three pi over two, three pi over two. And there we're going to have zero. So if we just look at the curve again, that's going to be this one at the bottom. And then we can clearly see the slope there is going to be zero. And the slope right up here is going to be the negative five square root of three over two. And yeah, positive five times the square root of three over two. So problem 1.6. We've taken sort of looked at all the different questions that you can get. And this is the last type of one that you can get. You're just going to use a different function here, polar coordinate curve, r equals two times. And then one plus cosine of theta. And we'll have a look at what it looks like. Or if you can't wait to just use the Wolfram language and do that. If you don't have that available during the exam, I mean, just take a couple of angles of theta there. And you don't have to do too many until you can see what's going to happen to the curve. So remember, we're going to have an extrema where the derivative is zero, unless we have this inflection point. But anyway, we we're going to have that. And the question already asks us for a maximum value. So we kind of think that it's going to go up to a maximum point and then come down. So at that point, you know, we're going to get a derivative that is equal to zero. So what we're going to what we're actually looking in for is where dy dx equals zero. And we remember the fact that let's do that in green. I haven't used green properly. So remember that is r times the sine of theta and x equals r times the cosine of theta. We remember that. So in the problem that we've been given here, as we're going to have r is that so that's two times. So this is put that that's two plus two times the cosine of theta. And we have to multiply that by the sine of theta. And here we're going to have two plus two times the cosine of theta times the cosine of theta. So that is what we have for for x and for y. And now what we have to do is we have to to take the derivative dy dx. And to do that, at least let's try to do that. Yeah, let's take the y d theta first. And we have we have this now. And what you can do is of course, you can just simplify this or just at least multiply it out. So I'm going to have two times the sine of theta. And we're going to have plus two times the sine of theta cosine of theta. So that's easy enough to simplify. And that's suppose let's do it here. The fact not dy d theta. That's just going to be why it's going to simplify down here. We have two times the sine of theta. And you remember this trigonometric identity, two times the sine of an angle times the cosine of the angle. Well, that is just the sine of twice that angle. So that's quite simple. And let's just multiply this one out. That's going to be two times the cosine of theta. And what we have here is plus two times cosine squared of theta. Now, that is a bit that is a bit of a difficult one. That's, we just have to sort of watch out for that one, not specifically now, but when we do integration, to get the integral of this, that's a bit difficult. So we have x here, at least as twice the cosine of theta plus twice times the cosine squared of theta. So we just have to watch out for those. So let's just, let me just take the eraser. There we go. Let's do it down here. So we need dy d theta. So let's have a look at what what dy d theta is. So that's going to be twice the cosine of that angle. And then plus two times the cosine of twice that angle. So that's simple, we're just using the chain rule there. And then that's going to be x, we're going to have negative two times the sine of theta. And then we're going to bring that to four. So that's going to be a four. So let's let's do that. That's going to be four times cosine of theta. But then also we have to do cosine of theta, which is negative the sine of negative the sine of theta. And what we need here, well, that is over dx d theta. Let's not forget that. So again, to do this, we remember we need this to be zero in the numerator and not zero, not zero in the denominator. So let's have a look at the Wolfram language again, see where this is going to happen. So there's problem 1.6. And we can see the two times one plus the cosine of theta. And we see this codaway pattern here. And we want to know where this is going to have a maximum. So that's going to be way up there. That's going to have a maximum. And what we've done here is I've just solved the numerator there or take the derivative there. So we see that we get the correct derivative there in the numerator. And then I'm just going to solve this. So I'm saying solve, use the solve function. I put my expression there and I'm saying where it is equal to zero, but you can't use equal because remember in a computer language, most computer languages equal means assignment. So I don't want to assign this to the value zero. I'm going to set it equal to zero. So I'm using two equal symbols, zero. And I want to comma solve for theta. And it gives me this sort of answer here. So theta is going to be negative pi over three plus two times pi times an integer and pi over three plus two times pi times an integer and pi plus two times pi with an integer. Remember, because this is the cyclic. So what I'm going to do here is just to solve for all of these to see all these numerators that I do get and all these denominators that I do get. And I'm using a union of making all three of these tables. And instead of using all the values of integers, I'm just using zero, one, two, three, four, five and six. So I'm getting the union of all of these. And that gives me this long list of values. And I'm going to do that the same with that with a denominator because we don't want for those to overlap with each other. So for us not to get any overlaps, I'm using the complement function. So numerator, comma, denominator is going to remove from the numerator all those values, the denominator ones. So that we don't have these duplicates. Think about how the complement works in set theory. So we can choose any one of these. So I'm going to choose theta equals three. And then I'm just going to work out the x coordinate and y coordinate. And we remember x equals r times the cosine of theta y equals r times the sine of theta. So I can just plug that in. That's my r sine of theta there. And I'm just using pi over three for my angle and pi over three for my angle. And I get three over two and three square root of three over two. So that is going to be this point here where it is at a maximum. So we're going to have this fact that y, x, comma, y, that's going to be three over two and three square root of three over two. That point there, let's have a look again at the Wolfram language. So that's going to be that point way up there. That's going to be, we can see it's about one and a half there right at the top for our x coordinate. Now y coordinate is going to be right about there, just under three. So three times square root of three divided by two. That very point way up there. But what we have to remember is that we're going to get a horizontal tangent line there. So if we take our function, we get the first derivative and set it to zero. We know that the numerator must be zero. The denominator must not be zero because if I take zero divided by a non-zero value, I'm going to get zero. And that is what we want. And all we have to do now is to find values for theta. And we just have to make sure that if we use that value, it can't also make the denominator zero. If I plug in into my dy dx, if I plug in theta values there, I don't want the denominator to be zero. Up here, we did it visually. We could see that the values were going to be zero, but they don't coincide. So no problems there picking any of the values in the numerator to make that derivative zero. Here, we had to go about it in slightly different way. And you can have a look at the code there. I'm using the union function. I'm using then the union of three different things. So there's my table function there. And then I'm saying comma another table function and comma another table function. So I'm creating these three lists and I want the union of them. So I don't get the repeats. And what I do there is minus pi over three plus two times pi times n. That's exactly what it is then. Negative pi over three plus two times pi times instead of c one, I'm using n because these are integers. And then I'm using the table function. So I've got a cycle over values of n there. And the values of n that I want to cycle over is zero, one, two, three. So that's how we constructed. Have a close look at that. So it's going to create this table and it's going to iterate n over zero, one, two, three. Plug it, plug those n's in there. So I'm going to get this whole bunch of values and I'm going to do that the same for the second one. Pi over three plus two pi n. Pi over three plus two pi n. This is, we have it there. Cycle through all those values of n and do it the last time. Pi plus that. And I just get the union of all of those. So I don't get any repeats. And that's all the possible values that I get that will make my numerator equal to zero. And we got these three values here by using the solve function. When is the numerator, which is the derivative dy d theta? When is that equal to zero? So solve this for me when the numerator is equal to zero and solve it for theta, please. Because sometimes you might have more than one variable in there. So get all these numerator values, or get all these denominator values. But I cannot include in my final selection something that will make the denominator. Some angle of theta that will make the denominator zero as well. We can't have that. And for that I then use the complement function. Numerator minus denominator. So it's going to remove all of those in the denominator. From my list of possible numerator values so that I know the ones that are left is only going to make my numerator zero. Not my denominator. And then we're just going to choose one that sort of makes sense for the problem, which is obviously going to be pi over three. And if we look up there, this angle up there from the origin way up there to where y is going to be a maximum. That's about pi over three. So sort of going to estimate that. And that means, well, I suppose if we take negative pi over three, we're going to land up here where y is a minimum. But we were asked for the maximum. So if we do that, it's easy enough for us just to remember then x equals r times the cosine of theta and y equals r times the sine of theta. Just plug our pi over three and we get the two coordinates. So twice I've run through the problem. I hope you get it. If not, leave a little message in the comment. And maybe I can get another way to explain this problem. But this is kind of a nice problem. If you get this kind of problem in the exams, this is more of a difficult one. And if you can solve that one, you really, this is sort of I would estimate for this level of calculus, this would be a difficult enough problem to show that you really know your work. So we move on to the second part of this lecture. We're just going to look at integration as far as polar coordinate curves are concerned. And what we have to have is an equation at least for integration. So in problem one, always 2.1 here derive an equation for the area of a polar coordinate curve on the interval with some starting angle to some end angle. Where we have this polar coordinate curve r equals some function of the angle. And we want this to be continuous and positive as far as this problem or setting up this problem is concerned. So if you have a look at the top right little figure there, we have a circle there with some radius. And we know what the area is of the whole circle by our squared. But we only have a fraction of that. We don't have the whole 2 pi there. So if we think about the area of this little bit, it's going to be pi r squared, which is the whole but only a fraction of the whole bit. So this is the angle we're dealing with, this is the angle theta. It's going to be theta, this fraction of the whole. So this fraction of the whole and then this would cancel out. So we really left with a half times r squared times theta. So that is going to be this area here, the purple area that we have here. It's some fraction of the whole, which we can see there. Now what we can do, I suppose this here is remember that r is some function of theta. So if we rewrite this, we would say the area there is going to be half times some function of theta, and all of this is squared times theta. That this and this here is exactly the same. So let's think about some curve that we have and that's in blue on the right bottom here. And that curve, we can also see if we take a tiny little fraction of it, that it goes from some angle to the next angle. So you see it going from theta i minus one to theta sub i. So we're just going from the one to the other and then that tiny little area will also be a tiny little fraction of, if this was part here, if this was part of some circle around the origin. So that's also that, that tiny little ai, the area of that tiny little bit. If we just think about the area of that tiny little bit, it is also going to be on this curve. Now if that's part of some bigger, we want from that angle to that angle, the area under the curve between those two, we're going to go from A to B, where we start theta, so theta, oops, that's not a theta. Where we start theta zero, we're going to start that, say for instance, at A and we're going to go all the way to theta, the end theta there, we're going to set that to B. And what we can always do is just to divide that up into equal little angles there. So that would have been the previous one and that's going to be the next one. So we have B minus A divided by N and that's this delta theta. So all these small little angles there are delta theta, delta theta, delta theta. So what we're going to have here is that this is going to be this function squared delta theta because we are just, it's just this tiny little one. And all we're going to do to get to all of them is that we just sum over all of these little i's. Now if we sum over all of them, that is just going to be approximately equal to. So that will be approximately equal to. It's going to be approximately equal to this f of theta, which is this r, remember, squared delta theta. And what we need to do if we sum over all of them is that we just have to do what we always do with this kind of problems. We're just going to let delta theta go to zero or N go to infinity. That I have an infinite number of these. So I have an infinite number of them. That means the limit as N goes to infinity of all of these tiny little. These tiny little angles that we have here, so that we now have this full area. It's going to be this limit and we go from i equals one to N. So we're taking all of these. So that's going to be the f of all these little theta i's. I should probably say they're all these little theta i's and we square all of them delta theta. So in the end that we have this limit, this is going to be the integral and going from a to b of a half times this angle or this function of this angle, which is this r. And remember we're squaring that d theta. So that's how we always set up these problems just starting and going smaller and smaller and smaller. And just noting the similarity between these two setups that we have here. So this little a sub i that we're getting there, that's approximately equal to. And if we make it smaller and smaller and smaller, of course we're going to get there. And we just have to sum of all of them. And I suppose the most important part here is just to think about this purple part here as this fraction so that we take pi r squared. But we multiply that by some fraction of the whole. And if we went all the way around two theta, that angle, that'll be two theta, two pi, I should say over two pi, and that's just one. So the whole thing would be pi r squared. And then we can just cancel out those and we left with this r squared theta. And we remember r is a function of theta when we're talking about polar coordinates. So that we can go from here to here, no problem. And as we divide this up into these delta thetas, we set up this tiny little one. We go to infinity as far as how many of these little things there are. And then we left with the integral. So that's the one we're going to remember. So let's just write that out in green that the area is going to be the definite integral and going from a to b of a half times this function of this angle, this r, and we square that d theta. And we're going to write that down a couple of times as we go through these problems just to sort of cement that. But even if you can't remember that, I think this whole idea of setting up this problem is quite intuitive. So here we are in problem 2.2 with our good old friend, the polar coordinate curve r equals the sine of 3 theta. We know what it looks like. But now we're interested in this area enclosed by this polar coordinate curve and going from an angle of theta being zero to theta being pi over three. So let's have a look at that in the Wolfram language. So there we can see our equation and we're going to do polar plot of sine of three times theta. Remember to put the little space there between the three and theta to indicate that's multiplication. And theta's going to start at zero and it's going to end at some value. And this value, we just want to manipulate this value to go from some star position to end position. So we're starting it here at 0.01. So that means it's starting at zero and initially it goes to 0.01, but then all the way up to three pi. So if we were to do that, we see the curve goes out and that's the first of the three little leaves that we saw there. So that's the area from zero to three pi that we want to know. What is the area under the curve of that? And we've used the integrate function there. We say half times sine of three times theta squared. And we're going from zero to pi over three. And if we use that, integrate that, we see it's pi over 12. So let's see if we can get to pi over 12. So how are we going to set this all up? So first of all, in green, we're going to remember that area, as far as these polar coordinate curves are concerned, that it's going to be the integral and going from a to b of a half times. And we have this function of theta squared d theta. And in this instance, let's see what we have. We have the fact that a equals zero. We have the b equals pi over three. And we have r equals the sine of three times theta. That means that r squared is going to be nine times the sine squared of three times theta. And we'll have to integrate that. So we just have to think about this a little bit. So let's just do this. Let's just do a little bit of side work. Let's choose a different color. So let's just have something like alpha being three theta. So that just makes life easier for us. We're going to forget the nine at the moment. And there's good reason why we should remember nine, because I don't know where that nine came from. Let's get rid of it there. Quite easy to make these little mistakes. So r squared is sine squared of three times theta. That looks a lot better. So what we have now is the sine squared of alpha. And we can rewrite that. There's another way to rewrite that, the sine squared of alpha. Remember, that's going to be one minus cosine squared of alpha. That's an easy one to do, because sine squared of alpha plus cosine squared of alpha, that's going to be equal to one. And sine squared of alpha, well, that's one minus. And there's a trigonometric identity for the cosine squared of an angle. And that is half plus half times the cosine of twice the angle. It's much easier for us to integrate twice an angle than it is to integrate the square of a function. So we have the sine squared of alpha. That's going to be one minus a half minus a half cosine of twice that angle. And sine squared of the angle. That is going to be half minus a half is a half minus a half cosine twice that angle. And now we just have to put in the fact that this angle alpha, that's actually three theta. So the sine squared of three times theta, well, that's going to be a half minus a half cosine of two times alpha, two times three, so that's six theta. So now we have everything really that we need, because the area that we're interested in, let's choose a darker color there. The area, and we've got it there in green, that's going to go from zero to pi over three. Pi over three over half, remember, that's a constant. So we can actually bring it outside of that. And we need, remember the f of theta in this instance for us, that was sine of three theta, so we need that squared. So f of theta all squared, that's the sine squared of three theta that we need, and we've just done that now. So it's half minus a half, so let's just put all of that. That's a half minus a half cosine of six theta d theta. And let's just do all of those, so I can just bring out that first half. So let's bring that out, and then we're going to have the integral and going from zero to pi over three of a half minus a half cosine of six theta. And all of this d theta. So there's a negative sign there, so we can just do this in two bits. So we're going to have the half here. And what we really have here is the integral going from zero to pi over three of a half d theta minus, we have another integral going from zero to pi over three. Pi over three, and we can bring that half out there cosine of six theta d theta. So that's what we have there, so it's going to be a half. Let's do all of that, this half d theta, that integral. So that's going to be a half times theta, and that goes from zero to pi over three minus another half. And now we just have to take this integral of cosine of six theta. So we know the sine of an angle, its derivative is the cosine. So we know that we're going to have the sine of six theta there. And then we'll have to take the derivative of six theta. So this better be a six in the front there. And that also goes from zero to pi over three. So that's just some nice, just remembering how all of these things work. So we still have the area that's a half times. So we're going to have this idea of a half times pi minus three minus zero. So we can just leave that alone. And a half times that, that just be a, just put 12 there. And then what we're going to have is the sine of six pi over three. Six pi over three, which is this two pi. And remember minus zero again, but we needn't worry about that. So a is going to be a half. And here we have pi over six minus. And nothing happens there because the sine of two pi, well, that's the zero. So we have that. And in the end, we have pi over 12, just as low frame language calculated for us, pi over half. It's actually quite a bit of work here and it's nice to remember, you know, some of your, your integral calculus for single variable. Because that's basically all we did here. And you can see this is quite easy as long as you can remember this equation here for taking that, for doing the integral. And of course this is finicky because you, you mustn't forget about this half and you have to carry that half the whole time. And in this instance, at least, very useful trig identity that you have to remember. Maybe we can do a video about how to drive that, but it's a very important derivative to remember. And or trig integral to remember at least because just taking cosine of some multiple of an angle, that's a much easier integral to take. You know, then to, to struggle with a square of some trigonometric function, as I said. And then it's just a matter of, you know, keeping track of this, of this sort of thing here so that you don't, so that you don't miss out on where these things are, you know, are supposed to go. Let's get cracking problem 2.3. We calculate the area enclosed by the cardoid r equals 1 plus the cosine of theta. So let's use the Wolfram language just to have a look, see what that looks like. So we go, where we go, problem 2.3 is 1 plus the cosine of that angle. And we're going all the way from 0 to 2 pi. So this whole area enclosed by this, this polar coordinate curve. And our limits of integration that's going to go from 0 to 2 pi. So let's first of all remember how we do this. And remember always with these things, the more you write it down, the more it becomes part of you and you're going to remember that. So that equals the half. And what I'm going to do from now is just bring the constant out. So it's a to b and we're going to take this function r, which is the function of theta, and we're going to square that d theta. So let's just do a little bit of work here on the right hand side. If we have r equals 1 plus the cosine of this angle. If we square this, that is going to be 1 plus the cosine of the angle times 1 plus the cosine of that angle. So r squared, what's that going to be? It's 1 plus 2 times the cosine of the angle. That's an easy one to do when we integrate that. And then plus cosine squared of that angle. So let's put it in green because, as I said, that's always going to be useful. This idea of cosine squared of an angle, this cosine squared of an angle. Remember that's equal to half plus half times the cosine of twice that angle. So that's a bit easier for us to deal with. So yeah, we're going to have plus a half, plus a half cosine of twice that angle. So in the end, if we simplify a little bit, we have 1 plus a half there. That's 3 over 2. And let's see, plus we have 2 times the cosine of theta. That's at least very easy to integrate. And in the end, another half times the cosine of twice that angle. And again, that's easy enough to integrate as well. We also just remember the fact that a is going to equal 0. And we have b is going to go 2 pi. So our limits of integration there. So let's do this. It's going to be half times the definite integral. We're going to go from 0 to 2 pi. And we have the square, and that is going to be 3 over 2. We're going to add to that twice the cosine of theta. And we're going to add to that a half times the cosine of 2 theta. And all we have to do now, that is with respect to theta, is not make any silly little mistakes. So we have the addition of three terms here, which means we can break this integral up. So that's going to be half is still way on the outside. So there's going to be the definite integral going from 0 to 2 pi of 3 over 2 d theta. And then plus we're going to bring that 2 out, because it's a constant, 0 to 2 pi, this definite integral of the cosine of theta d theta. And in the end, we can bring that half out as well. Definite integral going from 0 to 2 pi of the cosine of twice the angle d theta. And close all of that. So again, a equals a half, let's not forget that half. So the first one, it's just a constant d theta. So that's going to be 3 over 2 theta. And that's going to go from 0 to 2 pi, okay, plus twice. And what do we have there? It's the sine of theta as the antiderivative of the cosine of theta. And that goes from 0 to 2 pi. And then we have, eventually we have a half. And then remember it's the sine of twice that angle. It's antiderivative would be that, but then we also have the 2 there. So that's got to be a half. So think about it, take the derivative of a half sine twice the angle, the derivative of that is going to be cosine twice that angle. And that also has to go from 0 to 2 pi. And we close all of that because we cannot forget this half way up front here. So this first one is going to be 2 pi minus 0 as far as substituting as far as the theta is concerned there. So that's going to be 3 of 2, just 2 times pi there. So that one's easy. The sine of 2 pi, well that's 0 minus the sine of 0 which is also 0, so that term disappears. And the same thing is going to happen for that third term because we have the sine of 4 pi which is 0 and then minus the sine of 0 which is also 0. So that term disappears as well. So what we left with here is a half and we left with the 2s can go and that means we left with 3 pi and in the end that means we have 3 pi over 2. So once you get to do a couple of these, a bit of side work, but they are, they're quite exciting to do. And think about it, we have this very odd curve and we can calculate the exact area enclosed by that curve which is quite neat. Problem 2.4, calculate the area enclosed by the inner loop of the lemason or lemachon or the Turkish I suppose, we'll just say lemason. R equals one plus two times the cosine of an angle and we see the limits of integration and it's going to go from 5 pi over 6 to 7 pi over 6. This just means that there's going to be a lot of side work and there's going to be a lot of opportunity to make a little arithmetical errors because what you won't make an error with is just to remember that the area is going to be half times the definite integral and going from A to B and there's our A and there's our B of this function of theta R and we're going to square that d theta. That we won't forget and we have things right there. So it just means a lot of side work. So we have R equals one plus two times the cosine of this angle. By the way, let's have a look at just what this looks like. So there we go, problem 2.4 and what we really want is this little inner loop and hence we have the integral set by the angle going from 5 pi over 6 to 7 pi over 6. So let's see if we can do that so what we need is R squared of course. That's one plus two times the cosine of the angle times one plus two times the cosine of that angle. So R squared is going to be one plus four times the cosine of theta plus four times cosine squared of theta. Now we know that little integral there is a bit too difficult. So you're going to have one plus four times the cosine of the angle plus four times. Remember that's half plus a half cosine of twice the angle. So what we're going to have is R squared equals one plus four times the cosine of our angle plus that's going to be a two plus that's going to be a two cosine of twice the angle and eventually we get to R squared equals one plus one is three. One plus two is three I should say, four times the cosine of that angle plus we're going to have twice the cosine of twice the angle. So let's go about doing that. So we have a is the half and we're going from five pi over six. That's an ugly five, but remember what it's for seven pi over six and we have this three plus four times the cosine of the angle plus twice the cosine of twice that angle and that's all with respect to theta. So let's do a equals a half and this is where I suppose this is where all the little mistakes do come from. So we're going to have this integral in five pi over six to seven pi over six of three d theta and we have the integral of five pi over six to seven pi over six of, I suppose we can bring that four out, cosine of theta d theta and lastly we can bring that two out. Definitely integral five pi over six, seven pi over six and we have the cosine of twice the angle d theta and we close that all off there. So the area is going to be a half. We still have all of this. That integral there is going to be three times theta and that goes from five pi over six to seven pi over six and what is that? Let's just see because we're going to probably do that. So seven pi over six minus five pi over six. That's very easy. That's two pi over six and that's pi over three. We've got our one. So let's carry on plus four times. What's the antiderivative of the cosine of theta? Well, that's just the sine of theta and that's got to go from five pi over six. I keep on making those ugly fives. Seven pi over six and then we're going to have plus two and this is going to be a half. Sine of two theta. So take a half sine two theta. Take its first derivative. Let's go sine of two times that angle and that's also got to go from five pi over six to seven pi over six and we have got to close that because that's all got to be multiplied by this half. So if we do that, if we do this, we are set with the fact that it's three times that three that we have there times that. So that's just going to leave us with pi. So that just leaves us with pi and then we have to do this four times plus four times and then we have the sine of seven pi over six minus the sine of five pi over six. Great. And then that's just going to cancel out. So two times a half, that's just one. So we're going to have plus. We're going to have sine. So you sine of twice this angle. So that's 14 pi over six, 14 pi over six minus the sine of twice that angle. So that's 10 pi over six. Close that and all of that's going to be multiplied by a half and so where does that leave us? Still with a half. This is going to leave us with pi and I suppose what you have to do now is use the Wolfram language or take out your calculator because it's not so easy to remember what the sine of 14 pi over six is or 10 pi over six, even if you simplify those a little bit but if you do know great for you I'm going to just use the Wolfram language and you'll see that equates to, this is going to be minus five plus the square root of three. That's where we're going to end up with and I suppose we can just simplify things a little bit in as much as we can say a equals a half times we're still going to have pi we're still going to have positive the square root of three and we'll have a negative two. If we distribute that in, we can have a negative two at the end. So that's it. These problems are really designed just for you to go through a lot of exercise and just keep your head together so that you don't make an arithmetical error because that's where it's going to creep in and of course you might be required to remember what the signs are of all these angles or at least simplify them and remember what they are or you might be given them. I think it's just fair to give you access to a way to calculate those and in the end it's keeping your head together the problem itself and then as you can see is not that difficult. Let's have a look at the area enclosed by one loop of this, let me just get r squared equals cosine of two theta. Let's bring up the Wolfram language and here's our problem and remember this is a function of theta r but we have r squared so just to get r we just need the square root of cosine of two theta which is what we have there for polar plot and we're going to go from zero to somewhere and that somewhere is from 0.01 to two pi so we're just using this manipulate function, remember? So if we do that and we go all the way around so we'll see that's what it was but this is a trickier one than you think and remember with these, always with these calculus problems try to use symmetry and get the smallest possible area and multiply it because of symmetry and we see here we basically have these four quadrants and it's not as simple as that when you really sit down and you get to there it goes back and forth there's a lot happening there so let's just go from there slowly, slowly, slowly down to there, 0.8 somewhere there so that's pi over four and we can just do that integral from zero to pi over four and just multiply it by two because we just asked for one loop and if we were going to do both loops we would multiply it by four so that's just a little rule of thumb just do the smallest one you can and just multiply it because of the symmetric properties of the shape created by the curve so let's have a look at this so always remember we're going to write that down the area of these functions that are expressed in polar form that's going to be the integral of half of the f of theta squared d theta so let's get our blue pen there so let's do this and so what we're going to do we're going to do twice this area and we're going to go from zero to pi over four we saw from the plot that that was the smallest piece we can do we just multiply it by two and we have that area and then a half and then r squared and we've already given the function as r squared so this is going to be the cosine of two theta d theta so the area is going to be two times we can bring the half outside of the integral here so two times a half is this one so that's a different integral going from zero to pi over four of the cosine of two theta d theta so the area is going to be what is the antiderivative of the cosine of twice this angle but that's very easy that's half times the sine of twice the angle and if you take the derivative of this with respect to theta you get back to cosine of two theta so we know that's right and we go from zero to pi over four now try to do this by you know by not multiplying this by two and go from zero to pi over two or go from zero to pi and see what you know sort of see what happens and you can very quickly get to this idea of take the smallest possible area and just use the symmetric properties of this curve to get this area so this is the area going from half and so we're going to have sine of twice the angle and we're going to go from zero to pi over four so area is going to be a half times the sine of two times pi over four that's sine of pi over two minus the sine of that's going to be zero which is the zero so area is going to be the sine of pi over two is one and we just left with a half so that one loop if we just looked at the loop itself just the shape that is formed that the curve forms the area enclosed by that is going to be a half problem 2.6 calculate the area of one of the petals of the polar coordinate curve r equals the sine of two theta let's bring that up and there we go the polar plot of the sine of two theta and we're going to go from zero to two pi so we just want the area of one of these petals now in the problem I think we are given a hint that we have to take theta going from zero to pi over two so if you're using the Wolfram language put the manipulate function in there and then go from zero point zero one just something larger than zero to pi over two and you'll see one of the petals there so as always we're going to start with by just reminding ourselves what the equation is that we derived for this area we're going to go the definite integral of going from zero to some angle so zero to some angle of the half of our function squared with respect to that angle so if we just take our blue pen there so we're going to have the area is going to be this definite integral we were told and the hint from zero to pi over two so that's from zero to pi over two over half times we're going to have the sine of two theta all squared d theta so we've got to sort of first figure out how we're going to do this this what we're going to do with the sine two theta squared so that we can do this so let's just take a different color there usually the side work in this lavender color so what I'm going to do is just let let's do this let alpha equals two times theta so what we have here is the sine squared of alpha if we just have a look at this we're just plugging in alpha for two theta two times theta so that's the sine of alpha squared or the sine squared of alpha we remember from the most famous of the trigonometric identities that this is going to be one minus cosine squared of alpha so that's one way to do it you can of course just if you can't remember the double angle formula for this the one that I always remember is just the cosine you know what to do with the cosine squared and the sine squared is actually very easy and when we get to the individual C you've probably memorized it but let's go this long way around just a bit of exercise so the sine squared of alpha that's going to be one minus and remember what is this well the cosine squared of an angle that's half plus a half cosine of twice that angle so the sine squared of alpha that is one minus a half minus a half cosine of two alpha and sine squared of alpha one minus a half is a half so it's a half minus a half cosine of the double angle so if you didn't remember that the sine squared is one is a half minus a half where the cosine squared is a half plus a half it's very easy to derive as you can see there but we're not interested in alpha or we were interested in theta so that's going to be the sine squared of two theta that's going to be a half minus a half cosine of four theta so that is the integral that we're dealing with so let's just do this that's going to be the area of we're going to bring the half out definite integral going from zero to pi over two and we're going to have this half minus a half cosine of four theta d theta so we have two terms in there and we can take the integral of both of them so area is going to be a half let's just keep this half way out here so that's zero to pi over two zero to pi over two of a half d theta minus the integral and going from zero to pi over two of the cosine of well there's a half in there let's bring the half out there of four times the angle d theta and we close that so this is just an exercise really in not making silly mistakes here so that's going to be a half out there so this is going to be a half theta and this is going to go from zero to pi over two minus this is a half here and so we've just got to take the antiderivative of the cosine of four times an angle well that's going to be sine of four times that angle and a quarter in front and if you take the derivative of a quarter sine four times theta you're going to get to cosine four times theta and this also has to go then from zero to pi over two and all the way out there so the area here for us is going to be a half all the way still out here and then we've got this half this angle and we've got to go from pi over two minus zero so that's what we have there minus a half and we can bring the quarter out as well so that's going to be an eighth so then we're going to have sine of four times the angle and that's going to go from zero to pi over two and guess what's going to happen to that so the area is going to be this half and we have pi over four here and then we have this sine of this so let's just do that on the side here let's just grab another pen so I'm going to have sine of four times a half pi so that's two pi and we know that that's going to be zero and with a zero as well that's going to be the sine of zero they're both zeroes so this minus the eighth times zero nothing is going to be left there so we just have that that'd be the wrong color and eventually the area here is just one eighth pi or pi over eight that's right, they'd need a pi over eight so really just an exercise in remembering your trigonometric identities here and if you can't remember sine squared it's just easy to derive as I say somehow in my head this one sticks and I was sort of unsure about this one but it's easy to derive and it's really not that difficult to remember and of course just to make life easier for yourself just simplify the angle first so that you can deal with that problem 2.7 calculate the area inside r equals two plus cosine two times theta and where theta goes from zero to two pi let's bring up the Wolfram language and have a look at it and there we go two plus cosine of two times theta and we're going all the way from zero to two pi interesting shape and we need to calculate that area as always let's start by remembering that the area is going to be this definite integral and going from starting angle which is usually zero to some final angle times a half and we express our polar form of our equation there and we square that d theta so we're gonna probably have to do some side work here if we look at two plus cosine of twice the angle so let's give it a start so we're gonna go area equals we were told from zero to two pi we're going to have a half and then we're going to have two plus cosine of two theta squared d theta so let's do some some work on the side let's square that first so two times two is four and that's going to be plus four times the cosine of two theta and we're going to have plus cosine squared let's make that the positive cosine squared of two theta now just in previous problem 2.6 had a look at that let's have a quick look at that one because remember we substituted two theta for alpha and we remember we can remember what cosine squared is let's just do that here four plus four times cosine of two theta that'd be easy enough to integrate and what we're going to have here is remember let's just do one other color here if we have cosine squared of an angle remember that's going to be a half plus a half cosine of twice that angle but our angle is not alpha it's two times theta so that's going to give us four times there so that's going to be plus a half and plus a half cosine of four times that angle and four plus a half is nine over two plus four times the cosine of two theta plus a half times the cosine of four theta so that's quite you know there's three separate things there that we can integrate and that makes the integral not too difficult to do so let's let's get at it so area is going to be I'm going to bring this half out and we're just going to have all of these separate three separate integrals so we're going to go from zero to two pi just of nine over two d theta with respect to theta d theta plus let's just bring the four out that's going to go from zero to two pi and we have cosine of two theta d theta and another integral going from zero to pi I've got that half I can just squeeze it in there half cosine of four times and our angle with respect to our angle and close all of that up so the area is going to be a half so let's start here that's going to be nine over two theta going from zero to two pi plus four times and we've got to take the antiderivative of the cosine of two theta well we're going to have a half at least out there we're going to have the sine of two theta and that goes from zero to two pi and here we've got our half we're going to bring out a quarter there and we're going to have the sine of four times theta and that also has to go from zero to two pi and we have to close all our brackets now you can see what's going to happen to these terms sine of two theta two times theta and the angle is two pi so that's going to be four pi and that's going to be zero and the sine of zero zero and the same is going to happen here four times so eight times pi that's also going to be zero and the sine of four times zero that's zero sine of zero zero so those terms are all going to disappear at the end there so what are we going to be left with something really easy we've got the half here we've got the nine over two there and we've got the two pi minus zero there basically so that's going to be nine over four nine over four times two pi over one and that's going to simplify to nine pi over two so our area is just nine pi over two so a lot of work just to keep things together but in the end it really simplifies quite a bit as that our two term number two and three they just disappear on that interval that we have and there we have it there we go 2.8 calculate the area inside r equals cosine squared of theta over two so it's complicated by the fact that we have a problem with the angle and we've got a square already as far as our trigonometric functions concerned so let's let's let's try this we're going to remember that a equals the area equals from some starting angle which is zero to some final angle half and we have our angle our function at least but we've got a square that and d theta the other thing that we also going to remember let's just write that out cosine squared of some angle let's call this angle alpha remember that is half plus a half cosine of twice that angle so that's what we're going to use so let's just look at our cosine squared of theta and theta over two I should say so what's going to happen to that so that's going to be a half plus another half cosine of twice that angle but the angle's already theta over two two times that so that just leaves us with theta but we remember we've got a square this because we've got our function squared there so this is going to be a half times a half is a quarter a half times a half is a quarter times two that's again half times that's a that leaves us with a half and that's cosine of theta and plus a quarter cosine squared of theta so we've got to do something like cosine squared theta again so this is going to be a quarter plus a half cosine of theta plus a quarter here and this is going to be a half plus a half cosine of twice that angle so that leaves us with a quarter plus a half of cosine of that angle plus an eighth plus an eighth cosine of twice that angle quarter plus an eighth that's three over eight and plus a half cosine theta plus an eighth cosine twice the angle so it's as simple as that and what you can already see here is that the antiderivative of cosine is going to be sine and we've got the angle itself and we've got twice the angle but we're going from zero to two pi and that means the sine of those in the integral they're just going to disappear so if if you want to do that the long way please go ahead but what we're going to be left with of course is going from zero to two pi and so we're just going to have three eight d theta there we go three eight d theta because all these others as you see in all the previous problems but please write them out do them the long way they're all going to disappear so what we left with is a half times three over eight and we just have theta and we're going to go from zero to pi zero to two pi so area is going to be three over sixteen and times two pi so that's going to area is going to be three over eight pi this is right that a bit neater three pi over eight three pi over eight and bring up the orphan language just I didn't show you I think in this one just to see what that what the graph looks like of this polar coordinate function and please write out all the other terms here I'm being a tad lazy here but you've seen with all the other problems now if we take the antiderivative that's going to be the sign and for the the bounds of this integral those terms are just going to disappear calculate the area here in problem 2.9 swept out by this polar coordinate equation r equals the tangent of theta and theta gained from zero to pi over four let's have a look at what this looks like there we go I've got my tangent of my angle there and as we sweep it out from zero so they think about it we are obviously this is going to be about vector calculus so we'll get to vectors but think about this vector going straight down the x-axis so along the x-axis and then it curves up to pi over four pi over four 45 degrees so if I had the straight line going through there because what we really wonder when we what we've got to think about when we're doing this integral the area under the curve here that's not the area under this curve and certainly when you transform this into rectangular coordinates I can do that equation you can see it there try to do it yourself and the area under the curve would be that but that's not what this problem asks this problem actually asked the following so our curve is now down here in orange but it's this area remember we go counterclockwise from zero to pi over four so it is this area here the inside area on top of so if I had the straight line up here it's this inside area that we're talking about remember if we have other curves going all the way around we want this anti-clockwise movement as the angle gets gets larger so we're sweeping up this top side so let's have a look remember always area equals this integral over half our function of theta and we square all of that d theta so we're certainly going to going to need that other things that we might need is a trigonometric identity and if you can't remember the trigonometric identity let's just remind ourselves we have sine squared of some angle let's call it alpha plus cosine squared of some angle that equals one but if we divide both sides by cosine squared cosine squared and cosine squared of that angle we are left with the tangent squared of that angle plus one plus a little bit better and that's the secant squared of theta so there's your second one and since we have r equals the tangent of theta we've got a square that that's going to be the tangent squared of theta that we have to deal with so we can have tangent squared of theta equals secant squared of alpha what am i writing theta say let's just correct that before we go any further so where are we alpha we dealing with alphas here minus one so we're certainly going to do we certainly going to need some of that and we're gonna we're gonna need one more thing but let's have a let's let's have a look so we got this half we're going from zero to pi over four tangent squared of the angle and with respect to that angle so we're going to rewrite this as area equals half of the definite integral and going from zero to pi over four of secant squared theta minus one so that secant squared theta d theta we can say because we can also then say let's just put that there this integral zero to pi over four of one d theta because it's just two separate expressions so if they're all part of the integrand we can of course just do them separately take the integral separately so that's going to be a half and what is the antiderivative of the secant squared of theta and as that's what i said that's the other thing that we must remember what is d d theta of the tangent of theta well that's the secant squared of theta so what we're going to have here is the tangent oh let's just make a lighter blue there the tangent of the angle and that's going to go from zero to pi over four minus and then we're going to have theta and that's also going to go from zero to pi over four so what is the tangent of let's have a look at that what is the tangent let's do this first one so the tangent of pi over four well that's just one minus that's going to be a zero tangent of zero zero minus what we have here pi over four minus zero so that's what we have there so we have a equals a half and then we have one minus pi over four and let's simplify that that's going to be a half let's put that four there and that is going to be four minus pi and eventually we have that the area is going to be four minus pi over eight so another lovely little problem just to remind yourself of of your first year calculus and some trigonometric identities and just remember the very simple fact that the integral of the difference between two expressions that's equal to the taking this the integral separately problem 2.10 calculate the area inside the cardioid r equals one plus the cosine of theta and outside the circle r equals one let's have a look at those plots there we have it in the warframe language so it's this blue one which is the cardioid so it's going to be inside that but outside outside of this circle r equals one so what we're going to do is one minus the other one and we're going to make use of symmetry because if we go from zero that's this positive x axis to pi over 2 which is the positive y axis we do this up a half and then we just have to multiply by 2 for the bottom half but i hope you can see that this area that we're interested in here in between these two that's this this outside one minus the inside one let's see what that looks like as always area is going to be this integral of a half our function which is the function of theta square d theta and let's pick our blue pen so what we're going to have here we're going to use symmetry so that's going to be two times a half we're going from zero to pi over two and now we just have the one minus the other one so this is going to be one plus cosine of theta all squared minus the other one which is one all squared and that's d theta so the area is going to be equal two times a half that's this one so that's be simple zero to pi over two and we have here a one plus twice the cosine of theta so i'm just squaring this bit here plus cosine squared of theta minus one all of that with respect to theta and now i can just remember what to do with the cosine squared and i can say one minus one that goes away so this is this definite integral and going from zero to pi over two as i say one minus one that's zero we're going to have two cosine of theta and here we're going to have plus a half plus a half cosine of twice the angle all of that d theta so let's do that so area is going to equal well that's going to be two times the sine of theta that's the antiderivative and that's going to go from zero to pi over two plus this is going to be a half theta and that's going to go from zero to pi over two and that's going to be a half and this is going to be another half sine of twice the angle and that goes from zero to pi over two now i hope you can see what's going to happen to these because the sine of pi that's zero and the sine of zero zero so that term is going to drop away this first term let's have a look at that so the area the area here that's going to be twice what is the sine of pi over two well that's just one minus the zero sine of zero zero so that's there plus a half and here we have pi over two minus zero that's there and as we said that the last term is zero so this is going to be two plus pi over four two plus pi over four and you can eaten that up if you wanted to just to have a single fraction so i hope you enjoyed these couple of problems dealing with a calculus of functions that are expressed in polar form they're not really difficult to do just do a bunch of them as i say my one tip is this always going to be keep it very simple take the least amount of angle that you have to go to do your integral so from zero to the smallest possible and then just use symmetry multiply by two multiply by four and for the majority of these you see all you need to remember is about a bit of or some at least trigonometric identities and you might have to remember that the antiderivative of the secant square of an angle is the tangent of that angle that sort of thing