 This lesson is on area and polar coordinates and this of course is something that was not done in AB calculus So those of you who have taken AB calculus need to work on this very carefully some preliminaries Make sure you know how to stay points Grab and state angle values, which create a curve one of the polar graph curves You should also know how to go from rectangular coordinates to polar coordinates and back And this should be in a preliminary lesson or review for you The polar graphs that we normally work with are circles, Limissons, rose curves, luminous gates, and spirals and we will use many of these in our examples today So let's go on Determining the area and polar coordinates. Well, the formula is area is equal to The integral from a to b of one half r squared d theta Where does this come from? If we create a polar curve? Something like this and it can be any type of squiggly graph that is created with an r and a theta and We just take one portion of this let's say that portion and if I Break that up into many many portions or parts as we do in Riemann sums and Just take one of those parts and look at it We will see that the area of the sector that we're going to be working with is area is Equal to one half r squared and some sort of a theta But in this case if we don't take our beginning angle to be zero our theta Becomes a delta theta So when we put the Riemann sum component on this We will have the Riemann sum formula the limit as n approaches infinity of the sum sigma i Equals one to n of one half r sub i squared Delta theta sub i depending on which segment we are on That equals the integral from a to b of one half r Square d theta and that gave us the formula that we originally had up there So let's go on and try to do some problem-solving with this area under one leaf of a rose curve Given r is equal to two cosine three theta Determine the area under one leaf well the graph of this has the first leaf sitting on the x-axis Because it's cosine and it's two units long and then because it's three theta. We have three leaves So there's the second leaf. There's the third leaf Now the only other thing we need to do is determine what angle it takes for us to complete one graph let's go to our calculator and Put in our function Two cosine three theta and our window on this is from zero to pi And of course increments are point one radians, which normally you should have when you are doing these So let's check this out and see what it grabs like Sure enough the worm finished when the graph finished so pi is a good number for our theta So let's look at that So we are going from zero to pi for theta And that would give us a whole one, but we only want one leaf So we really want it to be pi over three so it goes from zero to pi over three but we can even be smarter than that and just take one small portion of this which will be a Zero to pi over six So if we do an integral from zero to pi over six of one half in front of it Of course and how many of these do we need well We need two of those to make one leaf and square our r So we'll have four cosine squared three theta d theta cleaning this up We get four zero to pi over six and remember cosine squared has an identity Which we use when we want to do our integrals by hand, which is one half times one plus Cosine six theta d theta cleaning this part up and taking the integral we get two Times theta plus one six sine six theta and we're going to go from zero to pi over Six and we substitute the pi over six in we get a pi over six for the theta when we substituted in here We get sine of pi which is zero and if we do the zeros in We get zero so we really end up with two times pi over six, which is pi over three for our answer So this isn't so bad the important thing is to know what it takes to draw out or sketch out One whole graph of these in this case the rose curve Let's go on to another problem Determine the area enclosed by the curve r squared is equal to four sine two theta Well, this is a limit escape and looks like this for the sign now What does it take to sketch this one out? Well, if you graph that one on your calculator You will find the theta is equal to two pi But I'm going to break this one down even further two Because I am just going to take one fourth of that and just go from zero To one fourth of two pi which is pi over two and then of course I'm going to multiply it by four So area is equal to one half times four because I'm doing the whole curve times the Integral from zero to pi over two because r is already squared here We can just put down four sine two theta and d theta that and in cleaning all this up We get eight times the integral of zero to pi over two of sine two theta d theta Take the integral we get eight integral of sine two theta is negative one half Cosine two theta and we go from zero to pi over two this becomes a negative four cosine of two times pi over two which is cosine of pi is negative one the cosine of zero minus One again or it's plus one, but it's minus plus one of course and that gives us a negative two times a negative four Which gives an eight square units? Well, let's go on and do a little bit more difficult type of problem this time We are going to determine the area between two curves so the problem reads determine the area between the curves inside of r is equal to two cosine two theta and Outside of r is equal to one So if we sketch that out again, we have a rose curve this time with four leaves Cosine starts on the x-axis and the whole distance for that pedal is two So we have four of those and then the circle r is equal to one is nothing more than a circle With the center of the circle at the origin So if we just look at a small piece of this We have the leaf and we have the circle and they intersect at some point so first of all let's find that point of intersection and then Decide what we're going to do with it so in deciding what that intersection is we make two cosine two theta equal to one or Cosine two theta is equal to one half or Two theta is equal to plus or minus pi over three or theta is equal to plus or minus Pi over six which means this is negative pi over six and that's positive pi over six and The area we want is this region up in there So let's manipulate this a little bit and say area is equal to one half and we want this to go from Zero to pi over six always use that zero when you can and now how many of these little intervals Do we have or we have actually eight of them two for each pedal four pedals and then we have two Cosine two theta Quantity squared because the region that we're taking is just the region inside of the two cosine two theta and D theta there and this is equal to four times 1.913 and that's equal to 7.652 Square units so again, let's look at this find the points of intersection Look at the area that you need to find and in this case It was only the area inside that curve r is equal to two cosine two theta And that's the one we used our last example is even more complicated Determine the area when r is equal to five sine theta and r is equal to two plus two sine theta Overlap so what does that area look like well r is equal to five sine theta is a circle that goes up and Is five units in diameter r is equal to two plus two sine theta is our Cardioid and looks like this it only goes up four units So it is lower than the circle the area that we are looking for is where they overlap which is in here But what we need to do again is break these two areas up because we have one that's coming from the sine and The other one that's coming from the cardioid so let's just focus in on that This area is coming from the sine and of course There's another area on the other side equal to that so whatever we do I'm just going to double and then the other area comes from the cardioid So we're going to have to add two areas together So our area is equal to one half and let's double that now times the integral Where do they intersect so again? We have to set the two equations equal to each other So five sine theta is equal to two plus two sine theta or three Sine theta is equal to two or sine theta is equal to two thirds And we find out that theta is equal to not an exact value for this one of course So it's zero point seven two nine seven So our first interval is from zero to zero point seven two nine Seven and we again are using the circle for this so we will put five Sine theta squared d theta for that one and we will add to that one half times two and go from zero point seven two nine seven Two where that one ends which is pi over two and this time use the cardioid two plus two Sine theta quantity squared d theta and we do have to do this one on a calculator at some point So why not just stick it in and do it all at once and eventually you will get an approximate answer of fourteen point nine one three square units Again on these where the areas intersect for two curves Make sure you find that point of intersection and note which areas you need either to add together or just use by itself This concludes your lesson on area and polar curves