 So in the previous video, we saw how we could find the area under a polar curve And we use the area of a of a circle sector to help us out there Well, what we want to find the area between two curves like this illustration we see below What if we want to find the area of the region that's between the circle r equals 3 sine theta and the Cardioid r equals 1 plus sine theta Well, if you want to find the area between the curves Basically, you're going to take a sector which is really just a pizza slice and then someone took a bite out of the Pizza slice, right? You're going to have two different radii. So you have your outer radius We'll just call that one r1 and then you have your inner radius We're gonna call it r2 the area of the outer circle Outer sector I should say that's going to be one half r1 squared theta and Then you subtract from that the area of the inner one one half r2 squared theta like so And so then when you eat the take a bite out of your pizza slice right here You get this sort of like trivial pursuit symbol now. This will just be one half R1 squared minus r2 squared times theta So you have a difference of squares when you want to find the area between the curves Now if you apply that if you take limits and Riemann sums and all that all that jazz You're gonna get the following the area between two polar curves You're gonna get one half the integral From a to b of the difference of squares the difference of the radii d theta So let's apply that to the context of our circle and our cardioid like we have here as the illustration shows us the Outer radius r1 is going to be the circle three sine theta The inner radius r2 is going to be the cardioid one plus sine theta and then taking the area between these two curves We get one half the integral will come back to the bounds in just a second We have to take the outer radius squared. So three sine theta Squared minus the inner radius squared one plus sine Theta squared and then times everything by d theta. We're going to integrate with respect to theta here So how do we identify the bounds? We have to figure out these things right here Well, the bounds are going to come about by determining these points of intersection So we go from here to here now one thing I'm going to utilize here is that this thing is symmetric along the y-axis So we just have to figure out what's this angle right here theta equals wet eves And then we're going to go from that angle to pi halves and that we can just we just take this region double it to get everything so the nice thing about doubling it is Well, it makes the top one easy. We just need to do a pi halves And then by doubling it that cancels out with the two that's there. That'll be nice But who's the one on the bottom? We're gonna have to solve this by intersection right so we take three sine theta Set this equal to one plus sine theta and now we want to solve this thing subtract Subtract sine from both sides minus sine so it cancels on the right minus sine So we end up with a two sine on the right equals one divide both sides by two we get sine theta equals one half and sine equals one half at pi sixth and Then the other one would be 5 pi 6 So one in the first quadrant over here So this is our pi sixth and then one on the other side 5 pi 6 But because of the symmetry argument we made we don't need the 5 pi 6 on the side We're just going to go from pi 6 to pi halves So simplifying what we got here the area between the two curves will equal the integral We don't need the one half because it canceled the two Pi halves to pot. Sorry pi 6 to pi halves If you take 3 sine squared and you square that you'll end up with a 9 sine squared like that with the 1 plus sine Quantity squared you have to foil that thing out. So you get 1 plus 2 plus 2 2 sine theta and then you're going to get a sine squared theta This last term is a like with the 9 sine squared. We combine those together 6 So we're going to end up with 8 sine squared theta We get a minus 2 sine theta and then we get a minus 1 Like so and this is almost ready for integration sine squared is not our friend right here We're going to replace sine squared with one half One minus cosine of 2 theta This one half cancels with it's part of the 8 leaving a 4 behind and so now let's integrate that thing So in terms of integration the antiderivative of the 1 will be theta. So you get a 4 theta Antiderivative the negative cosine theta that'll be a negative sine of 2 theta We'll also have to divide by 2 from the chain rule that combines with the 4 that's in play here So that leaves us a coefficient of 2 Antiderivative of negative 2 sine will be a negative I'm sorry. That should be a positive 2 cosine Always have to be careful on the signs there and the entire negative one is gonna be negative theta As we go from pi 6 to pi halves Admittedly you could combine the 4 pi with the sorry that 4 theta with the negative theta to give us a 3 theta And so as you start plugging these things in there be aware of that Plugging in a pi halves into the 3 theta will give us a 3 pi halves Plug it into the sine we're going to get negative 2 times sine of pi Sine of pi is 0. So I'll just disappear We also have to do plus 2 times cosine of pi halves conveniently, that's also a 0 this is sort of the convenience of using the pi halves Using the symmetry here. It's really nice. Then we subtract we're gonna take 3 times pi 6 Which admittedly there is some reduction going on there. This will become a pi halves. That's nice Then we do pi 6 here. We're gonna get negative 2 sine of pi thirds Be aware that sign of pi thirds that one's not gonna get any simplification That's just gonna be a root 3 over 2. The 2 is 2 here cancels You just get the negative square root of 3 and then lastly we have to do plus 2 times cosine of Pi 6 Cosine of pi 6th is likewise the square root of 3 over 2 That's kind of nice the 2 is here cancel and now you'll notice that We have a negative a negative square root of 3 right here, and then we have a positive one. Those are gonna cancel out and So what didn't cancel out what survived this massacre of cancellation? We end up with a 3 pi halves from the first group and then from the last group we got a negative Pi halves and so those was combined to give us a 2 pi halves. That is a pi And so the area between these two curves is exactly pi So when you're dealing with the area between two curves kind of thing of the washer method You need to have a difference of squares not not a square of a difference. There's there is a difference there, right? Pun intended of course, but if you use the formula that we saw before this difference of squares formula for the For the area between two curves, you'll be just fine