Find the area enclosed by the cardioid r = 2 − 2 sin θ r=2-2\sin\theta between θ = π 6 \theta=\frac{\pi}{6} and θ = 2 π 3 \theta=\frac{2\pi}{3} .

So in this problem, we're asked to find the area enclosed by the cardioid given r is equal to two minus two psi beta between an angle of pi over six and two pi over three. Now I can see that we are dealing with polar coordinates here because we have this r in terms of theta, and we can see that this curve is going to look something like this. I went ahead and sketched what this cardinality would look like, and we're trying to find the area of this region right here. Now recall our strategy for finding the area of polar regions. What we can do is use this equation. The area is equal to one half times the integral from alpha to beta of r squared integrated with respect to beta. Now we actually have everything we need given to us to set up this integral. So we have this one half, which will be out front, then we'll have alpha, which in this case is going to be our starting angle of pi over six. Then Then we're going to go all the way up to beta, which you can see is our finish angle here of two pi over three, and then this is going to be r squared. Now I can see that r is this two minus two sine beta. So that means we're going to have two minus two sine beta, and all of this is going to be squared and integrated with respect to beta. Now we have this integral set up, so at this point, it's just about solving and evaluating. So we'll keep this one half on the outside, and to make things more simple for myself, I'm going to evaluate everything that we have inside of brackets for the integrand. So that's gonna be this two minus two sine beta squared. Now what happens if I go ahead and I multiply this all out? So we have two minus two sin beta times two minus two sin beta. That's what happens when we take two minus two sin beta squared. And what is all this going to come out to? Well, we'll have two times two, which is four. Then we'll have two times negative two psi beta, which is negative four sine beta. Then we're going to have negative two sine beta times two, which is negative four sine beta. And then we're going to have negative two sine beta times negative two sine beta, which comes up to positive four sine squared beta. Now if I want to write this, I can combine these negative four sine beta. So you get four minus eight sine beta plus four sine squared beta. So that's what's gonna happen when we write this all out right here. So inside of here, we're going to have four for this integrand minus eight sine beta plus four sine squared theta, all integrated with respect to theta. Now at this point, it may seem like I can just go ahead and do the antiderivative. But if I go ahead and look ahead of what's gonna happen here, I can see that I could integrate this four with respect to theta. I can also integrate the sine in a relatively straightforward way, but this sine squared is going to cause problems because we have a sine raised to a power. Integrating that's gonna be a bit tricky. What I can do is use one of the power reducing identities here, where what I go ahead and do is take the sine squared beta, and I can recognize this is the same thing as one minus cosine two beta over two. This is an identity for sine squared beta. Notice this gets rid of that power of two when we write it out like this. Now I'm actually going to go ahead and do this off to the side here. So notice that this whole thing is four sine squared theta. So I'm going to write this in right here, so we have four sine squared theta. Now what I'm saying is that this whole sine squared theta can be equivalent to one minuteus cosine of two theta. So that means we're going to get four, and it's going to be times one minuteus cos two theta over two. Now I can take this four and I can divide it by this two here, and doing that is going to give me two, and then we'll have one minus cos two beta. I can go ahead and distribute this two into the parentheses. Giving me two minus two cosine of two beta. So that's what happens when we go ahead and write out this whole four times the sine squared of beta. And that just allows me to write it in for this integral a bit more easily. So I have one half, and it's gonna be the integral from pi over six to two pi over three. Now what I'm actually gonna do is since I'm taking this whole thing and I'm plugging it in right here, well, what I notice is that there's a constant here, which I can add to the four out front right there. So I can just go ahead and add the two and the four, which will give me six. Then we'll have minus eight sine theta, which is this next part right here. And then we don't have to worry about this too, so we just have everything remaining here, which is maybe this minus two cosine two theta. And all of that is integrated with respect to theta. So now we have this, and at at this point, well, I can see this is a place where I can actually run this antiderivative, because we have each of these terms right here that I can just integrate separately. So let's go ahead and do that. So keep this one half out front. Now the integral of six with respect to theta, that's just six theta. Now if I integrate eight sine of theta, that's gonna give me eight, and then the integral of sine, well, that's gonna be negative cosine. So we're gonna get plus eight, and that's gonna be cosine theta because it's gonna cancel with the negative sign that we have right there, since the integral of sine is negative cosine. And then we need to integrate this two cosine two beta. Now if you were to integrate this piece, what you'd wanna do is do a u sub on the two beta. Then you go ahead and take that derivative, and what you should get is that this integral of cosine two beta ends up coming to one half then sine two beta. The integral of cosine is sine, and this one half will cancel with that two. So all we're gonna have in here after we do this integral is this is going to be minus, because we have a minus right there, and that's going to be the sine of two theta. And then all of this is going to be bounded from pi over six to two pi over three. Now at this point, it's just a matter of plugging in the bounds, and I trust we could do the rest of this problem from here. So So we're gonna have one half, and then I'll go ahead and plug in the bounds here. So plugging in first my upper bound of two pi over three, we would get six times two pi over three, which is replacing the beta here, plus eight times the cosine of two pi over three. You're going to have minus the sine of two times two pi over three. And now we've gone ahead and plugged all of this in, and we're going to subtract off our lower bound of pi over six plugged in for theta. So that's going to give us six times pi over six. You're gonna have plus eight times the cosine of pi over six. Plugging the same for theta, then we're gonna have minus the sine, in this case, is gonna be of two times pi over six. And That's what happens when we plug everything in, and so this is what it's going to look like. Now at this point, there's a few things I can simplify here. You notice we have a six and a three. Now three can go to six two times, and two times two is four. So this whole thing is going to reduce to four pi. Now next, I can go ahead and simplify this by just combining the twos there. Two times two is four. So we end up getting is four pi over three. Now moving on from here, I can go ahead and cancel the sixes right about there. So I can see that these sixes are just going to cancel. So that's gonna leave me with just pi. And then we have this pi over six here. That's fine. And then we have a two and a six. Two over six reduces to one third, so this whole thing is just going to reduce inside the side here, two pi over three. So now that we have everything reduced at this point, what you can do is you can go ahead and fish each one of these solutions out on a unit circle. So you find the cosine of two pi over three, sine of four pi over three, and then you can find the cosine of pi over six, the sine of pi over three. You can plug in those results and then multiply everything out. And when you do this, your result should come out to and then don't forget this one half that you multiply at the end. Your result should come out to three pi over two minus two minus three times the square root of three over two. This is what everything's gonna come out to if you just multiply out all these values, or if you're allowed to use a calculator for this, we wanna get a decimal approximation, I can go ahead and plug this all into my calculator, and the result that I get when I do that is about 0.1143. And then this decimal keeps on going. So that's approximately what this is all equal to. So this being the exact answer and this being the decimal approximation would be the solution to our problem, and that right there is how we can find the area for this Cartier weight that we formed with this polar region. So I hope you found this video helpful, and that is how you can solve these types of problems when you're trying to find the area of polar regions. See you in the next video.