 In this video, we'll work through two examples of finding areas of polar curves. Consider the cardioid r equals 1 minus cosine of theta. Let's find the area of the region in the first quadrant within this cardioid. So let's get a sense first of what the curve looks like. Here's a graph. And the part of the polar curve that I'm interested in finding the area of is just that first quadrant piece. So one of our tasks is to identify what our limits of integration are. We know from earlier in this lesson that the area of a polar curve is the integral from theta 1 to theta 2 of 1 half r squared d theta. So we have our r, which in this case is 1 minus cosine theta. But the question is, what are our limits of integration? Well, the initial theta would be zero because that's the theta value where my cardioid begins being graphed in the first quadrant. It then enters the second quadrant at theta equals pi over 2. So our limits of integration will be zero to pi over 2. So we have our region that we're interested in. And we integrate from zero to pi over 2 of 1 half r squared d theta. Now r was 1 minus cosine of theta. So I'm going to expand that 1 minus 2 cosine theta plus cosine squared. Now in order to make this integration that much simpler, I'm actually going to use one of my trig identities, knowing that cosine of 2 theta is equal to 2 cosine squared theta minus 1. Knowing that, I can solve for cosine squared by adding 1 to the other side and then dividing by 2. So I have 1 half times 1 plus cosine of 2 theta. That expression is a bit easier to find the antiderivative of. So I'm going to pull this 1 half out. 1 half times the integral from zero to pi over 2 of 1 minus 2 cosine theta plus 1 half times 1 plus cosine of 2 theta d theta. This gives me 1 half times the integral from zero to pi over 2 of. I have this 1 half here. I'm going to add that to 1. That's 3 halves minus 2 cosine theta plus 1 half cosine of 2 theta. So we have this expression. And we integrate 3 halves theta minus 2 sine theta plus 1 fourth sine of 2 theta from zero to pi over 2. And that gives us, I'm going to distribute this half across. I get 3 fourths times pi over 2 minus sine of pi over 2 plus 1 eighth sine of 2 times pi over 2, which is pi. Now if I substitute 0 in for each of these theta's, I actually get 0 for this entire expression. So minus 0. And I get ultimately 3 pi over 8 minus 1 plus 1 eighth sine of pi is 0. So that's my area of the cardioid in just that first quadrant. Let's now consider the area of the region inside the cardioid r equals 4 plus 4 cosine theta and outside of the circle r equals 6. With any of these types of problems, we'll need to find the points where the two graphs intersect. One approach is to equate the two equations and solve for theta. But this won't always work, because while the curves may pass through the same points, they may not do so for the same theta values, or quote unquote at the same time. It's always a good idea to graph the curves and further investigate where the curves actually intersect. So let's take a look at the curves. The circle is pictured here and the cardioid here. And I'm interested in the area of the region inside the cardioid but outside the circle. So I'm looking here. So ultimately, I'm looking for the theta values where the two graphs intersect. So let's take care of that first. 4 plus 4 cosine theta, that's my first r, is equal to 6, or my second r. Let's check out when that happens. So I get 4 cosine theta equals 2, or cosine of theta equals 1 half. Now between 0 and 2 pi, this occurs at pi over 3 and 5 pi over 3. While we may want to put these as our limits of integration, we also recognize that the direction of these curves as their graphed as theta increases, these points of intersection don't align with that. Oftentimes what we need to do is rewrite our thetas to align with the order at which the intersection points actually occur in graphing the two curves. 5 pi over 3, we can also express as negative pi over 3. If we think of the theta value as sweeping between negative pi over 3 and pi over 3, that actually aligns with the graphing of these two curves. If we were to integrate from pi over 3 to 5 pi over 3, we'd be looking at areas in this direction. So we have our limits of integration. So here again, we have our little graph. We're interested in this area, where our theta here is negative pi over 3, and theta here is pi over 3. And I ultimately want to find the area within the cardioid but outside of the circle. So the area of the region can be obtained by subtracting the areas of the cardioid and the circle. So I'll take the area between negative pi over 3 and pi over 3 of the cardioid, which is 1 half r squared, which is 4 plus 4 cosine theta squared, and subtract out the part of the circle that we want to eliminate, because all we want to do is be left with this outer piece of the area in the cardioid. So the area of the circle between negative pi over 3 and pi over 3 is the integral from those bounds of 1 half times r, which is 6 in this case, squared. Let's expand this. We have integral from negative pi over 3 to pi over 3 of 1 half times 16 plus 32 cosine theta plus 16 cosine squared minus integral from negative pi over 3 to pi over 3 of 1 half times 36. Now because these integrals have the same limits of integration, I can actually combine them. This is negative pi over 3 to pi over 3 of 1 half times 16. We get 8 plus 16 cosine theta plus 8 cosine squared minus 18. We further clean that up, negative pi over 3 to pi over 3, of negative 10 plus 16 cosine plus 8 cosine squared. Change colors here just to make things a little bit clearer. Negative pi over 3 to pi over 3 of negative 10 plus 16 cosine theta plus 8 times the identity we used in the previous example, which was 1 half times 1 plus cosine of 2 theta. And that gives us integral from negative pi over 3 to pi over 3 of negative 10 plus 16 cosine theta plus 8 times 1 half times 1 plus 8 times 1 half times cosine 2 theta. We're going to clean this up a little bit from negative pi over 3 to pi over 3 of I have negative 10 plus 4, so I get negative 6 plus 16 cosine theta plus 4 cosine of 2 theta. Now the nice thing is we have some symmetry happening. We know that the polar curve, the cardioid and the circle intersect when theta is pi over 3. And negative pi over 3, we all can also recognize that there's some symmetry here. I could find the area of just this upper portion above the polar axis and double it. So I'm going to do that here. I'm going to multiply the integral from 0 to pi over 3 of our function plus 4 cosine of 2 theta by 2. I can't always do this. This one is special because of the symmetry that's involved in this region. And I would get exactly the same result if I were to strictly evaluate this integral. But for purposes of simplicity, we're going to multiply the integral from 0 to pi over 3 by 2. So let's integrate. We have 2 times negative 6 theta plus 16 sine theta plus 4 divided by 2, which is 2 sine of 2 theta. I divided by 2 to compensate for the chain rule that I would use in differentiating sine of 2 theta. This is from 0 to pi over 3. So I have 2 times negative 6 times pi over 3 plus 16 sine of pi over 3 plus 2 sine of 2 pi over 3. And then we recognize I would, of course, using fundamental theorem, I would subtract this function evaluated at 0. But what we notice is that value will be 0 because negative 6 times 0 is 0. Sine of 0 is 0. Sine of 2 times 0 is also 0. So this is our value. And that's going to give me negative 12 pi over 3 plus 32 times root 3 over 2 plus 4 2 times 2 times root 3 over 2, which gives me negative 4 pi plus 36 over 2 root 3, ultimately leaving us with negative 4 pi plus 18 root 3 as the area inside the cardioid but outside the circle.