 Welcome back to our lecture series, Math 1220, Calculus II for students at Southern Utah University. As usual, I am your professor today, Dr. Andrew Misseldine. This video is the first one in lecture 34, for which we're gonna be talking about integrals involving polar functions. And there's particularly two problems that we'll be interested in. We'll be interested in the area problem and the length problem. It turns out the length problem's gonna be very similar to how we handled tangent lines previously. That is, we're gonna treat the polar function as a parametric function and handle arc length like we did with parametric functions. Turns out area's a little bit differently because the parametric approach for area is pretty horrendous. And so we actually wanna sort of start from the drawing board when it comes to area problems for polar functions in the following way. In the past, we've used integrals to calculate areas, specifically calculate area under a curve, that is between a curve and the x-axis, or we've also found the area between two curves, right? Now in the context of the polar coordinates, there really is no x-axis to be taking the area between. Instead, it's more relevant to calculate the area between the function and the pole. As we can kind of see in this first diagram right here, we're interested in the area that's between the function's curve and the pole. So for example, if we wanna calculate the area between the polar curve, and so let's say this is given as f r equals f of theta here, illustrated as function f, let's find the area between the origin and the function as theta ranges from theta is equal to alpha to theta equals beta. So as you see this illustration right here, we have some interval of theta as we go from alpha to beta. We wanna find the area between the origin and the function. Now if we were to do this, like we've done in the past, what we would wanna do is we wanna take the interval of input, that is theta, and we wanna subdivide it into smaller pieces. Now subdividing theta into smaller pieces would look something like this, where as we see down below, we would wanna subdivide the intervals into evenly distributed pieces of theta. So just like in Cartesian coordinates, we're gonna subdivide the region into approximating quote-unquote rectangles. We wanna approximate this using rectangles, although rectangles might not be the exactly the correct thing to use here, but that's gonna be the idea. We wanna approximate this thing using subdivisions. So we subdivide the interval, so we have this interval from alpha to beta, and we're gonna do in subdivisions, in slices. Well, we want each slice to be equidistant, so we're gonna calculate it delta theta. This is gonna be beta minus alpha over n. Exactly like we've seen before, we want these to be equal slices. And so then we're gonna find the height, so for each of these subintervals, right, so there's some interval like here, like there's this theta i and this theta i minus one, we need to pick a representative, someone in the middle, a delegate to represent us, and we're gonna choose theta i star. And so then choosing theta i star, we're gonna figure out how big is the function at theta i star, well, that would be f of theta i star, or you could think of r i star right here. So r i star is gonna be the representative radius for each interval. But the area between the function and the origin, we wouldn't approximate this using a rectangle. In fact, it looks to be more like a triangle of some kind. We could re-approximate the area using a triangle. And actually a little bit better than a triangle, we're gonna approximate with a pizza slice, right? So let's take a sector of a circle whose radius is determined by r i star. Well, if you wanna find the area of a sector of a circle, that is the area of a pizza slice, you're gonna get that the area is equal to one half r squared theta. So what we mean here is r is the radius of the circle and theta is what percentage of the circle we're using, what angle is going on right here. So with that consideration, the area of the ith pizza slice, this would be one half r i star squared times delta theta, where i r star here would be one half f of theta i star squared delta theta. And we really kinda like this because as we start setting up Riemann sums, this is just one of the pizza slices. We of course have multiple pizza slices, we have to add together to approximate here. So what we're gonna see is that the area, the area under the polar curve will be approximately the sum as i goes from one to n of these approximate pizza slices, one half f of theta i star quantity squared times delta theta. And so then as we take the limit, we take the limit here as n goes to infinity, this Riemann sum will then turn into an integral where we wanna go from alpha to beta, that's the type of there, we wanna go from alpha to beta, integrate from alpha to beta, one half r squared d theta. And so area under a polar curve is a little bit different here, but if we remember this, it's just for a Cartesian under the curve, we get y times dx, it's length times width of a rectangle. We're using sectors here so we get one half r squared d theta. So with that adjustment, then we can actually calculate the area under a curve very nicely. And so this works out really great for the following situation. Take our four pedaled Daisy right here, R equals sine of two theta, what is the area of a single loop? That is the area of a single pedal, what would that then be? So like we saw on the previous screen, we're gonna get one half the integral from alpha to beta, one half r squared d theta, great. So let's first identify alpha and beta, what are the bounds here? We're gonna be going from the x-axis, plus of x-axis, so this is theta equals zero, and we'll be going to the y-axis here, so this is the first quadrant, theta equals pi halves. This is going to be our bounds. So we're gonna integrate, so one half, you can just bring the constant out, we're gonna go from zero to pi halves. Our radius, what's the radius? Well, that's just given this right here, this is our polar function. So we're gonna get a sine squared of two theta, and then we get d theta. So we're gonna integrate this thing with respect to theta. And so this is a trigonometric integral, but we have this skill set to do this. We need to do the half angle identity here, in which case sine squared, as we go from zero to pi halves, we can replace sine squared with one half, one minus cosine of four theta, d theta. And so now this thing is ready to go, the one halves combined, and this is Captain Planet, no, not really, but you get one fourth. Integrating the one minus cosine of four theta, you're going to get one fourth theta minus one fourth sine of four theta as you go from zero and pi halves. Now notice what happens when you plug in the pi halves. We're gonna plug in pi halves for theta, that's just it. When you plug it in for sine, you're gonna end up with sine of two pi, which sine of two pi is gonna go to zero. So you end up with one fourth times pi halves. Great, so there's gonna be a minus zero right there. Then we have when we plug in the zero, when you plug in zero for theta, that's gonna give you zero. When you just plug in zero to sine, you get sine of zero, which is also zero, right? In which case, everything's just gonna disappear. So we end up with just a pi halves when we plug it all in there. Times that by the one fourth and you end up with the area of a single petal will be pi eights. So as long as we use this formula right here, integrating isn't so bad here. It's really no worse than it was like we've seen before, but we can find the area under a polar curve in a very similar way as we can find the area under a Cartesian curve. The main difference here is that we don't measure area of polar functions using rectangles, we're gonna use pizza slices that is sectors of circles.