 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 Misaligned. We are going to start in this video lecture 31, which is going to focus on integration and how it's related to parametric curves. This will be a continuation of the discussion we began in lecture 30 and driving some examples from section 10.2 from James Stewart's Calculus textbook. So we talked about how integration is adapted for parametric curves. What about integration? Well, let's think about the types of problems that we did with integration, right? So the main problem was the area problem, that is the area under the curve. How does one find the area under the curve here? Well, in the usual setting, we had basically the following situation. The area a equals the integral from a to b of f of x dx, where y is described by a function of x. But it turns out this is actually how we adapt it to the parametric setting. Don't focus on f of x, focus instead on y. That is, as you think of these functions, right? You have the function, you cut the x-axis into little pieces, right? Then you look at a rectangle right here. The rectangle's thickness is gonna be a delta x, which is determined as a change of x here. The height of it is determined by a y-coordinate. So the area of a single rectangle is gonna be yi star times delta x. Taking the limit here, as we add all these things together, you sum this up, the area will be the integral of y dx, as you go from a to b. And that's what we see right here. The area in the curve is just the integral of y dx. Now, if our function is given by the parameterization x equals f of t, and y equals g of t, then we can substitute y with simply just a g of t, like so. That's a simple switch. How do we deal with the x here? Well, since x is f of t, we could take, by the chain rule, we could take the derivative of x with respect to t. This would give us that dx over dt equals f prime of t. And then, clear the denominators, we get dx equals f prime of t dt, like so. We make this substitution in, and that then becomes this boy right here. We take the integral, the integral of g of t times f prime of t dt. But these were x-coordinates before, we're gonna have to change them to t, the parameters. And so we have to look for values of parameter alpha, which produces x equals a. That'll be inside the f function. And we have to look for a parameter beta, which we put into the x function will give us a b. And so we can integrate, we're just integrating y dx, just adapt y and x into the parametric setting. That's all we're doing. So let's look at two examples of this. So our first example, let's take the area under one arch of a cycloid, which the cycloid remember has the parameterization x equals r theta minus sine theta, and y equals r one minus cosine theta. So to identify one arch, we have to find a parameter, well, we have the parameterization, we have to know the bounds, right? So x equals zero corresponds to theta equals zero, and x equals two pi r corresponds to theta equals two pi. So one arch is a complete rotation of the cartwheel. And so that's the parameterization we're working with. So area is going to equal the integral from zero to two pi r of y dx. But as we parameterize this, we're going to go from theta equals zero to theta equals two pi y is then given by the function above, we're going to get r times one minus cosine theta. And then for dx, we have to take the derivative of the x function, which case we're going to get r times one minus cosine theta d theta. And so putting things together, we're going to have an r square that comes out front. We integrate from zero to two pi. We have to integrate one minus cosine theta squared d theta. Foil that out. We're going to end up one minus two cosine theta. Those ones are ready to go. But then we're going to have a cosine squared theta d theta. We're going to want to use the half angle identity to take care of the cosine squared there, one half, one plus cosine up to theta. And so then if we integrate that thing, we still have the r squared out in front. And so the antiderivative of one is theta. Antiderivative of negative two cosine will be a negative two sine. Antiderivative of one half is going to be theta halves, and we should combine these like terms right here. And then antiderivative of one half cosine of two theta. Sometimes my twos get a little bit of careless here. Let me correct this to make sure no one is confused if you're watching this at home. This is cosine of two theta. And so finding the antiderivative of one half cosine of two theta is going to be one fourth sine of two theta as we go from zero to two pi. Now consider what happens when we start plugging these values in here. So when you plug in, so admittedly, if you take the theta plus theta halves, that's going to equal a three theta over two, just so we're clear about that. And so then plug in the two pi first, right? So when you plug in two pi for theta, the two will cancel on top and bottom. And so you end up with, you'll end up with just a three pi, like so. When you plug in two pi into sine, sine of two pi is zero, so that whole part's just going to disappear, so you're going to go minus zero. And when you plug in two pi into that sine, you're going to get sine of four pi, which is also a zero. So that will disappear. Plug in the zero everywhere, zero in for theta will make it disappear. Sine of zero is zero, and then sine of zero is zero, again. So we're going to minus a bunch of zeros, whoop-de-doo, in which case then this simplifies just to be kind of cool when you think about it, three pi r squared. That's a cute little number when we're nice and done here. And then notice, of course, pi r squared is the area of a circle. That's the area of the cartwheel that was rolling and produced the cycloid. So when you come back up to the picture right here, we claim that this area under the curve is three pi r squared. So if we took our cartwheel down, we melted it. We melted our cartwheel, and we took three of them. So we take one wheel, two wheels off of our hand cart, then steal a third wheel from another hand cart. We melt those things down. That area, we could then repaint the area of this one arch. Kind of cool how that worked out. Speaking of area of a circle, we could also use this parameterization to compute the area of this circle, right? Let's take the circle with the radius r. As we've seen before, this will be parameterized by x equals r cosine theta and y equals r sine theta. So one, the area of the circle, to get the whole circle, we have to do a complete rotation, which will be two pi, or go from zero to two pi, the area here, y dx. But y, we would replace with the y function, which is r sine theta. And then the derivative of x, we're going to get r, negative r sine theta d theta, like so. We can pull out the negative r squared, so we have to integrate sine squared theta d theta. This is very similar to what we did a moment ago. Using the half angle identity again, this time for sine squared, there should be a two right there. We get one half, one minus cosine of two theta this time, like so. And then finding the anti-derivative, we'll just pull out the one half there. And then the anti-derivative, we get theta. We're going to get minus one half, oh, I'm sorry, we should get in that case a, no, that's right, negative one half sine of two theta as we go from zero to two pi. And the calculation is very similar as it was before. Plug in two pi for theta at six round, plug in two pi into sine. You can get sine of four pi, so it's just going to vanish. When you plug in the zeros here, all that stuff will vanish. So we end up with negative r squared over two times two pi, which simplifies to be negative pi r squared, which that almost seems right, just where did the negative sign come from? And that's because these area problems, it depends on orientation, right? If we integrate backwards compared to forwards, you can get a positive or negative. So in terms of area, like we've seen previously, we just want to take absolute values of these things. And we end up with pi r squared, the area there. And that's because if we just integrated it the other way, we get a negative sign because of the orientation. And that's not meant to be such a big deal. Now, I want to mention here that we are able to calculate the area of the curve without using a trigonometric substitution because we've done this exact problem before, except when we did it, we did it the first time, we had to, the area was twice the area of the semicircle, which went from the upper semicircle, this one up here, which went from negative r to r. And we had to integrate the square root of r squared minus x squared dx. And so when you go about doing this one, you have to do a trigonometric substitution. And when you do a trigonometric substitution, you end up getting something that looks a whole lot like this guy right here. It's not exactly the same, or maybe it is, but it's you get basically the exact same thing right here. And so this idea of switching from a function to a parameterization, I want you to think of as really just a, is a type of like U substitution, type of substitution or a trigonometric substitution in this case, this parameterization is switching from the variables x and y into a new variable. And so that's going to feel like a substitution. So that's a pretty cool observation, I think, that doing integrals with parameterizations is really just the substitution method we'd seen before, such as trigonometric substitutions, U substitutions, ratio substitutions, all of those things right there. So turns out we're really not doing anything that different, just a new type of substitution.