 In this video, we're gonna conclude lecture 31 with a very short example about how other things can adapt, other integrals can adapt to the parametric setting. In the previous video, we saw how we can adapt arc length to the parametric setting. And if we can do that, turns out we can do area of a surface of revolution because surface area of revolution depends on arc length. If we can adapt the arc length, we're good to go. You just have to take two pi times the radius that you're spinning around your axis times the arc length. And we know how to adapt arc length here. Arc length ds is gonna equal the square root of dx over dt squared plus dy over dt squared dt. We're gonna use that here. So can we find the surface area of a sphere whose radius is given by r? We wanna prove that this is equal to four pi r squared. So the surface area of a sphere is just four times the area of the great circle. Now, how we're gonna do that is we're gonna take a semicircle and we're gonna rotate this semicircle around, rotate the semicircle around the x-axis right here. And so this is gonna be the circle given as x squared plus y squared equals r squared or we could do this as y equals the square root of r squared minus x squared. Now, that's how we do it with Cartesian coordinates, but if we wanna do this with parametric coordinates, we wanna use the parameterization that x equals r cosine of theta and y equals r sine of theta, like so. And to rotate the semicircle here, we just have to allow theta to range from zero to pi. That'll give us this semicircle that's oriented in this direction right there. And so the radius, as we rotate this thing around, think of a typical cross-section, the radius here will be the y-coordinate, which is given by this function right here. So as we start setting up the formula for surface area, s equals the integral two pi, we're gonna integrate from zero to pi with our parameter. Like I said, the radius of revolution here is the y-coordinate, which is gonna be an r sine theta. And then we have to do ds, which ds, like from the formula above, we're gonna take the derivative of the x-coordinate, which is going to be, and we're gonna square that thing, the derivative is gonna be negative r sine theta squared plus the derivative of the y-coordinate, which is gonna be r cosine theta squared d theta. I can fit that all in there. And so let's try to simplify this thing. Now this quantity we actually saw in the previous video, this thing right here is actually gonna turn out to be a constant. You're gonna have an r squared times sine square plus cosine square, that sine square plus sine square is a one, so you're gonna get the square root of r square, which is just an r, all right? So if you bring that out with the other r, that's right here. We end up with a two pi r squared out in front as we integrate from zero to pi of sine theta d theta. That thing simplifies very nicely in this context. Anti-derivative of sine, we get negative cosine theta as you go from zero to pi. And so you're gonna end up with negative negative one minus a negative one. Let me double check that. So when we plug in the pi, cosine of pi is negative one, like so. And so you're gonna get a negative in front and then you're going to get, you have a negative in front, so you get negative one, you get there, okay? So then the next part is when you subtract, you're gonna get a one right there, but it's negative cosine, so you get negative one right. So, and so this will all turn out just to be a positive two. So we get two pi r squared times two, and thus we get four pi r squared, like we predicted it would be. And again, these type of integrations, the integrals we've been doing with these parametric coordinates, with like these circles, and it's like we did areas of a circle, circumference of a circle, we rotated a semicircle to get the surface area of a sphere. All of these ones we could have done earlier, but it would have required a trigonometric substitution. And trigonometric substitution is just a special type of parameterization. We're gonna parameterize using trigonometric functions that we can predict are gonna be useful. All we're doing now in this section, section 10.1 and 10.2, is just taking more general parameterizations besides the trigonometric substitutions we saw before. Parametric equations are really, really cool, and we can do some, we can do the same type of characters problems with them. And like I said before, this is very much a precursor to what one does in multivariable calculus, because these parameterizations are just functions which take a single real variable input and they output two real variables. We take in t and we output an x comma y. In multivariable calculus, we can adapt this. We can take a single parameter in and we could take three parameters out, x, y, and z, but we could also take multiple parameters in. We could take like an s and t that comes in that outputs as an x, y, z or something. So this would take two variables in, three variables out. And it turns out the same analysis with integrals and derivatives can be very, very similar in that setting.