 Welcome back, everyone. In our previous video, we saw a quite long derivation of the surface area formula. It's important we understand where this thing comes from. But it's also now important we see how we can use it. It turns out using it's pretty benign compared to some other situations. So what we're going to do in this example is take the curve y equals the square root of 4 minus x squared. So be aware that this right here defines a semicircle in the upper half plane of the xy plane. But let's only go from negative 1 to 1. This is a semicircle of radius 2. And if we just go from negative 1 to 1, notice we're only going to get part of the semicircle. We just get a portion of the semicircular arch. And if you spin this thing around the x-axis, that's what our goal is going to be right here. You make this band. It kind of looks like a ring to me. And so we want to find the surface area of this ring that we formed by rotating the arc of the circle associated to x squared plus y squared equals 4. Alright, so that's what we're trying to do right here. Now be aware that I want to just mention that this image here is taken courtesy of James Stewart's calculus textbook. The surface area formula, as we saw before, is going to be the integral of 2 pi r ds, where ds is our arc length formula. We have seen before that arc length ds, it'll look like the square root of dx squared plus dy squared, which can be easily adapted for dx or dy, whichever you prefer. That's actually why I like to write this formula in this manner, because we can adapt ds however we want. So if we think of a typical slice here, so we would wrap this thing around the x-axis and make little bands like this. So thinking about a little slice, because again, we would picture this as a bunch of little line segments that we approximate this thing. What would be the radius of one of these sections right here as we rotate around the x-axis? A typical radius r is actually going to be given by the y-coordinate. y, which we saw before, y, which is our radius here, was the square root of 4 minus x squared. Because we can describe the y in terms of our radius, since we can describe the radius in terms of y and the y is given as this, this does kind of tell us that maybe we want to integrate with respect to x. Particularly the thing that really tells me I want to integrate with respect to x are these bounds right here. x sits between 1 and negative 1. That kind of informs me that I want to integrate with respect to x in this calculation. So the surface area is going to equal the integral from negative 1 to 1. We're going to get 2 pi y ds. And like we saw before, these are my x-coordinates, x equals 1 to x equals negative 1 there. So we get negative 1 to 1, 2 pi. We're going to get the square root of, well I'm not going to be able to fit all that in there. Let's move this down below. We're going to get the integral from negative 1 to 1, 2 pi, the square root of 4 minus x squared. Now we're going to use the format of ds that's helpful for here to integrate with respect to x. We're going to use the format to take the square root of 1 plus y prime squared dx. So what we have to do is we're going to take the square root of 1 plus, so we need to figure out what is actually the square root of y right here, not the square root of y, what's the derivative of y. So using this formula right here, y prime, y prime by the chain rule, you're going to get 2 times the square root on the bottom, 4 minus x squared. Then the derivative of that, the derivative of the inner function, 4 minus x squared, negative 2x right there, the force, not the force, the 2 cancels out, 2 cancels out there. You have a negative sign and this is what we need to square. So we're going to square that. We're going to get an x squared on top, and then when you square the square root, you get a 4 minus x squared on the bottom. This is what we get right here. Now when you're looking at this thing, you might be like, but Professor Messedon, you said that surface areas are a lot easier than arc lengths. This looks intense. Well, maybe at first, but let's do a little bit of algebraic simplification. It turns out that's going to be our saving grace right here. This one turns out to be really, really nice. I promise you that. So looking at this, we're going to rewrite this one as a fraction. And so the fraction we're going to get here is we're going to rewrite this as 4 minus x squared on top, 4 minus x squared on bottom. We then add to this the second one, x squared over 4 minus x squared. This all sits inside the square root now. So what you're going to notice is the following is that the negative x squared will cancel with the x squared. That leaves us now. Let's look at this. This simplification is going to be important for us. We're going to have 2 pi, the integral from negative 1 to 1, the square root of 4 minus x squared. We're then going to get the square root of 4 over 4 minus x squared, dx. In this situation, now we have a square root of 4 minus x squared. They cancel with the square root of 4 minus x squared right there. We now have a square root of 4, which itself is just 2. And so then we're going to end up with 2 pi times the integral from negative 1 to 1 of 2, dx. Told you, this wasn't so bad at all. Integrating this thing, you end up with a 4 pi times x as you're going to get from negative 1 to 1. This will give us 4 pi and then you get 1 plus 1. And so in the end, you end up with an 8 pi as our final result. So yeah, the integral did look quite intimidating. But like I said, the r interacting with the dx actually can lead to a simpler calculation. And this example was chosen because it did simplify so beautifully here. So what looked like a really intimidating integral, if you just have faith in yourself that you can do it, and you also have faith in me that I'm not going to give you examples that are just too insane, right? I think you can do these calculations, I promise you that. We'll do another example in the next video. Take a look at that.