 So in this video, we're going to find the center of mass of a semicircle of radius r. Notice that this is a semicircle. If I was looking for a circle, it'd be a much easier problem, but we're going to focus on the semicircle here. So we've learned previously that we can find the center of mass by the following formulas, x bar and y bar look like the following. x bar is one over the area times the integral from a to b, x f of x dx and then y bar is equal to one over the area, the integral from a to b of one half f of x squared dx. Now here the area is the area of the region, which that can typically be found by taking the integral from a to b of f of x dx. As this is a semicircle, I'm just going to use the formula one half pi r squared. The radius is r, so therefore this actually gives us the area here. I don't need integration to do that. I mean, because integration is a very powerful tool, but remember like any tool, we only want to use it if it's the best tool for the problem. And oftentimes that is the case, but in the case of a semicircle, we can find the area much simpler from just standard formulas we're aware of. And so if we apply this to x bar, x bar is going to be one over one half pi r squared. We get that. We get the integral. In this case, we're going to integrate from the left side to the right side. That's where the x ranges. So we get from negative r to r. We're going to get x times our function. Well, since this is a semicircle, we can represent the semicircle by the formula y equals the square root of r squared minus x squared. We plug that in for f of x right here. Like so. And so this is the integral that we want to compute at this moment. Now, I noticed that there are, that we're over a symmetric interval. This is negative r to r. So I'm going to want to be looking for, do I have a function that is even or odd? Well, this semicircular function here, the square root of r squared minus x squared, if I replace negative, if I replace x with negative x, we'll see that that part's even, but this part's odd. And when you combine those together, the total is an odd function. What this tells me is that this will equate to be zero. And so I don't actually want to go find an anti-driven. We're going to see this as zero just by using the symmetry here. And that kind of makes sense that given the semicircle we see over here, it does have a line of symmetry, which is the y-axis. And one thing that you should know is that centroids always lie on lines of symmetry. If a region has a line of symmetry, that means the mass on one side is equal to the mass on the other side. And therefore, the axis of symmetry will contain the centroid there. So because the y-axis was a line of symmetry, I actually didn't have to do any calculation whatsoever. I would have known from using the line of symmetry that x bar was equal to zero. And that was a consequence of this thing being an odd function. y bar, on the other hand, is a little bit more complicated because we don't know, we don't know, we don't have a horizontal line of symmetry. That's what I'm trying to say here. So we have the one over the area, which is going to be two over pi r squared. We're going to go from negative r to r. And then we have one half f of x squared. So our function y equals r squared of r squared minus x squared. We're going to square that, which is actually quite fortunate for the situation dx right here. There's a nice cancellation here. The one half here cancels with the two right there. And the square cancels with the square root. So we end up with, with those simplifications, one over pi r squared, the integral from negative r to r of r squared minus x squared dx. So that's a pretty good place to be. Also, symmetry does apply here that this time we have an even function that as you integrate from negative r to r, r squared minus x squared, there's the same amount to the left as this to the right. And I'm not referring to this picture over here. I'm referring to this function specifically. If you replace x with a negative x, you'll get the exact same function here. And so by symmetry, we can simplify this integral to look like two over pi r squared integral from zero to r of r squared minus x squared dx. So integrating our antiderivatives, we're going to get r squared x minus x cubed over three as you go from zero to r. Plug it in zero makes just everything disappear, which is really nice. Plug in an r cubed. So we get two over pi r squared. We're going to have an r cubed here. Sorry, we're not, we're not plugging in an r cube. We're plugging in an r, but some r cubes will emerge. You'll get an r cubed minus an r cubed over three. If you factor out the r cubed, we're going to get two r cubed over pi r squared, which is nice. And then one minus a third. There are square of the bottom will cancel with most of the r's right there. And then if you take one minus a third, that's a two thirds. So we get two over pi, sorry, two r over pi times two thirds. And so in the end, the y-coordinate y bar of the center mass should be four r over three pi, which we can write that as a percentage of the radius r. We found it as a formula. And if we come back up here, this diagram here is drawn on the scale, which whatever the radius turns out to be, the center mass would be this location right here.