 Hello, and welcome to a screencast today about finding the volume of a solid of revolution. So today we're going to take the region bounded by the cube root of x, the x-axis, the y-axis, and the line x equals 8, and we're going to revolve that around the x-axis. All right, so I know this is a lot of information and it's easy to get overwhelmed, so let's just go ahead and draw out our region piece by piece. So starting with our function, we've got the cube root of x, and hopefully that's been one of the toolbox functions for you. So cube root of x kind of looks like that, x cubed, but inverted. The x-axis, so that's down here, the y-axis, which is up here, and then the line x equals 8, so that would be right here. Okay, let me color this region in just so we have an idea of what we've got there. Okay, fabulous. Now we're going to be taking this and we're going to be revolving this region around the x-axis. All right, if you're anything like me, picturing three dimensions is not the easiest thing. So what I tend to do, let me grab a different color, is I tend to take my region and I'll reflect it over wherever it is that I'm revolving, so kind of like make it go down here, and then try to make that look three-dimensional. Okay, so it kind of look like that because it's going to be going around that. So we're taking this and we're swinging it around this axis, and then it's going to make this kind of a bowl shape. All right, if my picture isn't good enough here for you or my description, a colleague of mine was nice enough to graph this out for me. So this is what the region is going to look like. Okay, so it's a bowl. All right, fantastic. So now we have to figure out how we're going to slice this thing because we need to be able to take slices and we're going to add up all those slices in order to get the full volume of the solid. Right, so slicing, that needs to happen perpendicular to whatever axis it is that we're revolving around. So in this case, my slice is actually going to look like this circle that's over here. So my slice is going to go just like that. That's pretty good. And it's going to be like a really thin cylinder, a disc, or whatever you want to call it there. So we need to figure out what the volume of that cylinder disc is. So the volume of our disc. And to do that, you know, it's a disc or a cylinder. So it's pi r squared h. So the pi is in there. So the radius of this, I'm going to grab another color. The radius is going to be then, you know, it's half of the size because, you know, it's a circle. So it's going to be half your diameter. So it's going to be what goes from the x-axis all the way up to our function. So in this case, the radius is just our function. Kind of handy. Not always going to be the case just for the record. So the cube root of x squared. Now the height of this, and when I say height, in this case, it's actually the width. So here the width is going to be delta x because we're slicing vertically. So we know that we're going to be doing everything with respect to x. All right, so hopefully this makes sense to you. So we've got the volume of one of our disks. So now we've got to think about what if I were to able to make infinitely many of these little disks starting at this point over here and ending up at this point over here. Well, the idea of that then leads us into an integral. So whenever we do an infinite number of something and we're going to be adding all that stuff up, then like I said, that gives us an integral. So this is going to become the integral. Now how do we figure out our endpoints? Now that comes from where we're starting and stopping. So in this case we're starting at x equals zero. So that's our bottom limit of integration. And we're stopping at x equals eight, our top limit of integration. Right, then I'm just going to do the volume of my disk. So that's going to be pi, the cube root of x squared. And then as always delta x becomes dx in your integral. Okay, now this is actually a fairly easy integral to do. I'm not going to show you the details. I'll trust that you guys can handle that part. But just to give you a little bit of hints on this one, I always factor the pi out. That's just my instinct. And then rewriting this integrand as a power instead of a power and a root, I think is going to be helpful. So that's going to be x to the two thirds dx. So eventually when you're going to do this out, you're going to end up with 19.2 pi. Right, thank you for watching.