Shell method for rotating around horizontal line

What we're going to do in this video is take the function y is equal to the cube root of x and then rotate this around the x-axis. And if we do that, we get a solid of revolution that looks like that. And we're doing it between x is equal to 0 and x is equal to 8. And you get something that looks like this. And you could find the volume of this actually quite easily using the disk method. But just to show you that you could do it an alternate way, we're going to use the shell method. But we're going to use the shell method now to rotate around a horizontal line, and specifically the x-axis. So how would we do this? Well, what we want to do, what you could imagine, is constructing a rectangle that looks like this. Let me do it in this salmon color. So you have a rectangle that looks something like that. Its depth or its height, you could say, is dy. And then its length right over here is going to be 8 minus whatever x value this is. Let me make this clear. The width is going to be 8 minus whatever x value this is right over here. And you might already realize that if this is going to be dy, we're going to be integrating with respect to y over an interval of y. And so we really want to have everything in terms of y. So this x value, whatever it is, as a function of y. So if y is equal to the cube root of x, we can cube both sides, and we can get x is equal to y to the third power. These are equivalent statements right over here. And so, this distance right over here is going to be 8 minus y to the third, if we were to express it in terms of y. And then when you rotate that thing around the x-axis, it's going to construct the outside of the cylinder or shell, as we like to call it. And I'll try my best to draw that shell. So it will look something like this. So you're going to have a shell that looks like this. And there you go. And actually, this shares a common boundary right over here. So you're going to have a shell that looks something like that. So that hopefully helps a little bit. Let me give it some depth. You'll have a shell that looks something like that. If we could figure out the volume of that shell, which is really going to be the area of the outer surface area times the depth. And then if we were to sum up all of those shells, this shell is for a particular y in our interval. If we were to sum up over all of the y's in our interval, all the volumes of the shells, then we have the volume of this figure. So once again, how do we figure out the volume of a shell? Well, we can figure out the circumference of I guess the left or the right, in this case-- the left or the right of our cylinder. If we can figure out that circumference, that circumference is equal to 2 pi times the radius. And what is the radius? The radius of these things are just going to be your y value. That's this distance right over here. That's just going to be the y value. So it's going to be equal to 2 pi times y. And then if we want the area of the outer surface, we just multiply the circumference times the width of our cylinder. So let me write it over here. So outer surface area is going to be equal to our 2 pi y times our 8 minus y to the third. 8 minus y to the third is just this length. You multiply that times circumference. You get the outer surface area. Now, if you want to find the volume of this one shell, it's going to be the outer surface area. 2 pi y times 8 minus y to the third times the depth, times this dy right over here. I'll do the dy in purple. So that's the volume of one shell. If we want to find the volume of the entire solid of revolution, we have to sum all these up and then take the limit as they become infinitely thin, and we have an infinite number of these shells. So we're going to take the sum from, and remember, we're dealing in y. So the volume is going to be equal to, so what's our interval in terms of y? So y definitely starts off at 0, and when x is equal to 8, what is y? Well, 8 to the 1/3 power is just 2. So y is 2, this value right over here. Let me make it a little bit clearer. This value right over here is 2. So y goes from 0 to 2, and we've set up our integral. And this one looks pretty straightforward. So I think we can crank through it in this video. So this is going to be equal to, we can take out the 2 pi, 2 pi times the definite integral from 0 to 2. Let's multiply the y of 8y minus y to the fourth. All of that dy. This is going to be equal to 2 pi times the anti-derivative of this business. Anti-derivative of 8y is 4y squared. Anti-derivative of 1 negative y to the fourth is negative y to the fifth over 5. And we're going to evaluate it at 0 and 2. So this is going to be equal to 2 pi times, we evaluate this business at 2, 2 squared is 4 times 4 is 16. 16 minus 2 to the fifth is 32. So minus 32/5. And then, you evaluate this stuff at 0, you just get the 0. So that's what we're left with. And now, we just have to simplify this thing a little bit. So let's see. 16 over 5, this part right over here. 16 is the same thing, I should say, as 80/5. And from that, we are subtracting 32/5. And so, that is equal to 48/5. So all of this business is equal to 48/5. Did I do that right? 80 minus 30 is 50, and then minus another 2 gets us to 48. So this is 48/5 times 2 pi. And now we deserve our drum roll. 48 times 2 is 96. Let me do this in a new color, just to emphasize that we're at the end. 48 times 2 is 96 times pi over 5. So once again, this is something that you could have solved using the disk method in terms of x. And we're just showing that you could also solve it in the shell method in terms of y. The shell or the hollowed out cylinder method, whatever you want to call it, in terms of y.