 We'll come back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I will be your professor today, Dr. Andrew Misseldine. In this lecture, which is the first video of a couple of lectures here for lecture four, we were actually going to start our conversation about using integrals to calculate volumes of three-dimensional solids, right? And so consider the following diagram that you can see in front of you right here. So this image right here is taken courtesy from James Stewart's Calculus textbook here, section 6.2 about volume. And so imagine you have a two-dimensional region that you see maybe right here, right? So we have some function in this example. We're taking the function y equals the square root of x, but we have any two-dimensional region we want. We're looking at the region that's below the curve here. And let's make it go from x equals 1 to x equals 0 to 1 here. And so we get this slice. And so imagine we take this two-dimensional region and we were to rotate it around the x-axis. That is, allow it to spin through three-dimensional space. And what would happen, what type of shape would we create if we allowed this thing to spin? And what would the span of such a thing be? We'll try to visualize that spinning that goes on in your mind right there. This square root region on the left would spin and make this spice drop-like shape. You know, these type of things we like to get at Christmas time, right? The greens and red ones. We would get this three-dimensional object that's created from spinning this thing around. Another example you can kind of think of is if you have like a coin of some kind, a penny, a quarter. It looks, if you look at it straight on, just like a circle, but if we were to spin it on a table or on the floor, the blur as that coin is spinning would make what would look like a spherical shape of some kind. And so this is the topic we want to talk about in this lecture here today. These ideas of solids of revolution. Could we take a two-dimensional region like one illustrator here, spin it around the x-axis or the y-axis or whatever axis we want to, let's spin it and form these solids of revolution. And so this illustration is just an example of the more general principle that we want to explore right here. So let me bring up the proposition and give some explanation onto what we mean by this so-called disc method. So suppose we have some function y equals f of x, like so. And the graph of this thing might look something like the following. Here's our x-axis. Here's the y-axis. And maybe f looks something like the following. So this is our function f of x. And so we're going to pick two numbers in the domain of this function, the one on the left we're going to call it a and the one on the right we're going to call it b. So then the function f, the x-axis, and the lines x equals a, that's a vertical line, and the line x equals b, this forms a region in the plane. So we're talking about this region right here, bounded between f, x-axis, a and b there, like so. So we've seen how to use integrals to find the area of such a region. We've done that in the past. Now what we want to do is we want to suppose that this region is rotated around the x-axis. So we spin this region. What type of object would we get in three dimensions? This so-called solid of revolution. Well, it turns out we can use the integration to find the volume of this solid revolution we just created. And the way that we approach it is going to be very similar to how we found area of this same region. So what we're going to do is we're going to take the domain a to b, and we're going to subdivide it into smaller pieces, in which case we have an x1, an x2, all the way up to some xi minus 1, some xi, continue up to b, right? And as always, b is just xn, and a is just the number x0, like so. So we've subdivided this thing into little slices right here. So when we consider the area problem, we would then try to form a rectangle using these thicknesses. And to make life easier, we're going to make all of these tick marks, x1, x2, x3, x4, whatever. We're going to make them the same distance away from each other. So we're going to take that distance to be delta x. Well, delta x has the same meaning as it did previously, right? And so think of this in the same way that we would be when we're trying to approximate the area under the curve using rectangles. But this time, if we put a rectangle in there, so we might think of a rectangle like this, we're approximating the curve using all these different rectangles. Well, the thickness of this rectangle is going to be delta x. The height of this rectangle is going to be some value f of xi star, right? So xi star is just a representative who lives inside of this interval, xi star. And now we're going to consider what happens if we rotate just this rectangle around the x-axis. Well, if we allow the rectangle to spin, if we allow that rectangle to spin, try to imagine what's going to happen. If you allow that rectangle to spin, you're going to form this object that we call a disk. So that's a poor drawing. Let me try this one more time. Something like this. Essentially, it's a cylinder. Oh, that's a horrible drawing. I'm sorry. It's a cylinder. I mean, we take that rectangle, we spin it around the x-axis, you form this cylindrical shape. And so that's kind of what you want to see from this figure from above, right? If you take the rectangle associated to some specific x-value and you allow that to spin around the x-axis, you're going to form this disk-like shape. And this disk right here is none other than just a cylinder. And so one thing that's going to be important to us is that we need to know how to calculate the volume of a cylinder. Now, the standard volume formula for a cylinder follows volume equals pi r squared. So you look at the area of the circle associated to that cylinder and then you times it by height. How tall or how thick is that cylinder? Now, in our purpose here, the pi is just a constant which will stick around for the whole journey. r here is the radius. The radius of the cylinder, this value right here, this is none other than the height of the rectangle that we see over here. And so that radius would then be f of xi star, the height of the rectangle and we're going to square it. And then the thickness, the thickness of the rectangle, sorry, the thickness of the height of the cylinder of this disk would be the thickness of the rectangle, which as we saw before, that's none other than delta x. And so the volume of one of these disks, volume i, the volume of the ith disk would be this pi f of xi star squared times delta x. And so that gives you the volume of a single cylinder. But what happens if we start doing this for every single interval that's in consideration? We take this rectangle and we spin it. We take this rectangle and we spin it. We take this rectangle and we spin it. And we get all of these, all these different disk stacked on top each other, in this case stacked side by side by side. Each of those disk would have a volume, which is pi f of xi star squared times delta x. This quantity you see right here, this is a single volume, vi. Then in order to get the approximate volume of the solid of revolution, we would take the sum of all of these volumes, the volumes of each of these disk individually. We add these together. Now just like with the area problem, this only gives us an approximation of the volume of the solid. To improve the approximation, we take more and more and more subdivisions. We take thinner and thinner and thinner rectangles. And what's the best choice? Well, if we allow in to go towards infinity, realize that we take this limit, then this Riemann sum right here, when we take the limit would become an integral. And that integral would give us the true value, the true volume of this solid revolution. And so this right here is commonly referred to as the disk method for finding the volume of a solid of revolution. You take the integral from a to b of pi f of x squared dx. Now f of x here is taken on the role of this f of xi star. When your subdivisions get infinitely thin, we no longer need to choose a delegate because the only number in that infinitely small interval would be x itself. You can also notice that the delta x, which is the thickness of our disk, becomes delta dx, this infinitesimally small slice there. It's super small. And it is part of the dimensions of this consideration here. Now one could try to memorize this integral when it comes to the disk method. That's perfectly great. But another way of just thinking about it is just coming back to the volume of a disk. The volume of a disk is just the pi times the radius squared times its thickness. These things might adapt. This formula, which only works in specific situations, isn't as useful as recognizing this formula you see right here. Pi times radius squared times thickness. And we take the integral of things that look like this. We take the integral of this and that'll give us a method of calculating the volume of these solids of revolution. So let's do an example of this exactly right here. Let's find the volume of the solid revolution formed by rotating about the x-axis, the region bounded by the line y equals x plus one. And then also the coordinates x equals one and x equals four. So you see the region, the two-dimensional region that we're trying to rotate right here, this right trapezoid you see right here. We want to rotate this around the x-axis. And so if we did that, it would form this kind of looks like a volcano, right? This is often referred to geometry as a frustum. But we take this trapezoid and spin it around the x-axis, we're going to get this like volcano-like shape. What would be the volume of said volcano? Well, based upon the disk method we saw before, the idea is you slice this volcano into all these little pieces. So you have one slice here, another slice here, another slice here. And these things are not circles, they're disk, but just they're so thin, we can't really tell much of the difference because it's like looking at a piece of paper. It's thickness is super, super thin, even though the other dimensions are much, much larger. So we want to calculate the volume of this frustum here. And so we're going to apply the disk method we saw on the previous slide here. So the volume is going to equal the integral. Well, what are our bounds? This thing is going to go from one to four. It's given to us. We're going to get pi times the radius. The radius of a generic disk is what we're looking for right here. Now in the diagram you see this orange rectangle that's indicated here to show you what a representative cross-section would look like. As we slice this thing open, one of the cross-sections would look something like this orange rectangle. And so for the disk method, the radius of the disk is going to be the height of this rectangle, which is going to be our function f of x. So we end up with this x plus one squared, remember to square the radius. And then the thickness of a disk is dx right here. So this right here will give us the integral that calculates the volume of this solid. And so we get pi. Pi is just a constant. So oftentimes we just take it out of the integral. In order to integrate x plus one squared, we could try maybe like a u substitution or we could just do, we could just foil it out. There's a couple of different options. You could just foil this thing out and you end up with x squared plus 2x plus one. You can integrate something like that. You could also try some approach of u substitution where you take this x plus one and you would take u equals x plus one, hence du equals dx. Now, if you do that, you're going to want to change the bounds, right? Because when x is one, that actually would mean u equals 2. And when x equals 4, that would mean u equals 5. So whatever approach, those are two natural approaches to this. So when I'm just going to take the foil approach here, so we take this x squared plus 2x plus 1 dx. I mean, I already multiplied it out. It seems kind of straightforward here. And so my antiderivative would look like x cubed over 3 plus x squared plus x as you go from one to four. And so plug in the numbers one and four. And I'm just going to keep the pi out in front of everything. I'm not going to bother distributing it through here. So we're going to end up with 64 over 3. 64 is 4 cubed plus 16 plus 4. And then we're going to get minus a third, minus one, and minus one, like so. So trying to combine these things together. Notice if we take 64 thirds and subtract a third from that. That ends up with 63 thirds, which is nice because 63 is divisible by 3. 16 plus 4 is a 20. And then minus 2, we have to consider from that. And like I said, 3 goes into 63 21 times. So we end up with 21, 20 minus 2. So 21 and 20 is going to be 41 minus 2 is 39. So we end up with 39 pi as the volume of this region. So take a look into our next video. We're going to do some more examples of calculating volumes of solids revolution using this disc method. I'll see you then.