 Suppose we want to find the volume of a cylinder, well a cylinder we could consider as a solid revolution after all. I mean if we take the x-axis right here and we take some type of rectangle sitting above the x-axis, the height of the rectangle, we think of as the radius of the cylinder and then the length of the rectangle would be the height of the said cylinder right? And we rotate this around the x-axis. This would form a cylindrical type object right? We would get something that looks like the following. Apologize for my crudeness there, but the height of the rectangle of the cylinder will be h and the radius of the cylinder would likewise be r. But what's the problem with this type of reasoning? We can we certainly can do that but when you think about it the disc method and the washer method although this would be disc in this context, the disc method is based upon the fact that we know how to find the volume of a cylinder pi r squared h right? And so it would kind of be circular reasoning, sorry no pun intended there, that in order to find the volume of a cylinder we have to already know the volume of a cylinder. So if we wanted to sort of take a different perspective to find the volume of the cylinder that perhaps avoids this logical problem, it might be useful to think of a circle, sorry to think of a cylinder instead as a stack of circles that are sitting on top of each other. Imagine like you had a long tower of like pennies or coins that are sitting on top of each other. If you put these things on top of each other you can make a cylinder and thus a cylinder is essentially we just take the cylinder and you can slice it into smaller and smaller and smaller and smaller things as well, smaller prisms so to speak. Where this prism you're going to take the area of the circle and then you're going to times it by the thickness which in this case would be like a dx or dy depending on how you orient it. So what I mean here is we can think of the cylinder as a stack of circles that is the cylinder as a continuum of cross-sectional areas. In particular if we subdivide the cylinder into these cross-sectional discs with uniform thickness we can then calculate the volume of the cylinder by integrating this area times the thickness of each of these discs as x ranges from A to B. And if you think about that the area of a circle is typically this pi r squared dx right where for a cylinder the radius doesn't change as you go from one circle to another this would then give us this value pi h squared pi r squared h kind of like so. And so this the strategy of using cross-sectional slicing to calculate the volume of a solid of any kind can be very useful and this also allows us to move away from solids of revolution. So take a look at this picture right here which is taking courtesy of James Stewart's calculus textbook. I think this is from the seventh edition although it's in most of the editions of the textbook. Imagine we have some solid you look at this picture right here there is no rotational symmetry going on here but the idea is whatever whatever the shape turns out to be we can cross-sectionally slice it, slice it, slice it, slice it in the way that we're doing right here in which case we then get this cross-sectional area like this. And we can think of the solid as just a continuum of all of these different cross-sectional slices like if we were to take an apple or a banana we just slice into those little pieces here. The whole together forms original solid but we can study these cross-sectional slicing individually. Now imagine we could come up with a function a of x that could capture the area of each cross-sectional slice. Well then if we took the integral of that area function with respect to the thickness dx that would give us the volume. Now the disc and washer method we've seen before is a specialization of this. For the disc method your area function a of x here is just pi r squared where the radius here is just a function of x. Well we can approach other problems where the area function doesn't necessarily have to come from this rotational this rotational consideration as well. And so let's look at an example of this and yes I'm going to look at a sphere in this situation. A sphere we could find the volume as a solid of revolution but we rotate a semicircle around the x axis. But let's try it with this cross-sectional approach right we could think of a sphere much like a cylinder a sphere is gonna be a stack of circles but the circles will change their radius dependent on where they are in the tower. These ones are positioned horizontally again these are also images taken from James Stewart's calculus textbook and so we can try to make a prediction on what the volume of a sphere would be as thinking of these stacks this this this continuum of circles. You see three approximations of the sphere in front of you right here. There's an approximation with five discs ten discs and 20 discs as well and it dole converge towards the true volume of the sphere as you get more and more and more disc. So we'll take for granted we have the volume of a cylinder which we could do by this cross-sectional slicing like we saw on the previous slide but if we treat this as these cross-sectional circles these discs what would the volume look like? The volume would look like the integral from the integral of ax dx like so and so the idea we have is the following we start off with a circle of radius r circle of radius r like so and we're gonna slice this thing perpendicular to the x axis let's say that the center of the circle is zero zero and so we take these perpendicular slices like this right and so if we were to take any one of these slices and look from a different angle it would look like a circle but if you turn it it would look like it's just way through thin right the thickness here it's just gonna be a dx so what's gonna be the area of this circle the area of the circle is gonna be pi r squared but the radius of this circle depends where it is along the x-axis so if we go along the x-axis here we'll add a specific moment x the y-coordinate up here so you have this point x comma y the y-coordinate is going to give us the radius of this circle so this circle is gonna have an area of pi y squared and so we would integrate pi y squared dx like so and then how do our x values change we go from x equals negative r which is the left side of the circle and then we come over here to the right side x equals r and so now we have to consider how do we represent oh I wrote a lot an r when I meant to wrote a y earlier so this right here should be a pi y squared like so as x ranges from negative r to capital r how do we represent y as a function of x well this is where that semi-circle approach comes into play here if we think of the unit circle as x squared plus y squared I guess it's not the unit circle because the radius is r here if we solve for y we're gonna get the square root of r squared minus x squared like so we're gonna insert that in for y squared and if you're worried that having a square root it's gonna make the anti-derivative difficult while we're squaring the square root so that's gonna disappear giving us the integral from negative r to r we get pi times r squared minus x squared dx and this is the thing that we're gonna try to integrate now for the sake of symmetry because I mean the sphere is symmetrical right if we take the y-axis as this line of symmetry right here the volume of one side is equal to the volume of the other so using that symmetry we can actually rewrite this integral as 2 pi I brought the pi out the integral from 0 to r of r squared minus x squared dx like so basically we're using the fact that we're integrating a cement an even function along a symmetric integral an interval and so if we integrate this the integral of r squared with respect to x is just gonna be an r squared x because we're treating r squared as constant the anti-derivative of negative x squared with respect to x will be negative x cubed over 3 as we go from 0 to r plug it in the r there we get 2 pi we're gonna get an r cubed minus r cubed over 3 factor out the common factor of r cubed we get 2 pi r cubed and we're left behind a 1 minus a third 1 minus a third is 2 thirds and this then gives us the standard formula for a 4 thirds pi r cubed which is again the standard formula for a sphere which we can use we can we found this formula using this idea of cross sectional slicing it's pretty neat isn't it