 Let us do another example of finding the volume of solid using the technique of cross-sectional slicing. And this time we're going to consider the volume of a pyramid. So we want to find the standard volume formula for a pyramid whose base length is this number L and the height of the pyramid is H. As you can see right here, just to make things a little bit more clear in terms of the diagram here, which this diagram is taken from James Stewart's calculus textbook, the height is this dimension right here and then the length is going to be this dimension right here. So this is our L value. You can see then a two-dimensional diagram of this thing right here. Now, this is going to be very similar to how we tried to compute the volume of a cone that we did in a previous lecture here, but we can't use the technique of solids of revolution. The washer disc method wouldn't be appropriate because there's no type of rotational symmetry here. But what we can see, like the diagram illustrates, is that we can't think of this pyramid as a stack of squares where the squares, these cross-sectional squares, are getting smaller and smaller and smaller as you move from the base of the pyramid up to the apex over here. Now for convenience, I'm going to align my pyramid so that the apex coincides with the origin 00 and then the height of the pyramid is aligned with the x-axis where the x-axis is this altitude of the period. It passes through the center of all the respective squares that we see there. So if we want to calculate the area of this thing, the area is going to be the integral of the, sorry, not the area of this thing, official JK right there, my goof. We want to find the volume of this thing. The volume is going to be the integral of the area times dx. And so let's first figure out what bounds do we have for the x-coordinate. Well, like we said over here, it goes from the origin x equals 0 all the way up to this value right here, which is x equals h. So we're going to integrate from 0 to h. For a lot of these volume problems, we can actually position the solid in such a way that the geometry is simplified, like the reason we placed the apex at the origin is so that we could integrate with respect to 0 here, which would be very desirable from an arithmetic point of view. Well, now continuing on, the area of each of these cross-sections, we have to find the area of a square. And so if we, much like the previous example we've seen, if we want to find the area of a square, let's say that the side length of one of these squares is s, then the area is going to be an s-squared, but we have to represent the side length s with respect to the variable x, so because we're integrating with respect to x. And that's where this two-dimensional picture comes into play here. Look at a typical cross-section. This is the side of our square. Now if we look at this, and we look at this point right here, it has coordinates x, y. y is given by some generic x here. The side length s is going to equal two times the y-coordinate. So that's a good step in the right direction. Our volume is going to look like the integral from 0 to h. We're going to get two y-quantity-squared, so we get four y-squared dx, but we still have the same predicament. How do we represent y in terms of the variable x? Well, if we consider this side length right here, this is a line which would have the form y equals mx right here. Typically, it's y equals mx plus b, but as it goes through the origin, the y-intercept is just 0. So let's identify the slope of this line right here. The standard slope formula comes into play y, or m equals y2 minus y1 over x2 minus x1. We're basically just making a proportions argument, a similar triangle type argument, that the square will get proportionally bigger the farther away it gets from the apex. So we'll use two points we know on this line. There's the origin 0, 0, and then there's this point p right here, which would have the coordinates, x-coordinates h, and the y-coordinate would be half of l, right? Because the entire thing is l, so we're gonna get half of l, l over 2. And so putting those things in there, we're gonna get 1 half l minus 0 over h minus 0. That simplifies just to be l over 2h, what you see right there. And so then putting that together, we're gonna put this l over 2h in for y right there. So plugging that in, the 2 on the bottom was gonna square to become a 4, it'll cancel with the 4 that's in the numerator. And so we end up with the integral from 0 to h. We're gonna get an l squared over h squared when you plug that thing in and simplify. Like so. Oh, I forgot the x, I'm sorry. I knew something was missing, my spider sense was tingling. When you plug in, because this right here, the l over 2h was just a slope, we plug that back into the function y equals l over 2hx. You plug that in, so you get an l squared on top, you'll get a 4h squared on bottom and x squared as well. So here we have the following. The l squared over h squared is a constant, we can factor out of the integral. We're left with just an x squared. The antiderivative of x squared is going to be, let me write this first, we're gonna get x cubed over 3, plug in 0, plug in h. And so we end up with an l squared h cubed over 3h squared. There's some h's on top and bottom, they cancel, so h squared cancels with this right there, leaving just 1h in the numerator. And then this reproduces the traditional pyramid volume formula. We're gonna get the volume of a pyramid as 1 third, the length squared times its height. Which notice the length squared, this is just the area formula of the base. And so for, honestly, what our exercise shows us is that if we take any pyramid, right, you could have any blobby blob base you want, but if this comes to a point, this gives you this pyramid, and then we get that the volume of this pyramid is gonna just be whatever this is right here, your volume will just be 1 third the area times h, where h is the height of this pyramid right here. Which if you think of this more generalized pyramid, right, because you have like square pyramids and triangular pyramids and what have you, you can think of a cone, actually, as a special case of this as well, where the area of the circle at the bottom, right, is pi r squared, and then it comes to a point where the volume of a cone is 1 third pi r squared h. And so this technique of cross-sectional slicing is a little bit more challenging, maybe it's a lot more challenging than it knows, but it is a more challenging approach to appreciate and understand how to stuff these integrals, but has much wider reach than the technique of the solids of revolution that we had seen in previous parts of this lecture here.