 In this video, we provide the solution to question number 11 for practice exam number one for math 12-20, in which case we're gonna have to calculate the volume of a solid and the solid is described in the following manner. The base of a solid is a disk in the XY plane with a boundary curve of X squared plus Y squared equals one, so that's the unit circle. Cross sections perpendicular to the X axis are given as squares. You need to set up the integral to find the volume of the solid, but we do not have to evaluate the integral. So let's try to imagine what's going on here. We have this circle, right? And then we're gonna be stacking the squares on top of the circle. And so this is what a typical cross section would look like, okay? So if we look at it from one perspective, let's add in the X and Y axes. And then our circle, I know I'm gonna do a bad job drawing my circle, that's good enough. We get the unit circle, so this is X squared plus Y squared equals one. Our cross sectional squares are supposed to be perpendicular to the X axis. So they're gonna go up and down like so. So this is what a typical cross section would look like. If we look at it, there's a little bit of thickness to it. The thickness is gonna be a DX. If we pull it out and look at it front on, it's a square for which, given the sides of the square, the area of the square is gonna be A squared like so. So as we calculate the volume of the region, we're gonna integrate this area function S squared and we're gonna do it with respect to the DX, like so. We then can ask ourselves, what are the bounds of this calculation here? Well, X can go all the way to the far left of the circle that would happen at X equals negative R. The rectangles can also come all the way to the right hand side of the circle, which would coincide with the X coordinate of X equals R. So that's the integral that we're trying to compute right here. But in the present form, I couldn't really do much with that. I can't integrate S with respect to X. So I need to improve upon that. Now, one thing we can utilize is symmetry here because the volume of the solid to the right of the Y axis is gonna be half of the total one. So I could rewrite this as two times the integral from zero to R of S squared DX, which again, that's a very optional step here because I don't actually need to evaluate this thing. And I should also mention that I'm doing this sort of generically where the radiuses are. Specifically though, this circle is a circle with radius one. So I can actually do better than that. This would be one and this would be negative one, like so. And so with regard to our integral, we can put in a specific bound. Well, R is good for this one, we need to have the specific bound of one right there. So how do we deal with the S squared? Well, when we look at the square, the side length, this right here is an S. So if I were to make that label on my diagram right here, this is equal to S. But the distance from the X axis to the top of the circle, that's just the Y coordinate of this point right here. So in all reality, the top half is Y, the bottom half is likewise Y. We get this observation that S equals two times Y. So if we plug that into our integral, we end up with two times the integral from zero to one of two Y squared DX, which is the same thing, of course, as two times the integral of four Y squared DY from zero to one. You can combine, of course, the two and the four together to form an eight. So eight times the integral from zero to one of Y squared DY, excuse me, DX. Not sure where that DY came from. Let's fix that up here as well. Make it accurate. So this is the other thing. So we are still integrating the spectra X, but now we have a Y. The S gives the geometric area for a square, but now we are able to change it from the dimensions of the square to actual a coordinate in the plane here, the Y. We can do one better, and that's where this equation is also gonna come in play again. If I solve this equation for Y, notice that you end up with Y squared is equal to the square root of one minus, sorry, one minus X squared. Then when you take the square root, you're gonna get Y equal to square root of one minus X squared. There is a plus or minus, but we actually, by symmetry, are only working with the right-hand side of this semicircle here. But honestly, it doesn't even matter since I don't even need Y. I actually need Y squared. Y squared equals one minus X squared. So when you put that together, we're gonna get eight times the integral from zero to one. The Y squared becomes a one minus X squared DX, and therefore, this is the integral we're looking for. We don't have to evaluate it. Not that it's too hard to do, but the instructions say we don't have to evaluate it. We get that. Now, of course, if we didn't use the symmetry from beforehand, we would get that the volume is equal to four times the integral from negative one to one of one minus X squared DX. That would also be considered as an acceptable solution.