 In this video, we're gonna discuss the solution to question 14 from the practice midterm exam for calculus two math 12-20. And we have here a base of a solid. It's gonna be a disc in the xy plane with boundary curve x squared plus y squared equals one. So we just stop there for a moment. We have a circle as our base. And then cross-sections perpendicular to the x-axis are gonna be squares. So we're gonna be stacking, you know, if we look at these different cross-sections, we're gonna be stacking squares on top of these things, like so. And so we're gonna build a solid based upon these things. And we wanna set up the integral to find the volume of this thing, do not evaluate. This is just a setup type question, do not evaluate it. I wanna look at this from different perspectives. If I look on the circle from the top, right, we have the x-axis and the y-axis. And a typical cross-section is gonna look like this piece right here, all right? And some important things to notice about this thing is that this cross-section is gonna have a thickness of x. And so we're ultimately gonna wanna integrate this thing with respect to x. Now, if we look at our square from a different perspective, we're gonna get the following. That kind of looks like a square, right? In which case, some things to note here is that if the length of the square is s, s and s, the area of the square is gonna be s squared. The reason this is significant is in order to find the volume using cross-sectional slicing, we're gonna integrate from a to b, the area function with respect to x. Don't forget the dx here, because that's the thickness of our slice times the cross-sectional area there. So we see here that we're gonna integrate, we're gonna be integrating s squared dx. Now, as we're integrating with respect to x, we one could sort of determine what are the bounds of x. The cross-sections can go anywhere from this side of the circle to this side of the circle, are looking from on top, we go all the way over here to here, which the one on the left is a negative one, the one on the right is a positive one. So we're gonna integrate from negative one to one, s squared dx. I should also mention that we can use symmetry here, that the volume of the right side is just gonna be half of the total volume. So we actually could write this as two times the integral from zero to one of s squared dx. And honestly, if you had stopped here, you'd get some good partial credit on this problem, because this is a good analysis so far. But how can we improve upon it? We have to integrate with respect to x. How do we write s in terms of sx here? So something's to note here that while this side right here is s, that's also this side right here. And so if we take a point right here, whose coordinates are x comma y, this distance s, which is the whole thing, this is half of s is gonna be y. And so we see that s is equal to two y, which tells us the area will equal four y squared. That's a great next step. So notice what we now have, is that our integral could be written as, we're gonna get two times the integral from zero to one, of four times y squared dx. Or in other words, we get eight times the integral from zero to one of y squared dx. That's a step in the right direction, but we still have to represent y in terms of x. But that's where the original equation comes into play here. We haven't really used it yet at this moment. x squared plus y squared equals one. If you solve for y, you're gonna see that y equals the square root of one minus x squared. Well, technically there's a plus or minus, but even technically we don't actually need y. We need y squared, which is gonna equal one minus x squared. So making that substitution here, we then get eight times the integral from zero to one of one minus x squared dx. And so this right here then gives us the integral that we're looking for. This integral would then give us the area, the volume of this curve. And so this is one interpretation. If you didn't use symmetry, of course, it instead would look like four times the integral from negative one to one. You still get one minus x squared dx. That part didn't change. So one of those would be the integrals we're looking for here. Again, you're just supposed to set it up. You don't have to compute it. And so one could then stop at this moment.