 Welcome back everyone. In this video, we're going to find the hydrostatic force on one end of a cylindrical drum with radius of three feet if that drum is submerged in the water 10 feet deep. And so you can see right here an illustration of what's going on. We have a cylindrical drum, which if we just look at the side of it, that is a circle for our purposes right here. The radius of said circle is going to be three feet, like so, and it has been submerged 10 feet into the water. So some things that are important to notice here is that while the drum has gone 10 feet into the water, the top of the drum is actually only going to be, what is that, four feet in the water? All right, because the diameter of this circle is going to be six feet there. And if we're interested in the center of the circle, which we'll see in a little bit, but we will be, the center of the circles actually is submerged 17 feet into the water. So that's going to give us some significance to us here. So something to remember about hydrostatic force type problems is that with hydrostatic force, if we're looking for the force, this is going to be the integral of pressure times area, where with pressure pressure is always going to be the same basic thing. We have to take the density multiplied by the depth right here. Now, you'll notice that the units in discussion here, we're talking about feet and pounds. So we don't have to worry about acceleration due to gravity because the measurement of pounds is already a force that's built into it. And so here, the the density is going to be pounds per cubic volume. And that is pounds per cubic foot, so I meant to say here. And so for water, that's going to be 62.5 pounds per, oops, sorry about that, pounds per cubic foot. And then the depth, well, this is going to depend on how we choose our variable X. There's a lot of sense in placing X equals zero at the very top and orienting it downwards. That would be perfectly acceptable and there's a lot of sense for doing that. And this one, I do want to orient to things a little bit differently and some for reasons that will be clear in just a moment. I actually want to choose the center of the drum to be my origin, zero, zero. The X axis will then be horizontally going through it. And then the Y axis will be vertically going through this point as well, as you can see already here on the diagram. Right? So that's going to be our, that's going to be our, what X equals, that's what zero, zero is going to mean in this situation. The advantage of doing that is then we can represent the circle as the equation X squared plus Y squared equals nine. Nine, of course, being three squared, the square root there. If we solve for X, if we solve for X here, we're going to get that X equals the square root of Y minus Y squared. We could equally solve for Y as well and do something from like that. But why we care about that is we're going to be focusing on this side of the circle, this semicircle is going to be of interest to us. Because as we think of a typical cross-section, you can see one illustrated right here. With your coordinate system, this cross-section is going to have a thickness of delta Y. This is kind of getting forward towards, going forward towards the area here, area, because these cross-sections are always going to be rectangles. The area is going to be length times width and the length is still to be determined, but the width is going to be a dy in this situation. Because of the way we set the variable, the thickness of a rectangle is going to be dy or delta X or delta Y if you prefer. So how does one determine, how does one determine the length of this thing? Well, the length, which of course is this value right here, if we use the symmetry of the circle, the length, well, if we take this point X comma Y right here. So X is the distance from the Y axis to this point right here, this is our X value. The entire length, the entire length is going to be a 2X like so. And so making that observation, we're going to get 2X dy. Since we have to integrate with respect to Y, we have to write X in terms of Y, and that's why this observation is critical for us. We're going to get that the area is going to equal 2 times the square root of 9 minus Y squared dy. So that's where we get the area. And so this area function, this length function, it's going to be not that one. I'm sorry, this one right here, this is going to be much more complicated if we chose the coordinates as to be anything other than we did. So this is the main motivator why we chose this one. So coming back to our pressure problem then, we have the density of 62.5 pounds per cubic foot, but what about the depth? So the depth is the distance from the surface of the water to a typical cross section right here. Now be aware that if we go from the X axis up, this is what Y means right here. And so what's going to be this distance above? We have to take Y minus from something, but what is that something? That something can be 7. And the reason is 7 is the depth of the center of the circle. So as we go above the center of the circle, we're going to lose depth, right? So if we go to the very top, this should be the depth of 4. 4 of course being 7 minus 3 at the very top. At the very bottom, the depth should be 10. And notice, of course, 10 is 7 minus negative 3. That is 7 plus 3. And so 7 minus Y is what we mean by the depth right here. And so that's where we get our depth. And that's where we get for not choosing the surface of the water to be zero in terms of our coordinate system. There's trade-offs that come from here, but in terms of integration, it'll be much easier here to utilize the center of the circle. So our force, like we said before, it's the integral of pressure times area. Pressure is going to be 62.5 times 7 minus Y. The area is 2 times the square root of 9 minus Y squared dy. And in terms of integration, we can go from the top of the circle, which is Y equals 3 to the bottom of the circle, which is Y equals negative 3. So negative 3 to 3. And so this is the integral. This is this is the heavy lifting part of this problem. If we can get right here, that's really, really, really good. But we have to also compute this bad boy. How are we going to do that? So notice that there's a negative 3 and a 3. We're very tempted to be like, oh, I could use symmetry to help me out here. And you can't but you can't just use it immediately, right? In terms of the picture, if you go from negative 3, if you go from negative 3 to 3, you can't just you can't just take the top half of the semicircle because the bottom half of the semicircle has dramatically more pressure than the one at the top. The force is not going to be symmetric in that regard. We did use the right side to help us describe the length. The pressure on the right side of the plate is the same as the left side. But we can't use the symmetry on top and bottom here. But we can we still can't use symmetry. You just have to be a little bit more careful. So first of all, let's see. First of all, let's take 62. Well, let's distribute this. That's what I want to do first. So distribute these things onto the Y and onto the 7, the 62 as well. So when we do that, we're going to get 62 times 62.5 times that by 7 times that by 2. We'll compute that in just a moment. That that that's a detail we don't need yet. And then you're going to get the square root of 9 minus Y squared dy. And then that's that that's the first thing we get from this distribution. Minus you're going to get 62.5 times 2 times the integral from negative 3 to 3 of Y times the square root of 9 minus Y squared dy. So let's look at these two integrals for a second. Now, this first one, I want to point out to you, this first one is going to be an even function. That is, this function is symmetric with respect to the Y axis. You can replace, well, in this case, it's symmetric with the X axis, mind you. But if you replace your variable with a negative Y, it'll simplify. This is just the graph on the semicircle. You're going to get even symmetry there. And as such, you actually get 62.5 times 7 times 4 as you integrate from 0 to 3, the square root of 9 minus Y squared dy. You get that one right there. Then with the second one, though, I want to mention that this function right here is actually odd. Right? It's an odd function. We can see this by recognizing you have an even function right here times an odd function. An odd times an even here is going to give us an odd function, right? Or just another idea, if you just replace, if you replace Y with a negative Y, the negative Y squared will absorb its negative sign, but this one will be left out. And therefore, this is an odd function. The critical thing about observing that is that if this is an odd function, odd functions integrated across a symmetric interval are always 0. So no anti-derivatives actually necessary here. We're going to end up with a minus 0. That is a super nice simplification. And so then going forward, going forward from here, I mean, 62.5 times 7 times 4, that's equal to 1,750. Going forward with the integral, if you want to, we could do some type of trig substitution. This is a good one to take Y equals 3 sine theta. And therefore, dy equals 3 cosine theta d theta. And then notice also the square root of 9 minus Y squared, that will equal 3 cosine. And so if you make that substitution, you'll get the integral, you're going to get a 3 cosine squared theta d theta. The square root becomes a 3 theta. And then the dy also becomes a 3 theta. So actually I meant to say 9 cosine squared. If you switch the bounds, right, when Y equals 0, that means sine will have to equal 0, which is 0 right there. When Y equals 3, you can divide sides by 3 in this equation, you get sine equals 1, which happens at pi halves. Like so. In order to do cosine, you're going to have to use the half angle identity. So you get 1750 times 9, integral from 0 to pi halves. You're going to get 1 half, 1 plus cosine of 2 theta d theta. It's anti-derivative. It's going to look like theta plus 1 half cosine, not cosine, sorry, sine of 2 theta. B, and this goes from 0 to pi halves. Notice when you plug in 0, theta will be 0. Sine of 0 is itself 0, so it's going to disappear. When you plug in pi halves, there will be a pi halves that comes out from the theta. And then when you plug it in here, you're going to get sine of 2 pi halves, which is sine of pi, which is 0. So everything's going to vanish, everything's going to vanish from here, except you're going to get this 1750 times 9 over 2, and you're going to times that by pi halves. And so in the end, that's going to give you 7,875 pi over 2, which is approximately 12,370 pounds of force crushing this thing. Now, admittedly, we could have come back a while ago. I want to mention that we calculated this anti-derivative using a trig substitution back here, but I also want to comment that an alternative approach is that if you recognize this integral is really just the area of a quarter circle, right? Like if you come back up to the one above it, this one right here, this is just the, we're just calculating the area of a semicircle, and therefore the area would be one-half pi or squared, that is 9 pi halves, in which case if you multiply that by this friend right here, you'd have got the same thing. So there was a nice little shortcut in that case. I did go through the trigometric substitution, just as extra practice for people watching these videos, because oftentimes we don't detect it, but in this very special case, there are some simplifications you could do to help you out here.