 In this video, we are gonna discuss the solution to question nine from the practice midterm exam in Calculus II, Math 1220. And we have in front of us a hydrostatic force problem. A trough is filled with a liquid of density of 840 kilograms per meter cubed. That feels like a lot. What is this, like mercury or something? I don't know. So we have a trough that's filled with a liquid of density, 840 kilograms per meter cubed. The ends of the trough are equilateral triangles. So this is actually gonna form for us our plate for which it's gonna be submerged in this liquid. It's not water, but honestly, the fact that we're using a different liquid other than water will change the density, but it will make much of a consequence other than that, right? So our density value here we see is gonna be rho is 840. I'll show you how that compensates in a moment. But we have equilateral triangles and the vertex is at the bottom, like you see right here. And we see that the side lengths are eight meters long. So it's eight meters across the top. It's eight meters diagonally right here as well. So we wanna set up the integral for the hydrostatic force on one end of the trough. And we're gonna use acceleration for gravity to be 9.8 meters per second squared. This is a setup, do not solve the integral type thing. So when it comes to hydrostatic force, the guiding principle you wanna use here is that the hydrostatic force is gonna equal the integral of pressure times the area of a cross section where a typical cross section is gonna look something like this right here. Now, because it didn't say anything about a variable, we can define our variable to whatever we want. And I think the easiest setting is always gonna be set X equals zero to be the top of the water. This thing is completely filled here. It said the trough is filled. So it's all the way to the top. So start at the top X equals zero and point the positive X direction downward. What this accomplishes for us is that we'll note pressure is gonna equal our, it's gonna enter this, the pressure is gonna equal this delta times the depth D here where delta is these constant coefficients that show up here. It's gonna incorporate the density of the liquid which is 840 kilograms per meters cubed. It's also gonna incorporate the acceleration due to gravity. So delta here, delta is gonna equal gravity times the density row, right? This is when you use scientific units and if you're using British or American units, you don't need to multiply by the acceleration due to gravity because in that situation, you would have pounds per cubic foot and pounds is already a force. It's not a mass which has a subtle difference here. So you only have to multiply by 9.8 for scientific units. And honestly, because this is not a physics class I don't expect you to memorize this number. I'll actually tell you to multiply by 9.8 basically. If you don't see the 9.8, don't multiply it. And so getting back to pressure here, you're gonna have 9.8 times 840 times the depth. Well, if you set X equals zero to the be the top of the water and you point X downward, then depth will just equal X, which is really nice. Depth will just equal X. And so this gives us our pressure right here 9.8 times 840 times X, which if you wanna multiply those things together, you can, I don't need you to. If you leave a fact, it's perfectly fine. This gives us the pressure. Now in terms of area, right? How do we deal with the area? This is actually the hard part of the problem. Well, we have a rectangle over here which is our cross-section. It's a rectangle. Its width is gonna be a DX. Its length is gonna be, well, a length L. That's kind of the hard part here. Area will be length times DX. This is typically the hard part of a hydrostatic force problem because the cross-sections are always rectangles and the thickness will always be a DX or DY depending on what you named the variable. So how do we describe L right here? Well, to begin with, what I wanna mention here is that we do have this equilateral triangle. The height of the triangle is gonna be relevant for our discussion here. The height of the triangle. This is a right triangle, which one angle is 60 degrees. The other angle will be 30 degrees. This is a 30, 60, 90 triangle. You can use some trigonometry with cosines of 30 degrees, 60 degrees or a cosine of 30 degrees, 60 degrees. But if we know that this length here, my picture's kinda getting messy. I'm gonna draw it somewhere else here. If we cut the triangle in half where this length is eight and this is, again, our 30, 60, 90 triangle, then my symmetry at the top would actually be four. And then this side over here is gonna be four times the square root of three. Again, you can use a cosine argument or a tangent or something, so you can use some so-cato arguments to figure out the other side. Or that's one that many people just memorize as this 30, 60, 90 triangle. We do know the height of the triangle is four root three. The other thing that's important here is we're gonna set up a similar triangle, right? We're gonna set this triangle up, which is this big one, with a little triangle right here. Which this little triangle is gonna look like the following. We're gonna have half of L, one half L right here. And then I also wanna mention that notice, the way we set up X, the distance from the top of the plate to the cross section, that's what we mean by X. Which means that the remainder is gonna be the total height with this four root three minus X. And that's what's gonna go right here, four root three minus X. And so if we try to set up a proportion argument going on there, see what we're gonna get. We're gonna get L halves, it corresponds to four. This will sit above four root three minus X, which will correspond to four root three. So cross multiply here. We're gonna end up with two root three L is equal to 16 root three minus four X. And so divide by two root three on both sides. We end up with, get a little crowded here in my space. In the end, we end up with an L is equal to, let's see, we're gonna get eight minus two over square three X, that's our L value. In which case we then want to put that in for, we're gonna put that in for our L. That's what we had there, like so. And so putting this together, right? Force, the hydroxych force is pressure times area. So we have our area right here. We have now, sorry, we had the area right there and then the pressure's right there. So putting those all together, we're gonna integrate the pressure, which was 9.8 times 840 times X, that's the pressure. And then we're gonna times that by eight square root of three minus two X DX. That was the length times the width right there. Just double checking our math right there. That looks pretty good. Oh, I'm sorry, I put the square root of three in the wrong spot. Let me fix that. Eight minus X over the square root of three DX. That was the L value we found below. Oops, sorry, there was a two there as well. Apparently I can't even copy my own answers down. Eight minus two X over the square root of three. Right, that's good, that's good. So the only thing we have left to do is identify the bounds, right? And so looking at our diagram right here, what are the acceptable values of X? We could have X at the very top, which like we said before, that's X equals zero. And at the very bottom, that would correspond to X equals four root three. And so those are gonna be the bounds we set up here. We're gonna go from zero to four root three. So there's a lot of details going on here, but this would give us the correct antiderivative. We don't need to simplify anything really here and multiply them because we're not trying to evaluate. Just want you to set it up. So things I'm gonna look for on this one is I do wanna see the correct bounds. This thing should go from zero to four root three, right? I wanna make sure that the pressure is there, 9.8 times 840 times X. I wanna make sure that the area is correct and that includes the differential DX. If you're missing a DX there, that's gonna be a forfeiture of a point. So make sure you have your differential as it's part of the integral. Now this is just one example of how one could set up this integral. We based it on the assumption that X equals zero is the top of the plate. Another possibility would be to set X equals zero at the bottom of the plate. I think that sets up a slightly more complicated integral, but it is still correct. And so I will accept any correct integrals that you can come up with. It depends on your coordinate system. And as such, it can be very useful to actually specify with a diagram what is your coordinate system that makes it easier for me as I'm grading it. And therefore a happy grader makes a happy student. Am I right? Am I right? And that gives us an example of one of hydrostatic force. This one's a little bit more technical than others, but it was deliberately chosen for this practice exam to give you one that would stretch you maybe a little bit beyond what you saw in the homework perhaps.