 Okay, we are rolling. Today is lecture 17, math 241. We are in the middle of a problem or at least we have started with a problem. It is on page 480 in the text and it is problem 18. So let me read it again and we'll pick it up at the point where we left off. Hemispherical tank shown as full of water given that the water weighs 62.5 pounds per cubic foot. This is right, we didn't finish this, right? Thank you. Okay and I do want to adapt this problem too so we can see another type of problem that we haven't encountered yet. Given that water weighs 62.5 pounds per cubic foot, find the work required to pump the water out of the tank. So basically to the top of the tank. And that's what we'll adapt when we do a kind of a secondary problem when we're done. So here's where we were. We had the diagram. We decided to put the origin of our axis system here which made the equation a little bit simpler. You don't have to do that. You could certainly put the origin here and that would change the equation a little bit because now the center of the circle, the half circle that we have would not be at the origin but it would be pretty easy to work with. But this makes the equation a little easier. So we have a representative piece of this. This is not what I wanted. Not good. This is a work problem, right? Yeah. Work required to pump the water out of the tank. Okay and what is this set up as? This is set up as a pressure, isn't it? Isn't this the density of the stuff? Right? This is the depth of that. Okay so we don't have a tank that has a half circle for the end. Problem 18 is actually a work problem. Let's finish this one, okay? Which is not really problem 18. Glad we kind of got to this point and then picked up from here but this isn't really what we want. So let's say that we have some kind of a water trough that has a half circle for its end. That's what this problem is, right here. How much pressure is there on the end? So this is a pressure problem. Then we'll switch horses and do the work problem. Very glad we looked at this again. That's associated with getting all the slices of water to the top of the tank. Alright there's the density of the stuff that's doing the pressing. It's water. This is the depth because this distance from here to here is y. Which is a negative number. We want to negate that so that it's a positive number of units deep. And this is the area of that horizontal strip. So this is not a work problem but a pressure problem, right here. So we got this and we were out of time. So what are some things that can be moved and or changed so we can finish the problem? It's going to be dy. So y's are okay. That's okay. That's not okay on the pressure problem. So we need to solve for x. So there's x squared and we need x which is the actually plus or minus but we want the principal value. So we will put that value in here. So this is technically not problem 18 but a diagram that's similar to it. Working on the pressure on the end of this trough. Alright we can take 62.5 out in front. We can take the 2 out in front. Let's go ahead and take the negative out in front. So what do we have? Y and this radical and dy. Does that look correct for this pressure problem? You didn't get your test back today and you're like desperate to know what you got. I should, people leave me alone this afternoon. I should be able to finish those this afternoon and you can come by either before 5 today or tomorrow since we're not meeting tomorrow. It's very convenient that that y is in the integrand because we've got a y squared under the radical so that's our u, right. So du would be negative 2y dy so it's very important that that y is there if we don't want to use some approximating technique. So we need a negative 2 and a negative 1 half. I guess we already had a negative 2 right? We pulled it out front Chandler. We had the negative 2 there then that's fine right? Yes because these two knock each other out. So we've got a 62.5 out in front. Those two are gone. This is actually now going to be a definite integral but not changing the limits from negative 5 to 0. So we're going to have a what? U to the 1 half and everything else is du. So that looks like a u to the 3 halves over 3 halves. It's going to be a definite integral. There's what u represented for a few moments to the 3 halves over 3 halves or times 2 thirds. Now that we've got it back in terms of y we can bring those in. Does that look correct? The 3 halves over 3 halves bring that division by 3 halves multiplication by 2 thirds bring that back. What is not going to be the case here? A lot of times when we put in 0 we get 0. Not going to be the case here. When you put in 0 you get 25 minus 0 which is 25. Square root of 25 is 5, 5 cubed is 125. So we're not going to get 0 when we put 0 in. Any question about the evaluation? We did integrate but we haven't evaluated it yet. So this is not the work problem. This is a pressure. If the end of the trough or tank is a half of a circle. Would it change if the end of the trough was a spherical thing? Yes. So let's suppose if you can visualize this kind of looks like the water trough thing when you put this half circle on the end but if this is really half of a sphere that's going to change things and we don't have the two dimensional horizontal strips that we need to be able to do to do this problem. Yeah, we can't do that one. So that image, I guess, so if you took like a propane tank and just sliced it right down the middle you wouldn't have just a half circle at the end. You'd actually have a half of a sphere or a quarter of a sphere at the end and this method would not work because we need rectangles right here. We could probably come close but we're not going to be able to get it exactly. Alright, now let's do actually problem 18 since I kind of jumped there at the diagram and we didn't really solve the problem. Alright, so we have a hemispherical tank. It's full of water given that water weighs 62 and a half pounds per cubic foot. Find the work. What do we need for work? We don't need density, depth, and area of the horizontal strips. That's pressure for work. We need force and distance. Okay, so one of our slices that we're going to look at will look something like this. So we need to decide how much water is in that slice and what is that slice? What would you describe it as? Cylinder, right? That's a cylinder. So we need the volume of a cylinder which is pi r squared h. That's how much stuff is in it. How much water we need to exert a force equal to its weight. We're going to take that volume and multiply by the density to get how much force we're going to need to move it. So it looks like the radius would be what? So for our point that's on the edge, the edge being this half circle, that would be the x value. Is that correct? Of that point, the distance from the y axis over would be its x value. So if the radius is x, we're going to need to know what was the radius 5. So x squared plus y squared equals 25. We're going to need to possibly know what x is. It's going to be determined by if we're integrating with respect to x or with respect to y. So we're going to have a pi x. What is the height or thickness of this slice of water that we're going to try to move to the top of the tank? dy. So that delta y or dy. So that says we're going to have to get rid of x, right? Now very conveniently this x is squared so we can solve for x squared and then we're done. We can just replace something for x squared. So we'll be able to throw that in there for x squared. So that's the amount of water in that slice. How much force is it going to take to move that amount of water? We need to do what with this? We don't need volume. We want weight, right? Force. How much force is it going to take to move that amount of water? So we need to multiply this by the density, 62 and a half. So this x is in feet. So there's feet squared. Delta y is in feet. So we got feet squared times feet, which is feet cubed. So we need pounds per cubic foot. So the cubic feet units knock each other out. So the force would be pi x squared dy times 62.5 pounds per cubic foot. So now we're in pounds, which is what we want because that's the force required to move it. How far does this need to travel to get to the top of the tank? This distance, right? And we already decided in our other problem that that distance is really what? Y. That's the y value, or the negative of the y value. The y value is negative. We want to go in the opposite direction to that. So the distance will be negative y. And we do want y values because we're integrating with respect to y. So if we're integrating force times distance, then we're in business, right? So we're in business, right? So we're in business, right? And we're in business, right? So what are we doing basically except our limits of integration? We're replacing the x. Right, we've got to get rid of that. But what are our limits? Where do we get our first slice of water in terms of y? Negative five. At negative five, down here. In our last slice of water, it's the top of this to the pressure problem that we did, but it is a different problem. So we'll get rid of x squared. So for x squared, we'll put in 25 minus y squared right here. Let's move some things around so the pi can come out front. 62.5 can come out front. Do we need the negative in here? I'll keep it. For x squared, we'll put in 25 minus y squared. Probably means we're going to need the negative, right? Here's a negative y and a dy, negative 5 to 0. Don't absolutely need the negative though, right? Because since this is just to the first, we'd probably distribute the negative y to these pieces and then integrate each piece and evaluate from negative 5 to 0. I guess you could use a U substitution, but let's just distribute and integrate each piece. So 62.5 pi, negative 5 to 0. What's that going to be? Y cubed minus 25y dy. Pretty easy integration and evaluation problem. And our answer should be, this is work. So our units that we accumulated in this problem are what? Force was we decided in pounds, right? And our distance was in feet, foot pounds. And I know we haven't done enough examples yet with kilograms, cubic meters, the density of water with respect to those units, 1,000 kilograms per cubic meter. So that's our next example problem. We'll have that in it. Anybody questions or issues or discussion about how to get from here to a final answer? And I said I wanted to change this problem. I've already changed it once. In fact, we didn't do the problem the first time. So let's do the work problem. And let me change the last sentence. Suppose the last sentence said, find the work required to pump the water. Now, the way it reads here is out of the tank. So we get it to the top of the tank, and we don't care where it goes from there. But it's out of the top of the tank. Let's say we want to move it to a height five feet above the top of the tank. We've got this pipe up there, and it's going to take it over here and dump it somewhere else. We don't want to just get it to the top of the tank. We want to get it up here. How will that change our setup? Bounds change from negative 5 to 5? How do we determine our bounds? What is it that determines negative 5 to 0? Because that's where the water is. Those are our slices of water. Where do we get our first slice of water? Right down here. Where do we get our last slice of water? Will that change if we're going to take it to a height of 5 feet? No, there's no more water there. Our slices of water still go from negative 5 to 0. Change the distance. So instead of just taking it to here, we're going to take it all the way up to here, as far as the distance that each slice of water travels. So the force required to move each slice of water is exactly the same. The description of that is exactly the same. The limits are exactly the same, because this is where they start, the slices. This is where they stop, but the distance will change. So instead of being negative y, it's going to be negative y plus 5. So and there are some problems like that in here. I don't know if they made it to a web assign or not, but there are problems like that in this problem set, where you're not just taking it to the top of the tank. You're taking it to a place other than the top of the tank. So that would affect this problem right here. So the distance that each slice would have to travel would be negative y plus 5, if we're taking it to a height of 5 feet above the top of the tank. Everything else is the same. All right, let's get to kilograms and meters, cubic meters, density of water with respect to those units. Problem 30. I think I kind of started with problem 30, and then we jumped back to 18, but let's go back. This is a pressure problem, hydrostatic force against the gate. So we're talking about the pressure that's on a gate. There's a picture in the book. I'll attempt to redraw that picture up here. A vertical dam has a semicircular gate as shown in the figure. Find the hydrostatic force against the gate. All right, so here's what the picture looks like. So the water level is right here. This is 12 meters. Here's the gate down here, and this is 4 meters. OK, and let me read the problem again. A vertical dam, so this whole thing is the dam, has a semicircular gate, so that little gate to release water is at the bottom. As shown in the figure, find the hydrostatic force against the gate. So that's our picture. Here's the gate. I'm not going to shade it because it's going to get crowded anyway in a hurry. So what are some of the things that, when we read the problem, look at the diagram, make our own picture, what would come next? OK, we get to choose where the Cartesian axis system is, so let's make a choice. I mean the x-axis. OK, x-axis down here, and y-axis. Right down here. So why will that potentially make things easier for us to have it set up this way? The equation of the circle. OK, we've got a circle, or at least a half circle, whose center is at the origin, and what's the radius of that circle? Two meters. Two meters. So we now have the equation of that semicircle with center at the origin. x squared plus y squared equals 4. Is that all right? Seems like a pretty good choice for the axis system. So we need little horizontal strips parallel to the surface of the water, and this is a pressure problem. So we need the density of the stuff doing the pressing. Now we've got metric units here, so we've got meters. We need the depth of each horizontal strip and the area of each horizontal strip. From our little picture here, we're going to need the area of this horizontal strip. So from the y-axis over to this point, what is from the y-axis over? That's an x, right? And we need two of those because the width of this is, it's a symmetrical picture, so we've got x on the other side as well. So it's 2x by what? Delta y? How about the depth? Here's what we need. What do we have? Well, we do know this. This is y. So from the x-axis up to that point on the circle is y. So how deep is that horizontal strip? Because this is 10. So if the whole thing is 10 from here all the way down to here, and this little piece right down here is y, then what we want is 10 minus y. Does that work? So it looks like with a dy or delta y, we're going to be able to keep the y, but we're going to have to get rid of the x. It's all right, not a big deal. Density. If we were in feet and pounds, it'd be 62.5 pounds per cubic foot. We're in meters, right? So do we have cubic meters? Can you track the units here? Depth isn't that in meters, right? 10 minus y meters? 2x, x is in meters and isn't delta y in meters, right? So we've got cubic meters. So this is the first time we've had this, but the density of water is in these units, 1,000 kilograms per cubic meters. So we're going to put that in here. Now, is there something else that's missing? What else do we have to do with the metric units since we really, before when we had the slice of water and we knew the weight of the water, we need the gravitational constant with respect to this system, right? It's built into the other one. When you have pounds, it's already, the gravitational constants kind of build into that. We have to build it in here. And what is that? 9.8. 9.8. So we need the gravitational. At any time we have this system, we have to throw that in ourselves, which is 9.8. What's the rest of that? 200 meters per second squared. So we've got to throw that in here also. So we've got our component pieces, I think. Well, not completely. What are the limits? So we look at our little horizontal strips. Where did we start to form them? And where is our last little horizontal strip? Right on the other side of the x-axis, which would be y equals 0. And we form them all the way up to y equal 2. So here's the reason, really, for looking at this problem. We haven't seen that yet in a problem, 1,000 kilograms per cubic meter. And we have seen this, but we haven't done a problem where we had to throw that in because we were dealing with pounds, and you don't need that when you're dealing with pounds, that gravitational constant. So we could bring both of these out front. We've got a 10 minus y kind of stuck with that. We've got a 2. We could bring that out front. We've got an x. What is x equal to? Rout of 4 minus y squared. So x squared is 4 minus y squared. So x is, and we want the positive, the principal square root. So we've got the 2 out front. We've got the x. Here it is. We've got delta y, which really becomes a dy. That gets a little messy. We better not leave it like this without at least talking about how we're going to integrate this. Nice that all that comes out front. Why is that messy? What's the issue with the integration of this problem? You have to use trig substitution. OK, if we, say, let u equal 4 minus y squared, then our du would be what? Negative 2y dy? Well, that doesn't really help us with the 10, does it? Do we need to distribute this thing to its component pieces there in the binomial? I think we do. I don't think we have a choice in this problem. So let's take this a little bit further. So we've got this radical, and then we've got that times minus y. Here's our substitution piece. I think the choice of u, let u equal 4 minus y squared, that's going to work here, right? What about this one? You say, well, I don't like the 10 there. I want to move the 10 out in front. It can go out in front of this one. But it can't go out in front of everything because there's not a 10 in this piece. Suggestions or recommendations for this? Possibly table of integrals. That would be a what? a squared minus u squared. Probably find that. But then you need a du in there, which would have a y in it. And we can't manufacture the y. Trig substitution would work. You could let y equal what? 2 sine theta. So we could use the table of integrals. We could use trig substitution. I don't know. Anything else? I'm kind of thinking of something else. We can get an answer here or here. Is this, and I'm not sure. I have to think this through while I'm saying it, is this the area of half of this circle? Is that right? That for a second. Yes. It is. Right? It's the other half. So this, really, you could do without calculus at all. This is the area of half of this circle. So what is the area of this circle? Pi times the radius squared. Pi times 2 squared 4 pi. So if you use the table or trig substitution, you're going to get the area of this or the evaluation of this definite integral to be 2 pi, I think. I think that is the case. But we have two pretty darn good methods if that is not something that you see, that that's the area of half of a circle. But I think that's going to be 2 pi. So this one, substitution. This one, use of the table of integrals. Look for the section that has the square root of a squared minus u squared. Or trig substitution for y put in 2 sine theta. Or say, I think I know what that is. That's the area of half of a circle. And then just go from there. But that's not the reason for the problem. The reason for the problem is to look at a problem that has these two numbers in it and why we need them. Logan? If you're distributing in a negative y in the second integral, should that be a plus sign? Yes. Yeah, we've got too many negatives. Yeah. Is that correct? So one of these can be made positive. So let's change that one. Thank you. It doesn't come out to 2 pi because that's a pi. It goes to pi. OK. So Katie did the integration. Now, use the table. No, he's my god. OK. OK. So the question is, oh, I see why. Because. OK. It's not a half circle. It's a quarter of a circle. OK. Because we're only going from 0 to 2. This, this is the x part of this, right? We solve for x. So we've got this half circle is what we've got with this. The positive where the x values are positive because this is the principal square root. So we're talking about this half circle. And we're talking about the area under that half circle from 0 to 2. So you figured it out and got pi. We're not talking about the entire half circle. We're talking about the part from 0 to 2. So it's really a quarter of a circle. So the area is, so that's correct. OK. I think I had one more problem I wanted to ask. Yeah, but I want you to set this one up yourself. And then we'll kind of check it together. This is problem 27. And then we can, Friday, we will cover moments and center of mass. And I think one day is sufficient for that. All right. So if you have your book, you might want to read it long. But I will read it a couple of times. There's not a diagram. In fact, that's one of the reasons I chose this problem is that you come up with your own diagram. The trough is filled with a liquid of density, 840 kilograms per cubic meter. So it's obviously not water, something other than water, less dense than water. The ends of the trough are equilateral triangles with sides 8 meters long. So the ends, so the sides of the equilateral triangle are 8 meters long. And the vertex is at the bottom. Find the hydrostatic force, the pressure, on one end of the trough. Set up a diagram, establish your axis system, and let's see if we can kind of set this thing up together. So that's a little bit of a hint as to what it looks like. Do not have a picture. I'll have some background music when we do this. I'll work on that. All right, got an axis system set up yet. Here's what we want, right? This guy right here, the end. I want to split this right down the middle. I bet you that's not a point of contention with anybody. But I bet you we have the x-axis in different positions. Down here? Is that most of you? Anybody have the x-axis up here? That would also work. I think I want to put my x-axis right down there. So these are 8, 8, and 8. Well, we don't really need the 8. We need a pair of fours up there. So we need to know that that's four over there. What's the coordinates of that point? It's not 4, 8, because this distance is 8, right? But do we have enough information to see how that distance is what we need, right? Four. So we go over four and up something. What would that something be? 6.93. Or exactly. This is a right triangle. It's 8 on the hypotenuse. It's 4 on one of the legs, right? So isn't the other leg 4 times the square root of 3? Yes. So it's a 30, 60, 90. Because one of the legs is half the hypotenuse. So that means that's a 30 degree angle. So that means that's a 60 degree angle. 6 point, whatever Daniel said, that's fine. Or exactly it'd be 4 squared to 3. Is that distance from here to here? All right, so I think we have our diagram. What do we need next? Pressure on the end of this trough. We do need the equation of that line. So we have two points on it. What's the slope of that line? 4 squared to 3 over 4. 4 squared to 3 over 4, which is square root of 3. Or if you have decimals, then you're going to have 1.732, whatever square root of 3 is. So the equation of that line, y minus 0 equals slope times x minus 0. So there's the equation. y equals square root of 3x. So don't we need our little, did it say it was full, filled with a liquid of density, 8 kilograms per cubic meter? So it is full. We need our little horizontal strip parallel to the surface. There's the surface. So tell me some things to write down for this integrand. 9.8. We're going to need a 9.8 in there because we're dealing with kilograms and meters. We're going to need an 8.40 because that's the density of whatever this stuff is that's pressing on the end of the water top. 6.928 minus 1. And we need that distance. Doubled. Is that correct? Because that'd be 2x wide and delta x tall. I mean delta y tall. And what else do we need? The depth, which is 6.9. The depth, which is really that distance, right? So what do we know about this picture? This entire distance is 4 squared to 3. And this distance from here to here is y. y? There we go. So that's actually kind of different from the way we've written it. We've written the depth, right? Density and then the depth and then the area. So we've got the area here. But I think we've got everything we need to integrand. How do you decide on the limits? Where's our first little horizontal strip right down here at the bottom in terms of y? y equals 0 all the way to y equals where are we at the top? 4 squared to 3. Delightful little interval there. And what's the only change we need to make? Right, x can change x to y. Get rid of x. What is x in terms of y? 1 over 3. x equals y over the square root of 3. Does that work? So let's at least get everything written out here. See if we've got any problems to contend with. Let's bring the 9.8 out in front, 840. So if it's water, this would be 1,000. But it's something other than water. We've got a 2. We can bring out front 4x. We're going to throw in y over the square root of 3. dy is going to be our delta y. Is that going to work? So two sentences of a problem. And then we make our own diagram. We decided where are the horizontal strips are. What's the area of those? How deep are they? What are the other constants that we need? Because we're dealing with cubic meters and kilograms. So we could, on this problem, just distribute this to these two pieces and integrate each piece. I don't think the integration is going to be nearly what we had on that other problem. That is pretty good timing. That is about what I thought we would do today. And we did it. So we'll start with moments and center of mass on Friday.