 Welcome back. Today is lecture 16, Math 241. We are in the middle of chapter 6, section 5, Applications of Integration. So we are doing work problems. We left off yesterday. Let me read the problem first. We'll get the diagram back up here and then we'll see where we are on this problem. A right circular conical tank of altitude 20 feet and radius of base 5 feet has its vertex at ground level and its axis vertical. The tank is full of water. Find the work done in pumping the water over the top of the tank. So we have this for our diagram. What we do is I really didn't say much about this yesterday, but you want to take the situation that you're handed, diagram it appropriately, but not just diagram it appropriately, but try to develop an axis system such that when you draw this thing it's going to be easy to figure out what an x value is and a little bit later in this problem, which we haven't used yet, what a y value is. So we can describe each slice in terms of its radius or thickness or the x value or the y value fairly readily and sometimes it changes problems drastically based on where you place your axis system. So on this diagram it was already placed for us, but it's nice and convenient that the y axis comes right down the center of the cone and the x axis is down here at the base or the vertex of the cone. So that's going to make describing or defining this line which serves as the edge of the cone, it's going to make that a little bit easier for us. So when we get to choose where to put the axis system that becomes a pretty important choice. So what do we have thus far on this problem? We have the diagram so that we can name this line, we have the slices of water, not just to see how much water is in them, but kind of what kind of force is going to be necessary to move that particular slice of water. So we do have the volume of that slice is pi r squared h, the radius is x. So from the y axis over to a point on this line, we don't know what that is yet, but we do know it's x, the height or thickness of each slice is delta y which will become dy in the integrand. So that's how much water there is in there and if we want to take then the volume times the density to figure out what kind of force is going to be necessary to move it. So we'll take that volume with the values that we decided the radius is x and the height is delta y or dy, multiply it by the density which is 62 and a half pounds per cubic foot. So where do we get the cubic feet? Let me redo this because we're kind of in a hurry at the end of class. X is a linear distance measured in feet so x is in feet and that's x squared. So that's x squared. I guess technically there's a square feet here also because the unit two feet gets squared so you have two square feet. Dy is also in feet, the thickness or delta y increments of y are also in feet. So there's our units over here. So there's feet squared times feet which is feet cubed which knock out with feet cubed here. So you can kind of keep track of your units and it should sound like a force which pounds kind of sounds like a force that's going to be required to move that particular and that represents any slice of water. So at this point we have in our integrand pi x squared dy and we may not like the order that things are in but this is kind of how we've gathered them thus far. So there's our force. What else do we need in the integrand if we're going to do a work problem? Force and distance. So we have to go back to our figure and see how far each slice of water needs to travel to get to the top of the tank. So we need distance out here. Now in some problems the distance is just going to be little increments of y or little delta y's or little delta x's but in this case each slice of water has to travel a different distance. This slice of water has to travel this distance to get to the top of the tank right or if you want to think of it as coming over here this distance that's the same. This slice of water down here has to travel that distance to get to the top of the tank. So distance is going to be variable and it's not just going to be some increment of x or increment of y. So this diagram is pretty nice. Now think of the fact that we're going to have to come up with this on our own but in fact that's probably better at this point for us to come up with this on our own. So we have this axis system that's coming down here. This point was the radius was five feet so that's over five and what? How tall was the tank 20 feet? So if we go over five units and up 20 units we're at that point. So we establish coordinates for that y. Why would we want coordinates for that point? Where's that going to be helpful? We need the slope of that line so we've got this point strategically placed down here at zero zero so we can find the equation. I don't know that we came up with that. It was on the diagram but we haven't really talked about that. So when we're drawing our own diagram that's why I'm kind of abandoning that canned little diagram and us coming up with our own. So here's our slice that we're talking about. We know how much force it's going to be required to move it. We know the radius of that is the x value of that point on the line. We know this distance right here is the y value of that point on the line but that's not how far we need to move that slice to get it to the top of the tank. We need to move that slice that distance to get it to the top of the tank. So this entire distance is 20. This distance from the point down to the x axis is y so this has got to be 20 minus y. Is that alright? So we come up with this stuff on our own. It's not going to be plastered on the figure that's in the book or provided for you on a homework or a test. So we have the volume, the density. So there's the force required to move that slice. How far does that slice have to go? There's how far it has to go and if you look at any other slice it will also travel. In other words if the y value is down here the y value is smaller because we're not very far above the x axis you still describe how far that slice goes by saying it's 20 minus the y value to get to the top of the tank. So there's distance. Now that we're here well this tank is full of water so where do we get our first slice of water in terms of y because we're going to integrate with respect to y at zero. So right on the other side of zero is our first slice of water we've got to move that to in fact that has to move probably the furthest distance even though there's not much there and since this is full of water and we're integrating with respect to y we integrate to what? To 20 right? So we get slices of water from y equals zero all the way up to the top of the tank which is y equals 20. So what can stay and what needs to be changed somehow? 20 minus y can stay right? Why can that stay? Because we're integrating with respect to y x squared's got to go right? But it's got to go in a way that we can describe x in terms of the interrelationship of the x and the y's where? On the side of the cone well the side of the cone if you look at it two dimensionally it's just a line. So we need the equation of that line. Now Nicole already said it was on the other diagram it was but when we do a problem we have to come up with this ourselves. So we have two points on that line zero zero and five twenty. So the slope is difference of y's over difference of x's twenty over five which is four and the line goes through the origin so it's y equals mx right? Or if you want to go back to this if it's not quite so simple as it is in this case. So there's the equation of that line. So what do we need? y equals 4x we want to get rid of x okay? So we need something that x is equal to on that line so x is equal to y divided by 4 so we make that substitution. Now we're going to move some things but right now for x let's put in y over 4 so the integrand kind of looks messy. Well let's clean it up. Any questions about the setup first of all before we go any further? Is it hot again in here? Let's try to open the door again Chandler see if that'll help of course my hot air isn't helping is it? Any questions about the setup? Volume? Density so volume times the density tells us the force necessary to move it this is how far each slice is going. All right let's clean things up what can we move and where? Pi can go out front. What else? 62.5 can go out front and one-sixteenth right? Don't we have a one-sixteenth? Because we've got y over 4 squared so that division by 4 4 squared in the denominator. Is that it? All right so what's left in the integrand? We've got a y squared and a dy is that all right? So things that we can take and farm out to the front it's our advantage to do it. I don't think the integration is necessarily the toughest part of the problem in fact I think you'd find it to be the easiest part of the problem. So we've got 20 y squared minus y cubed. What will our units be on this problem? It's work. What are our units that we're working with in this problem? Foot pounds right? Feet because each slice is going to travel so many feet to get to the top of the tank and then how what kind of force are we moving it with? So many pounds of force. So when we integrate and evaluate our answer will be in foot pounds. Anybody think that we need to spend our class time integrating and evaluating at this point in time? Jacob? I don't think so. Okay thank you for clarifying it was the 62.5 that was given to us that's the density of water in kind of normal units. Other things that we deal with in a problem they will tell us what the density is for example if we're dealing with some oil in a barrel would you expect it to be more dense or less dense than water? More dense. What happens when you put oil in water? It rises to the top so it'll actually be less dense we think it probably you know it's kind of nasty it's going to be more dense but it's actually less dense. So those kind of things will be given to you on the problem. Usually we deal with several problems that have water in them so that's usually not given in the problem. Anything else from this? Okay let's look at a I just kind of went through the list of problems and wanted to get several different types of problems and then we'll move on to force due to liquid pressure which is another application problem but let's try a couple other ones here these should go pretty quick I hope. So I'm on page 479. Number five a force of ten pounds is required to hold a spring stretched four inches beyond its natural length. Spring problem what do we need for a spring problem to be able to solve a spring problem that's I think probably a pretty fair test question would be a spring problem. Okay we need the spring constant so the force required to stretch or compress a spring is directly proportional to the distance that it stretched the spring. So there's our sentence a force of ten pounds is required to hold a spring stretched four inches beyond its natural length. So we've got pounds let's convert everything to feet as well so we'll have foot pounds of work when we're done so four inches would be a third right so the first sentence ten pounds stretches it four inches beyond its natural length so that's a third of a foot. So k is equal to 30 so our setup to the problem for the work problem if it is a spring problem force force is k times x we're going to analyze this or break it down into little incremental movements or movements are going to be delta x or if the spring is oriented differently delta y how much work is done in stretching it from its natural length to six inches beyond its natural length what are we going to call natural length zero to six inches beyond its natural length one half will that work for that problem look like it set up properly so the task then is to integrate and evaluate and our answer will be in what foot pounds right so again the question how much work is done in stretching it so foot pounds is our unit here all right thought that one would go kind of quickly this one I think will go quickly but not maybe as quickly as that one problem 11 a cable that weighs two pounds per foot is used to lift 800 pounds of coal up a mine shaft 500 feet deep find the work done a cable that weighs two pounds per foot is used to lift 800 pounds of coal up a mine shaft when you start to lift this not only are you lifting the 800 pounds of coal aren't you also lifting the cable that is 500 feet long right so the initial amount of work is not only at the 800 pounds that's the coal but we've got it 500 feet right at least if we're we've got this 500 foot mine shaft 500 feet of cable and it's two pounds per foot so we've got a thousand pounds of cable right so the cable actually weighs more than the than the coal does initially so when we're first starting to lift this we've got a to move what is this 1000 pounds 1000 pounds of cable and 800 pounds of coal initially for the first little increment we're going to have to put out what 1800 pounds of force to move that so what happens as we wind this cable on to some kind of a winch and there's less cable out there okay the amount of weight decreases therefore the amount of work required to move that or the amount of force required to move that weight decreases so when we first start we're going to start with and there are other ways to approach this problem but I like to look at it as when I start what do I need and then is it a decreasing force that's necessary decreasing by how much perlending your foot so for every foot that I bring this thing up there's one foot less of cable that we actually have to lift right and for every foot it decreases by two pounds so two pounds for every x x being the amount of linear feet that we have actually wound this on to the cable so right at the end you have very little cable out there and most of the force that's required is just to lift the coal right so it's a variable and decreasing force that's necessary 1800 to start things for every foot it decreases by two pounds and then we're going to crank this up or I've got it in terms of acts I guess it probably be better in terms of why right for this visual picture little delta y's at a time but I've got an x in here so we'll call them delta x's from the bottom of the mine shaft we'll call that position zero now does that make sense with our model that we have here when x is zero we're at the bottom of the mine shaft how much force is necessary 1800 pounds of force and then we want to continue this particular situation to x equals 500 right the mine shaft is 500 feet deep so when x is 500 what are we lifting so right at the top of the mine shaft all we're lifting is the coal which is 800 pounds so if you put 500 in here for the force part you get 800 which is kind of all that's left because the cables all wound on to some winch okay let's actually finish the problem because I know what the answer is supposed to be and I want to make sure we get that answer questions about this so here's our force variable decreasing force necessary in this problem in our little dx or delta x is the increments that we lifted probably would be better if this were y and this were dy for our visual image integrate it 1800 x the integral of 2x is x squared zero to 500 and when we plug in zero we get zero so I'm not going to do that 18 x 590 is that right and then four zeros so 900,000 500 squared would be 25 with four zeros so 250,000 so 900,000 250,000 650,000 I'm pretty sure that's the right answer what foot pounds right questions on that one okay let's switch horses a little bit to another kind of problem in this section they call it hydrostatic pressure and force so it is technically a pressure problem force due to liquid pressure so if you think about a tank full of water and we want to know how much pressure there is on the end of the tank or on the sides of the tank or on the bottom of the tank there are kind of three components that are going to determine what kind of pressure we have I don't think this diagrams from your book but it's it's a pretty good diagram and the fact that it's really breaks it down into the three components that determine how much pressure there is on some plate or some edge of a tank so the we want to subdivide this region that is either totally submerged or partially submerged so if you think about what's going to create pressure on this let's call it the end of some kind of a water trough or some kind of a tank the force on one of these rectangles and that's all we really need to describe is the force on one of the rectangles and then we'll add them all up with integral calculus first of all the density of the stuff doing the pressing is it water is it you know oil is it what's in the tank what's doing the pressing that certainly is a key component into determining the pressure on this plate or edge of a tank that rectangle that we're describing wouldn't it matter how deep that rectangle is as to the pressure that's on it the deeper we are the more pressure that's on right if you get near the top of the let's say the end of this container that has less pressure on it because it's near the top the one the rectangles near the bottom are going to have more pressure on them so the depth of that horizontal strip is certainly important and then what is what's a description of the area of that horizontal strip so these three components and if you can think of something else valid or viable that would you think would enter in go ahead and let me know what that is but these are the three that come to my mind what's the density of the stuff that's pressing on the end of the tank how deep does it go so those little strips or rectangles parallel to the surface at the bottom of the tank as well as all the others can we describe their depth that certainly enters into the pressure and then how much stuff is there on each of these rectangles so density will describe that that's easy the depth of each horizontal strip and the area of each of the horizontal strips if we can do that then we accumulate all of the different pressures on each of the strips by using integral calculus so if we want instead of one we want all of them what are we going to do we're going to integrate that from the bottom of this plate whatever it is to the top of the plate so we do want little horizontal strips parallel to the surface of the in this case the water so can we describe that strip the same way we can just drive this one and if we can describe them all the same way integral calculus is going to add them all up all right first example problem here so we have the end of a water trough it is a isosceles trapezoid sounds pretty good doesn't it sound like a good name isosceles trapezoid how do you spell isosceles that's not that's not necessarily an easy question is I don't know isosceles so why is that important I don't know so we've got a water trough that is eight feet long so this is the end of it okay so the water trough is actually out here this is the end of it does it really matter that the water trough is eight feet long would it change the problem if it were 10 feet long or 12 feet long or 32 feet long or a thousand feet long as far as the pressure on the end of the water trough when you go to the beach and you step your foot in the ocean I mean if it really mattered how wide that ocean was you know that we've got all this ocean you went in the water like what four feet from shore and then you've got all that other water all the way over to England I mean it ought to just blast your foot out of the water and just you know put you up Greenville somewhere right it doesn't matter that it's water all the way to England versus the you know the six feet between your foot and the shoreline so it really this doesn't matter this piece of the problem doesn't matter what matters is the shape of the end and how deep it is and is it full of water so we've got this isosceles trapezoid the description says that it's eight feet across the top okay so you can kind of see the points are already established here but we have to do that ourselves it's four feet across the bottom and sorry what did I say eight it's not a what is it it's six six feet across the top four feet across the bottom and it's four feet tall so we've got to come up with our own diagrams doesn't it seem logical to take the y-axis right down the middle of this isosceles trapezoid and then put the x axis across the bottom right so we draw our own figure if it's four feet across the bottom and we put our axis system in this fashion then we've split it so it's two units right two units left if it's six feet across the top then we're going to go three units to the right and three units to the left we really don't care about the area of the side what we care about in the mathematical model are the areas of these horizontal strips because the pressure on this horizontal strip is different than the pressure on this horizontal strip so that's what the integral calculus is going to do is add all of these pressures together so the area of the whole side really never enters in so if you know how to find the area of this strip well what is this strip it's a rectangle right then we're in business you don't have to know the area of a trapezoid and then to get the other point the other part of this it's four feet tall so to get to this point we're going to go over three and up four to get to this point we're going to go left three and up four so we establish our own coordinates why do we need coordinates in this problem right to find the line this line which serves as the edge of the water trough the end of the water trough but also we need this distance don't we from here to here why would we need that distance right there which is called x of i in the diagram don't we need the area of that rectangle and what would we do with this x value double it right because this rectangle is 2x long or wide how tall is it delta y some increment of y right so if we need the area of the rectangle we need to know what that distance is from there to there and how do we describe that if we don't have the equation of this line which serves as the edge of the trapezoid so we could go ahead and take the point three four and the point two zero and find the equation we don't need it yet but we're going to need it before we're finished with this problem so the slope is four minus zero over three minus two four now it doesn't go through the origin so we're not quite as easy as our last one with the cone so why let's go ahead and take the point two zero so y equals four x minus eight so what are the three pieces three components of the integrand that are going to determine how much pressure we have on this end of this water trough we need the density right good now there is some memorization to this stuff I know that there's always memorization to mathematics but you're not memorizing some real complicated formula here the density of the stuff doing the pressing the depth of each horizontal strip in the area of each horizontal strip those are the three components that determine the pressure on the end of this particular water trough density depth so the density of the stuff doing the pressing it's water 62.5 pounds per cubic foot the depth of each horizontal strip okay here's our diagram this entire distance is what four this distance is what do we call that from the x-axis up to a point that's its y value so what is the depth here's the surface of the water it's four minus y is that correct four is the entire distance y is this distance so what we want for this strip is four minus y now would that be true for this one as well so we kind of have to make it just a real quick check is this horizontal strip also four minus y deep it is right so that describes all of them so depth four minus y and then the area of each horizontal strip well it's x from here to here so that entire distance across here would be 2x so it's 2x wide and delta y tall that same all right those are the three components that are going to determine the pressure on that end of the water trough notice the length of the water trough has nothing to do with this problem the fact that it's 8 feet long or 10 or 12 or 20 what are the limits and how do you find limits of integration for a problem like this zero to four so how did you come up with zero to four why values so and the why values because we're looking at where did we generate our first horizontal strip right right on the other side of the x-axis which is y equals zero and we continue to form these horizontal strips to the top of the tank which is y equals four is that correct so it's kind of like the pieces how are the pieces generated where did we generate our first one where is our last one all right well let's doctor this integral up a little bit what things can be moved what things can stay here and or subtly make changes 62.5 get that taking care of okay that's got to go right x has got to go because we're going to integrate with respect to y so the delta y for the integrand becomes dy x is got to be replaced well what is x every x value in terms of y when because it's on that line that serves as the edge of the trapezoid boy that's pretty isn't it x is think Chandler you said it we solve this for x y plus 8 over 4 right not pretty but we don't have choice in the matter that's so x gets replaced by y plus 8 over 4 so we take care of x we've taken care of the delta y okay so right now actually I've already written the 4 down but let's go ahead and take care of it now we've got to actually a 2 over a 4 right so let me get rid of the 4 and this 2 over this 4 will come out as one half that good so we've taken care of multiplication by 2 and division by 4 we slid it out front and everything else needs to stay right 4 minus y y plus 8 we're integrating with respect to y so what is that going to be negative y squared how many y do we have negative 4 y and the constant is plus 32 any question first of all about where anything came from in the original integrate density depth of that horizontal strip in area of that horizontal strip or how this became this x had to go right everybody convinced of that we're trying to integrate with respect to y we've got to get rid of x and what is x in terms of y on this problem everybody feel confident you could integrate and evaluate from here okay so we've been doing that for quite a while now these kind of problems can get kind of bizarre okay now you say well right ice or an isosceles trapezoid that's already kind of bizarre but they you have to make choices as to where things are I don't know that we'll make it through this problem but let's look at on page 481 problem number 30 I do want to do this problem I don't know that I want us to do this problem right now let's let's table that till tomorrow but let's let's do another problem we'll come back to problem 30 let's set up the diagram just trying to get different situations trapezoid this one has a hemisphere 18 page 480 actually a hemispherical tank shown is full of water sorry I'm a little scattered today let's go back to this problem something just came to mind when I was reading that problem and we'll do we'll set up 18 how would this problem if I were organized I would have asked this earlier how would this problem change if instead of the water trough being full of water suppose it's not all the way up to the top which is that the forefoot mark suppose it's at the two foot mark okay would that be true integration from zero to two well the density is still water so the water is going to be the same so that's going to stay the same what about the depth because the top of the water the surface of the water is now not the top of the water trough it'd be at height 2 so each one would be 2 minus yd so that would change right but with the area of each little horizontal strip be the same they'd still be 2x wide by delta y tall and the edge of the water trough is still that same line where would we get our first horizontal strip at y equals 0 where would we get our last one at y equals 2 so we do have to pay attention is it full and if it's not full we'll adjust because the top of the surface of the stuff doing the pressing it's important to know where that is sorry I intended to do that left that out till I read the next problem alright 18 a hemispherical tank shown is full of water given that water weighs 62.5 pounds per cubic foot find the work required to pump the water out of the top of the tank so I don't have a picture of this and a lot of times even though if you do have a picture in your book you're going to have to draw your own picture so we have a hemisphere which if you look at it from the side looks like a semi circle right so there's our tank we get to choose so here's the surface it's full of water we get to choose where to put our axis system which is why I wanted us to look at this problem we could here's our here here than the choices okay we could come right down here with our y-axis and come right under here with our x-axis that's a possibility where we could come right down here with our x-axis and right here with our with our y-axis right here with our x-axis so what determines where you put your axes on your problem because that's how we're going to name things right don't we eventually have to give an equation for this edge of the tank wherever it is I know which one I want to work with if I'm going to have to name the equation if you're working with circles which this is a circle even though it's a hemisphere the tank itself is hemispherical when we look at it two dimensionally it looks like a circle what's the easiest equation to work with in terms of circles right just the old center at the origin right so which one of these two has center at the origin this one right so we get to choose where to put our axis system so let's come down through here with our y-axis to make the equation now it could still do the problem but I think this makes the problem easier if we put our x-axis at the top of the tank now it doesn't say this in the problem but it does have it labeled on the diagram that the radius is five feet okay so what's the equation of the edge of this semicircle is the center at the origin yes we we got to choose that so when we get to make that choice it makes our life easier so it's x squared plus y squared equals 25 so we'll probably need to solve it for x squared or solve it for y squared but there's the equation we need horizontal strips parallel to the surface we need the density of the stuff doing the pressing water we need the depth of each horizontal strip this is this is actually a little trickier than it isn't this y now from the x-axis down to this point is that correct isn't that how deep this strip actually is almost right it's negative the negative of that is how deep it is this is a negative values everybody agree with that so in order to make that positive we're going to have to negate it so the depth is actually negative y well the way I've got my axes this is the origin so I don't you mean you can mess with that if you want to but the way I've got my picture that is why which is negative so that in terms of depth would be negative of y I don't think you're going to go wrong by calling that why you might get a negative answer you just have to negate it and what's the area of that strip we need to know that distance which is x is that correct isn't that the x value of that point we need to double it so it's 2x by delta y so that means we're going to integrate with respect to why well obviously we are that's how we're forming the horizontal strips and what would be the limits where are we down here negative 5 all the way up to y equals 0 now if you had not used negative y but used y instead and gone from 0 to 5 those two would have compensated for one another and you'd still have a positive answer okay we're out of time we'll finish this up tomorrow and probably move on to moments in center of mass