Hi everyone and welcome back. We're going to continue our conversation on double integrals. So friendly reminder, what are double integrals? Well, now we're going to consider a multivariable function. So we have a function f (x, y). We're going to think of its graph in the x, y, z coordinate system or space and remember its graph has some default picture, the magic carpet, and we're going to consider some region D in the x, y plane. And we're interested in computing the volume over D that lives under the graph. And this volume is going to be the double integral over the region D of f(x, y). We can go dx dy, we can go dy dx, that is Fubini theorem says order doesn't matter. We talk about things in general, we tend to say da with the emphasis we're integrating now over some object with area compared to a line segment A to B. And interval as a subset of the real line, we do these double integrals, we compute them as iterated integrals. So we work our way from the inside out and the bounds, that go on, the interval sign, come from the region D, come from the floor. So we're going to practice finding the right bounce today, we're going to do a couple problems we're going to work away off in difficulty. So let's see how we do. As a warm up, let's do a volume computation. So as I warm up, let's compute the double integral over the region R of the function 9- y squared dA. Now when we write dA means we get to pick the extra dy, it doesn't matter. I do have to tell you what R is though, R of course is the region from [0, 4] to [0, 3], okay? So they won't always give you a picture when you get asked to compute these double integrals, but in your head if you want to draw the picture. If we know what the graph of 9- y squared looks like, we can kind of guess about it as well. But at least focus on what's happening on the floor and the x axis, we're going from 0 to 4, and on the y axis, we're going 0 to 3. So we have this nice rectangle on the floor that is our region here, or region R, all right. If you'd like, pause the video for a second and try to work this out. Think about what the bounds would be, so because they hand us the boundary we don't think too hard. It doesn't quite matter the order, I don't know if one way is easier than the other way. But we're going 0 to 4, they're going to give you this as always this is your x and this is your y. So it's here to force your x 0 to 3 is your y, and then you have 9- y squared. And since we go inside out 0 to 3 corresponding to the y and 0 to 4 corresponded to the x, so dx. So this integral, it's doable if you make a substitution involving a trig function, you absolutely can work this out. But I'd like to sort of propose an alternative way to do this by interpreting this integral as a volume, the graph of z = square root of 9-y squared. Again, that's square root, so the whole thing, this is something that you might be able to figure out. I know there's no x's in here. So x is free, so really is relationship between z and y. If I square both sides, I get z squared is 9-y squared, and if I move the y squared over, I get y squared + z squared = 9, which is 3 squared. Do you recognize what this is? This is a circle centered at the origin of radius 3. Remember, r squared is equal to the number and this is in the y, z plane. So if you label our diagram with y's and z's and x's, I have a circle of radius 3 number in the y direction. That was the circle radius 3, this was our point 3, I go all the way up 3 and over 3 on the y axis and of course it dips below as well. But since this function is the square root, the z is always positive. I really just want the top half of the circle x is free, which means I'm allowed to take this circle and sort of pull it along the x axis. Remember, x goes from 0 to 4, okay? So let's draw this part from 0 to 4 as best we can. And now I want you to realize in this very specific case, because the function is relatively easy, although it does take a little bit work to recognize this. You're staring at half of a cylinder, this is a circular cylinder. My picture doesn't quite do it justice but it is a circular cylinder sort of sliced in half. So this double integral because we know what it represents, it represents the volume under the graph within the cylinder. Now, the cylinder technically goes into the negative y direction. And since we're only interested in the region over the rectangle, I really just want a quarter of the cylinder. So the part that lives in the first octant, the part that is sort of facing us my picture is getting worse I think. But the part that lives where x is positive and y is positive over the region 0 to 4 times 0 to 3, that is actually one quarter of a cylinder. So imagine take something in the cylindrical light on the side, cut it flat, and then cut it down the middle again and you get one-fourth of the cylinder. So I'm trying to get the picture to do this, one-fourth the cylinder. The only reason why I stress this is because the cylinder is a nice shape who's volume is known. The volume of a cylinder I think of it as circles stacked is pi r squared times the height. Now the cylinder is turned on its side. So the height happens to be 4. So I have one-fourth pi, the radius we said was 3, so 3 squared is 9. And then the height again this is sort of turned on its side is how we're x goes. So from 0 to 4, the fours cancel and we get 9 pi. So once in a while, one of the strategies that I want you to think about as you approach and try to solve these double integrals is yes you can try techniques of integration. But if the function happens to be nice, and once in a while they are, maybe you get a sphere or something else you know, then you get to use the fact. And again remind you what this number actually is, if you're going to take a page to compute, you can compute it using volumes. Let's do another problem. All right, let's do an example. I call this example easy, but realize that this level of your calculus career, your past things that are considered easy to most other people. I only consider this easy in terms of the information that's given in the problem where we don't have to go off, and do a lot of work to even just set up the problem. We're asked to compute the double integral of d of the function y squared over dA. So whenever we get to pick the x or the y, the region D in question is the all points x, y where negative 1 let's think of the y. Let's do a little 1 and negative y- 2 less wrinkled x- y. So the bounds on x are given by lines and the bounds on y are given by constants. This is another way to hand you a region D in the x, y plane. And the graph we're after, like what the volume of this is, okay. So this is nice, what have y, this is an easy question, I guess I should really put this in quotes. Why this is easy, is because we're handing you the bounds they're handing to you on a silver platter, you're saying you just go plug in. So -1 to 1 on your y and -y- 2, 2y for the function y squared, because keep the order in mind we're going x first, right? So the inner integral corresponds to the bounce on x, and the outer integral corresponds to the bound on y. So we're going dx dy, the fact that we could just look at this region and the ways resented finding the bounds on the double integral, that's what makes this problem easy. Oftentimes, it's not given or there's more work to do. But here we say thank you very much, and we go off and compute these iterated integrals. So let's do the inside one, of course, x is the variable, so y squared is constant, so the integral just becomes xy squared. Now we're going to plug in x =- y-2, and x = y in and then I will substitute those in and then compute one more integral. When I do this, I get the integral from -1 to 1. Let's plug in x = y, I get y cubed minus now watch out for y- 0- y-2 times y squared dy, I think it's worth it to distribute. That's to get y cubed. Let's see, so minus, minus is positive y times y squared is + y cubed, and then + 2y squared dy. Let's combine like terms y cubed + y cubed is 2y cubed + 2y squared dy. I could factor out the 2 if I want to, but that means no harm, no foul to leave it in a couple of things here. This is a straight single variable integral 2y cubed is a nice odd function with symmetric bounds. So I can if you want to think about it, we can use a little trick here. So 2y cubed dy plus the integral of -1 to 1, 2y squared dy, this first integral to y cubed. This is an odd function. So remember odd function is one that has area, the same area above and below the graph. It's y cubed, and since I have symmetric bounds, I know this is going to be 0. I can save myself some work from writing that out. And then this other function, well that just makes my life a little easier. We get 2 cubed y cubed or two-thirds y cubed, and I plug in from -1 to 1 and that gives me two-thirds minus, minus two-thirds, and that is two-thirds plus two thirds, which is 4. Okay, hopefully that was okay. Again, what I'd like you to do for these examples, we're going to do another one is, pause the video, work it out on your own, see if you can do it, and then check your answer against mine. Let's do another one slightly harder this time. All right, so here's a tougher problem. Let's see if you can see why it's a medium sized problem. Find the volume of a given solid bounded by the cylinder, x squared + y squared is 1 and the planes y = z, x = 0, z = 0 all in the first octant. Right off the bat, this is a little more wilder or a description than what we've seen before. So what I'd like to do before we even go through this is realize the difficulty is going to come from setting up the double integral. To get my head around this, what I like to do when I can is try to draw a picture. I'm not the world's greatest artist, but we'll draw a nice big picture and try to get a handle of what actual object we're trying to find the volume of. So we have a cylinder x squared + y squared = 1. So normally that's a circle on any given day, that's a circle the unit circle. We're in the first octant so I'm only going to draw the part of the circle in the first octant on the x, y plane. I'm going to try to draw it nice and big and as best to scale as I can so centered at the origin, radius 1, again, this would normally go all the way around, but I just care about the part in the first octant and z is free. So that means I would take this circle and pull it up and down. The z axis system in the first doctor is not going to care about what happens below the x, y plane. I'm really going to focus on this one once again, I guess a quarter cylinder that would live right above the circle the unit circle in this octane. So for best I can draw, we have a basically quarter cylinder that's what x squared + y squared = 1 is. I also have the plain y = z, it was a new object, it's a plane. So the way I can think about this plane y = z, it's a line y = z. So if you think like when y is 1, z is 1, and when y is 2, z is 2, so you get this line, right to the origin. This is the line y = z, but they tell you it's a plane. This is a reminder because remember, x is free as well. So now what I have to do is I have to pull this line in the x direction. This is going to be a little tricky to draw here. I apologize in advance, but I have some lines, think of it like a slice. I'm slicing the cylinder, okay? So this is like a plane and I'm slicing this thing as best I can. Not a very good picture but I'm trying and I'm slicing maybe like, think of it like a wedge of cheese, maybe a piece of a cheese with a rounded edge or something like that. So I have this plane, x = 0 that is the y z plane. So that basically is the back wall and z = 0 is the x, y plane, that is the floor. So I really do care about the part below the plane, the purple plane in the picture and above this blue unit circle on the x, y plane. So I have this odd random piece of cheese here. How do I find the volume of this object? Well remember a double integral finds volumes. I'm trying to find a volume of this wedge of cheese, the volume would be the double integral. Remember the bounds come from the floor. So the floor if I just draw the floor for you like a bird's eye view of the floor, remember, at the end of the day, this was that blue unit circle on the plane in quadrant one. This is the equation normally is x squared + y squared = 1. But since I'm in the first quadrant, I really want to solve for y. This is y = 1- x squared and all of that under the square root. Do you see all this work that I have to do to get the bounds? This is what we're going for here. So I want to find the area of this region on the floor. I have this function now that I want to compute what is the function that is the roof of this graph? Well, that's coming from z = y. So now I have the function y and I have to go dA. So the function is not explicitly given the bounds are not explicitly given, I have to go through and find all these things. When I do that, I always want numbers on the outside. So x is going to be from 0 to 1, my sort of thing about walking the x direction, what's your lowest bound? What's your upper bound? Those are always numbers at the end, in the y direction, I start at 0 and I walk to the function, the curve, the square root of 1- x squared. My function my multivariable function is actually does y. And remember the order here matters are going dy, dx. So this setup again, just timing everything how much work we had to do to get this correctly. This setup is what we need to get this volume. Once we have this now we go off and prosper, dy of course becomes y squared over 2 with y is the variable. We're going to plug in from 0 to root 1- x squared or dx, when I do that I get the integral from 0 to 1, I can bring the one-half off front. I plug in root 1- x squared and squared and I get 1- x squared and I plug in 0 I get 0 of course, dx. And then this is a pretty simple integral. I leave the work to you here but you get one-third after you work that out. Again, integral of 1 is x an integral of x squared is x cubed over 3. Okay, the thing I care about, thing that you should realize takes the most effort is just setting up the problem once you have it set up correctly, then just go slow. Be careful integrate like usual working away inside out. You can imagine though, if you set up the integral incorrectly, everything else that follows is wrong. It's like cooking and miss measuring your ingredients in the beginning and following the recipe exactly. You messed up in the beginning, so it doesn't matter what happens after this is why math is so frustrating. This is why people get really annoyed and crumpled papers and throw books. So just go slow, take the time, draw a picture. Do as best you can, do better than my image there, but do the best you can to draw these things. Let's do one more. This was a medium question. I think the picture makes it medium. And the fact that it's relatively simple integral to do, makes it a medium question, but let's do one it's a little tougher. Let's do a hard question. Find for me, the double integral from 0 to 1 of x to 1, e to the y squared dy dx. Okay, pause the video for a second. Give this one a shot, to see if you can do it and and come up with a number as your answer and see if you can tell me why this one is hard. All right, you ready? This one may seem easier than the last one. You say it's not a word problem. I don't like word problems, they're giving us the bounds. What's the difficult part of it? Well, the difficult part of it, is either the y squared is a notoriously famous function that has no antiderivative in close form. So when you try to integrate this dy, with the y squared, you probably, ran into a mistake, if you integrated this and came up with some function, you're wrong. And you can check it by taking a derivative, this either the y squared is a classically famous function that has no close antiderivative. It's not y, either the y squared or something else you'd like whatever if you came up with a function, I've seen people do this, it is incorrect. What we're going to do here, so the problem is the fact that they're asking us to integrate dy. We're going to think of the region first, we're going to always understand the region. So y is going from x to 1, and x is going from 0 to 1. Let's just understand this region, some people say what do I need to draw the region, well, here's a good reason why, I get a good habit. x is going from 0 to 1. Let me have the line y = x, so that's good, diagonal line to the origin, and it goes right through 1. So it's like lower bound is x and then it caps off at 1. So really what I'm looking at here is the upper triangle in quadrant 1. And what I need to do is I need to switch dy dx, I need to use Fubini theorems that's why this one's hard because it almost looks too simple, and it's trying to trick you. I need to use Fubini theorem which says I'm allowed to switch dx and dy, but you can't just do it Willy-nilly, you can't just put 1 to the outside, your outer bound always has to be numbers. If you think about y that's true when you plug in at the very last step, you want some number back, you don't want an expression involving x. So I'm going to have to switch the bounds, the function doesn't change either the y squared, but now I need to go dx dy. So now let's ask ourselves what are the right values to put on these bounds? So I need the numbers, the numeric bounds on y. y, if you look at the smallest and maximum midpoint, the lowest and highest, it gets, so y goes from 0 to 1. And the x value, if you sort of draw a random line in the x direction, what is its lowest and highest points? What's this min and what's its max. It goes from 0 to the line x = y. So switching the order does require paying attention to switching the bounce a little bit like use substitution. When you make use substitution integral, you gotta be careful to switch the bounds. Now, I hope you realize this was the hardest step. Once we have this, we can actually integrate e to the y squared because now it's a massive constant. And we're integrating with respect to the variable x. When I do that, it just becomes x e to the y squared and I evaluate this from 0 to y, again, these are what I plug in for x and I keep my dy comes along for the ride. Keep going to integrate the 0 to 1 you get x = y. So this is ye to the y squared and only plugins 0 you get -0, all of this is dy. This is a different integral than the very first one I started with, because now I have this y here, which allows for a very nice u substitution, u = y squared. So du is 2y dy that allows me to plug in, I gotta change my balance when I square 0 and square 1 and get back exactly the same. And then y dy becomes du, that's at the cost of putting a one-half in front, I can live with that. And I just get back e to the u after the substitution. Well, what's the antiderivative e to the u? Doesn't get any easier than that, it's just e to the u. And I'm going to plug in from 0 to 1, I get one-half, either 1- e to the 0. And be careful, don't mess this part up. It's one-half, e the 1 is e and e to the 0 is 1. So final answer is e-1 over 2 as its closed form. That is a tough question because most people either don't know to even think about using Fubini theorem, the picture is not given intentionally. So my advice to you is always, always, always try to draw a picture. You never know when that's going to come in handy. So you think of Fubini theorem as a technique for difficult problems if the integration stuff, draw picture. And remember if you do switch the bounds x and y, you have to be careful and find the new bounds on all the integrals that appear. All right, go through these examples again, make sure you can do all the levels easy, medium and hard. And good luck on these and then all the other ones. Great job on this one, I'll see you next time.
