WEBVTT 00:00:01.000 --> 00:00:03.000 The following content is provided under a Creative 00:00:03.000 --> 00:00:05.000 Commons license. Your support will help MIT 00:00:05.000 --> 00:00:08.000 OpenCourseWare continue to offer high quality educational 00:00:08.000 --> 00:00:13.000 resources for free. To make a donation or to view 00:00:13.000 --> 00:00:18.000 additional materials from hundreds of MIT courses, 00:00:18.000 --> 00:00:23.000 visit MIT OpenCourseWare at ocw.mit.edu. 00:00:23.000 --> 00:00:27.000 So, basically the last few weeks, we've been doing 00:00:27.000 --> 00:00:33.000 derivatives. Now, we're going to integrals. 00:00:33.000 --> 00:00:46.000 So -- OK, so more precisely, we are going to be talking 00:00:46.000 --> 00:00:58.000 about double integrals. OK, so just to motivate the 00:00:58.000 --> 00:01:04.000 notion, let me just remind you that 00:01:04.000 --> 00:01:11.000 when you have a function of one variable -- -- say, 00:01:11.000 --> 00:01:16.000 f of x, and you take its integrals 00:01:16.000 --> 00:01:21.000 from, say, a to b of f of x dx, 00:01:21.000 --> 00:01:33.000 well, that corresponds to the area below the graph of f over 00:01:33.000 --> 00:01:43.000 the interval from a to b. OK, so the picture is something 00:01:43.000 --> 00:01:47.000 like you have a; you have b. 00:01:47.000 --> 00:01:55.000 You have the graph of f, and then what the integral 00:01:55.000 --> 00:02:02.000 measures is the area of this region. 00:02:02.000 --> 00:02:04.000 And, when we say the area of this region, of course, 00:02:04.000 --> 00:02:06.000 if f is positive, that's what happens. 00:02:06.000 --> 00:02:11.000 If f is negative, then we count negatively the 00:02:11.000 --> 00:02:15.000 area below the x axis. OK, so, now, 00:02:15.000 --> 00:02:19.000 when you have a function of two variables, then you can try to 00:02:19.000 --> 00:02:22.000 do the same thing. Namely, you can plot its graph. 00:02:22.000 --> 00:02:24.000 Its graph will be a surface in space. 00:02:24.000 --> 00:02:28.000 And then, we can try to look for the volume below the graph. 00:02:28.000 --> 00:02:35.000 And that's what we will call the double integral of the 00:02:35.000 --> 00:02:44.000 function over a certain region. OK, so let's say that we have a 00:02:44.000 --> 00:02:51.000 function of two variables, x and y. 00:02:51.000 --> 00:03:00.000 Then, we'll look at the volume that's below the graph z equals 00:03:00.000 --> 00:03:06.000 f of xy. OK, so, let's draw a picture 00:03:06.000 --> 00:03:10.000 for what this means. I have a function of x and y. 00:03:10.000 --> 00:03:16.000 I can draw its graph. The graph will be the surface 00:03:16.000 --> 00:03:20.000 with equation z equals f of x and y. 00:03:20.000 --> 00:03:24.000 And, well, I have to decide where I will integrate the 00:03:24.000 --> 00:03:26.000 function. So, for that, 00:03:26.000 --> 00:03:30.000 I will choose some region in the xy plane. 00:03:30.000 --> 00:03:39.000 And, I will integrate the function on that region. 00:03:39.000 --> 00:03:46.000 So, it's over a region, R, in the xy plane. 00:03:46.000 --> 00:03:51.000 So, I have this region R and I look at the piece of the graph 00:03:51.000 --> 00:03:56.000 that is above this region. And, we'll try to compute the 00:03:56.000 --> 00:04:02.000 volume of this solid here. OK, that's what the double 00:04:02.000 --> 00:04:08.000 integral will measure. So, we'll call that the double 00:04:08.000 --> 00:04:13.000 integral of our region, R, of f of xy dA and I will 00:04:13.000 --> 00:04:18.000 have to explain what the notation means. 00:04:18.000 --> 00:04:22.000 So, dA here stands for a piece of area. 00:04:22.000 --> 00:04:26.000 A stands for area. And, well, it's a double 00:04:26.000 --> 00:04:28.000 integral. So, that's why we have two 00:04:28.000 --> 00:04:30.000 integral signs. And, we'll have to indicate 00:04:30.000 --> 00:04:33.000 somehow the region over which we are integrating. 00:04:33.000 --> 00:04:37.000 OK, we'll come up with more concrete notations when we see 00:04:37.000 --> 00:04:39.000 how to actually compute these things. 00:04:39.000 --> 00:04:44.000 That's the basic definition. OK, so actually, 00:04:44.000 --> 00:04:45.000 how do we define it, that's not really much of a 00:04:45.000 --> 00:04:49.000 definition yet. How do we actually define this 00:04:49.000 --> 00:04:52.000 rigorously? Well, remember, 00:04:52.000 --> 00:04:56.000 the integral in one variable, you probably saw a definition 00:04:56.000 --> 00:04:58.000 where you take your integral from a to b, 00:04:58.000 --> 00:05:01.000 and you cut it into little pieces. 00:05:01.000 --> 00:05:03.000 And then, for each little piece, you take the value of a 00:05:03.000 --> 00:05:06.000 function, and you multiply by the width of a piece. 00:05:06.000 --> 00:05:11.000 That gives you a rectangular slice, and then you sum all of 00:05:11.000 --> 00:05:14.000 these rectangular slices together. 00:05:14.000 --> 00:05:17.000 So, here we'll do the same thing. 00:05:17.000 --> 00:05:23.000 So, well, let me put a picture up and explain what it does. 00:05:23.000 --> 00:05:26.000 So, we're going to cut our origin into little pieces, 00:05:26.000 --> 00:05:30.000 say, little rectangles or actually anything we want. 00:05:30.000 --> 00:05:35.000 And then, for each piece, with the small area, delta A, 00:05:35.000 --> 00:05:39.000 we'll take the area delta a times the value of a function in 00:05:39.000 --> 00:05:43.000 there that will give us the volume of a small box that sits 00:05:43.000 --> 00:05:46.000 under the graph. And then, we'll add all these 00:05:46.000 --> 00:05:49.000 boxes together. That gives us an estimate of a 00:05:49.000 --> 00:05:51.000 volume. And then, to get actually the 00:05:51.000 --> 00:05:53.000 integral, the integral will be defined as 00:05:53.000 --> 00:05:56.000 a limit as we subdivide into smaller and smaller boxes, 00:05:56.000 --> 00:05:59.000 and we sum more and more pieces, 00:05:59.000 --> 00:06:02.000 OK? So, actually, 00:06:02.000 --> 00:06:11.000 what we do, oh, I still have a board here. 00:06:11.000 --> 00:06:20.000 So, the actual definition involves cutting R into small 00:06:20.000 --> 00:06:29.000 pieces of area that's called delta A or maybe delta Ai, 00:06:29.000 --> 00:06:36.000 the area of the i'th piece. And then, OK, 00:06:36.000 --> 00:06:42.000 so maybe in the xy plane, we have our region, 00:06:42.000 --> 00:06:47.000 and we'll cut it may be using some grade. 00:06:47.000 --> 00:06:49.000 OK, and then we'll have each small piece. 00:06:49.000 --> 00:07:00.000 Each small piece will have area delta Ai and it will be at some 00:07:00.000 --> 00:07:06.000 point, let's call it xi, yi ... 00:07:06.000 --> 00:07:13.000 yi, xi. And then, we'll consider the 00:07:13.000 --> 00:07:18.000 sum over all the pieces of f at that point, xi, 00:07:18.000 --> 00:07:22.000 yi times the area of a small piece. 00:07:22.000 --> 00:07:27.000 So, what that corresponds to in the three-dimensional picture is 00:07:27.000 --> 00:07:31.000 just I sum the volumes of all of these little columns that sit 00:07:31.000 --> 00:07:36.000 under the graph. OK, and then, 00:07:36.000 --> 00:07:49.000 so what I do is actually I take the limit as the size of the 00:07:49.000 --> 00:07:57.000 pieces tends to zero. So, I have more and more 00:07:57.000 --> 00:08:03.000 smaller and smaller pieces. And, that gives me the double 00:08:03.000 --> 00:08:07.000 integral. OK, so that's not a very good 00:08:07.000 --> 00:08:13.000 sentence, but whatever. So, OK, so that's the 00:08:13.000 --> 00:08:16.000 definition. Of course, we will have to see 00:08:16.000 --> 00:08:19.000 how to compute it. We don't actually compute it. 00:08:19.000 --> 00:08:21.000 When you compute an integral in single variable calculus, 00:08:21.000 --> 00:08:23.000 you don't do that. You don't cut into little 00:08:23.000 --> 00:08:25.000 pieces and sum the pieces together. 00:08:25.000 --> 00:08:28.000 You've learned how to integrate functions using various 00:08:28.000 --> 00:08:29.000 formulas, and similarly here, 00:08:29.000 --> 00:08:32.000 we'll learn how to actually compute these things without 00:08:32.000 --> 00:08:34.000 doing that cutting into small pieces. 00:08:34.000 --> 00:08:38.000 OK, any questions first about the concept, or what the 00:08:38.000 --> 00:08:48.000 definition is? Yes? 00:08:48.000 --> 00:08:50.000 Well, so we'll have to learn which tricks work, 00:08:50.000 --> 00:08:52.000 and how exactly. But, so what we'll do actually 00:08:52.000 --> 00:08:56.000 is we'll reduce the calculation of a double integral to two 00:08:56.000 --> 00:08:58.000 calculations of single integrals. 00:08:58.000 --> 00:09:00.000 And so, for V, certainly, all the tricks 00:09:00.000 --> 00:09:03.000 you've learned in single variable calculus will come in 00:09:03.000 --> 00:09:05.000 handy. OK, so, 00:09:05.000 --> 00:09:07.000 yeah that's a strong suggestion that if you've forgotten 00:09:07.000 --> 00:09:09.000 everything about single variable calculus, 00:09:09.000 --> 00:09:17.000 now would be a good time to actually brush up on integrals. 00:09:17.000 --> 00:09:21.000 The usual integrals, and the usual substitution 00:09:21.000 --> 00:09:26.000 tricks and easy trig in particular, these would be very 00:09:26.000 --> 00:09:28.000 useful. OK, so, yeah, 00:09:28.000 --> 00:09:31.000 how do we compute these things? That's what we would have to 00:09:31.000 --> 00:09:37.000 come up with. And, well, 00:09:37.000 --> 00:09:39.000 going back to what we did with derivatives, 00:09:39.000 --> 00:09:42.000 to understand variations of functions and derivatives, 00:09:42.000 --> 00:09:45.000 what we did was really we took slices parallel to an axis or 00:09:45.000 --> 00:09:46.000 another one. So, in fact, 00:09:46.000 --> 00:09:50.000 here, the key is also the same. So, what we are going to do is 00:09:50.000 --> 00:09:52.000 instead of cutting into a lot of small boxes like that and 00:09:52.000 --> 00:09:57.000 summing completely at random, we will actually somehow scan 00:09:57.000 --> 00:10:01.000 through our region by parallel planes, 00:10:01.000 --> 00:10:03.000 OK? So, let me put up, 00:10:03.000 --> 00:10:08.000 actually, a slightly different picture up here. 00:10:08.000 --> 00:10:11.000 So, what I'm going to do is I'm going to take planes, 00:10:11.000 --> 00:10:14.000 say in this picture, parallel to the yz plane. 00:10:14.000 --> 00:10:18.000 I'll take a moving plane that scans from the back to the front 00:10:18.000 --> 00:10:23.000 or from the front to the back. So, that means I set the value 00:10:23.000 --> 00:10:27.000 of x, and I look at the slice, x equals x0, 00:10:27.000 --> 00:10:31.000 and then I will do that for all values of x0. 00:10:31.000 --> 00:10:35.000 So, now in each slice, well, I get what looks a lot 00:10:35.000 --> 00:10:40.000 like a single variable integral. OK, and that integral will tell 00:10:40.000 --> 00:10:44.000 me, what is the area in this? Well, I guess it's supposed to 00:10:44.000 --> 00:10:48.000 be green, but it all comes as black, so, let's say the black 00:10:48.000 --> 00:10:51.000 shaded slice. And then, when I add all of 00:10:51.000 --> 00:10:55.000 these areas together, as the value of x changes, 00:10:55.000 --> 00:11:00.000 I will get the volume. OK, let me try to explain that 00:11:00.000 --> 00:11:07.000 again. So, to compute this integral, 00:11:07.000 --> 00:11:16.000 what we do is actually we take slices. 00:11:16.000 --> 00:11:29.000 So, let's consider, let's call s of x the area of a 00:11:29.000 --> 00:11:41.000 slice, well, by a plane parallel to the yz plane. 00:11:41.000 --> 00:11:46.000 OK, so on the picture, s of x is just the area of this 00:11:46.000 --> 00:11:51.000 thing in the vertical wall. Now, if you sum all of these, 00:11:51.000 --> 00:11:55.000 well, why does that work? So, if you take the origin 00:11:55.000 --> 00:11:59.000 between two parallel slices that are very close to each other, 00:11:59.000 --> 00:12:01.000 what's the volume in these two things? 00:12:01.000 --> 00:12:05.000 Well, it's essentially s of x times the thickness of this very 00:12:05.000 --> 00:12:09.000 thin slice, and the thickness would be 00:12:09.000 --> 00:12:16.000 delta x0 dx if you take a limit with more and more slices. 00:12:16.000 --> 00:12:25.000 OK, so the volume will be the integral of s of x dx from, 00:12:25.000 --> 00:12:32.000 well, what should be the range for x? 00:12:32.000 --> 00:12:35.000 Well, we would have to start at the very lowest value of x that 00:12:35.000 --> 00:12:39.000 ever happens in our origin, and we'd have to go all the way 00:12:39.000 --> 00:12:43.000 to the very largest value of x, from the very far back to the 00:12:43.000 --> 00:12:48.000 very far front. So, in this picture, 00:12:48.000 --> 00:12:57.000 we probably start over here at the back, and we'd end over here 00:12:57.000 --> 00:13:05.000 at the front. So, let me just say from the 00:13:05.000 --> 00:13:13.000 minimum, x, to the maximum x. And now, how do we find S of x? 00:13:13.000 --> 00:13:17.000 Well, S of x will be actually again an integral. 00:13:17.000 --> 00:13:20.000 But now, it's an integral of the variable, 00:13:20.000 --> 00:13:25.000 y, because when we look at this slice, what changes from left to 00:13:25.000 --> 00:13:35.000 right is y. So, well let me actually write 00:13:35.000 --> 00:13:40.000 that down. For a given, 00:13:40.000 --> 00:13:48.000 x, the area S of x you can compute as an integral of f of 00:13:48.000 --> 00:13:52.000 x, y dy. OK, well, now x is a constant, 00:13:52.000 --> 00:13:56.000 and y will be the variable of integration. 00:13:56.000 --> 00:14:00.000 What's the range for y? Well, it's from the leftmost 00:14:00.000 --> 00:14:06.000 point here to the rightmost point here on the given slice. 00:14:06.000 --> 00:14:10.000 So, there is a big catch here. That's a very important thing 00:14:10.000 --> 00:14:12.000 to remember. What is the range of 00:14:12.000 --> 00:14:14.000 integration? The range of integration for y 00:14:14.000 --> 00:14:18.000 depends actually on x. See, if I take the slice that's 00:14:18.000 --> 00:14:22.000 pictured on that diagram, then the range for y goes all 00:14:22.000 --> 00:14:26.000 the way from the very left to the very right. 00:14:26.000 --> 00:14:30.000 But, if I take a slice that, say, near the very front, 00:14:30.000 --> 00:14:32.000 then in fact, only a very small segment of it 00:14:32.000 --> 00:14:36.000 will be in my region. So, the range of values for y 00:14:36.000 --> 00:14:41.000 will be much less. Let me actually draw a 2D 00:14:41.000 --> 00:14:45.000 picture for that. So, remember, 00:14:45.000 --> 00:14:51.000 we fix x, so, sorry, so we fix a value of x. 00:14:51.000 --> 00:14:55.000 OK, and for a given value of x, what we will do is we'll slice 00:14:55.000 --> 00:14:59.000 our graph by this plane parallel to the yz plane. 00:14:59.000 --> 00:15:02.000 So, now we mention the graph is sitting above that. 00:15:02.000 --> 00:15:06.000 OK, that's the region R. We have the region, 00:15:06.000 --> 00:15:12.000 R, and I have the graph of a function above this region, 00:15:12.000 --> 00:15:15.000 R. And, I'm trying to find the 00:15:15.000 --> 00:15:20.000 area between this segment and the graph above it in this 00:15:20.000 --> 00:15:24.000 vertical plane. Well, to do that, 00:15:24.000 --> 00:15:33.000 I have to integrate from y going from here to here. 00:15:33.000 --> 00:15:37.000 I want the area of a piece that sits above this red segment. 00:15:37.000 --> 00:15:40.000 And, so in particular, the endpoints, 00:15:40.000 --> 00:15:43.000 the extreme values for y depend on x because, 00:15:43.000 --> 00:15:47.000 see, if I slice here instead, 00:15:47.000 --> 00:15:54.000 well, my bounds for y will be smaller. 00:15:54.000 --> 00:15:59.000 OK, so now, if I put the two things together, 00:15:59.000 --> 00:16:15.000 what I will get -- -- is actually a formula where I have 00:16:15.000 --> 00:16:30.000 to integrate -- -- over x -- -- an integral over y. 00:16:30.000 --> 00:16:42.000 OK, and so this is called an iterated integral because we 00:16:42.000 --> 00:16:54.000 iterate twice the process of taking an integral. 00:16:54.000 --> 00:16:57.000 OK, so again, what's important to realize 00:16:57.000 --> 00:16:59.000 here, I mean, I'm going to say that 00:16:59.000 --> 00:17:02.000 several times over the next few days but that's because it's the 00:17:02.000 --> 00:17:05.000 single most important thing to remember about double integrals, 00:17:05.000 --> 00:17:09.000 the bounds here are just going to be numbers, 00:17:09.000 --> 00:17:12.000 OK, because the question I'm asking myself here is, 00:17:12.000 --> 00:17:15.000 what is the first value of x by which I might want to slice, 00:17:15.000 --> 00:17:21.000 and what is the last value of x? Which range of x do I want to 00:17:21.000 --> 00:17:25.000 look at to take my red slices? And, the answer is I would go 00:17:25.000 --> 00:17:29.000 all the way from here, that's my first slice, 00:17:29.000 --> 00:17:34.000 to somewhere here. That's my last slice. 00:17:34.000 --> 00:17:38.000 For any value in between these, I will have some red segment, 00:17:38.000 --> 00:17:40.000 and I will want to integrate over that that. 00:17:40.000 --> 00:17:45.000 On the other hand here, the bounds will depend on the 00:17:45.000 --> 00:17:47.000 outer variable, x, 00:17:47.000 --> 00:17:54.000 because at a fixed value of x, what the values of y will be 00:17:54.000 --> 00:18:01.000 depends on x in general. OK, so I think we should do 00:18:01.000 --> 00:18:11.000 lots of examples to convince ourselves and see how it works. 00:18:11.000 --> 00:18:15.000 Yeah, it's called an iterated integral because first we 00:18:15.000 --> 00:18:18.000 integrated over y, and then we integrate again 00:18:18.000 --> 00:18:20.000 over x, OK? So, we can do that, 00:18:20.000 --> 00:18:23.000 well, I mean, y depends on x or x depends, 00:18:23.000 --> 00:18:27.000 no, actually x and y vary independently of each other 00:18:27.000 --> 00:18:31.000 inside here. What is more complicated is how 00:18:31.000 --> 00:18:34.000 the bounds on y depend on x. But actually, 00:18:34.000 --> 00:18:37.000 you could also do the other way around: first integrate over x, 00:18:37.000 --> 00:18:39.000 and then over y, and then the bounds for x will 00:18:39.000 --> 00:18:42.000 depend on y. We'll see that on an example. 00:18:42.000 --> 00:18:53.000 Yes? So, for y, I'm using the range 00:18:53.000 --> 00:18:58.000 of values for y that corresponds to the given value of x, 00:18:58.000 --> 00:19:00.000 OK? Remember, this is just like a 00:19:00.000 --> 00:19:04.000 plot in the xy plane. Above that, we have the graph. 00:19:04.000 --> 00:19:06.000 Maybe I should draw a picture here instead. 00:19:06.000 --> 00:19:09.000 For a given value of x, so that's a given slice, 00:19:09.000 --> 00:19:13.000 I have a range of values for y, that is, 00:19:13.000 --> 00:19:16.000 from this picture at the leftmost point on that slice to 00:19:16.000 --> 00:19:19.000 the rightmost point on that slice. 00:19:19.000 --> 00:19:24.000 So, where start and where I stop depends on the value of x. 00:19:24.000 --> 00:19:33.000 Does that make sense? OK. 00:19:33.000 --> 00:19:45.000 OK, no more questions? OK, so let's do our first 00:19:45.000 --> 00:19:52.000 example. So, let's say that we want to 00:19:52.000 --> 00:20:00.000 integrate the function 1-x^2-y^2 over the region defined by x 00:20:00.000 --> 00:20:06.000 between 0 and 1, and y between 0 and 1. 00:20:06.000 --> 00:20:09.000 So, what does that mean geometrically? 00:20:09.000 --> 00:20:13.000 Well, z = 1-x^2-y^2, and it's a variation on, 00:20:13.000 --> 00:20:17.000 actually I think we plotted that one, right? 00:20:17.000 --> 00:20:22.000 That was our first example of a function of two variables 00:20:22.000 --> 00:20:24.000 possibly. And, so, we saw that the graph 00:20:24.000 --> 00:20:27.000 is this paraboloid pointing downwards. 00:20:27.000 --> 00:20:31.000 OK, it's what you get by taking a parabola and rotating it. 00:20:31.000 --> 00:20:35.000 And now, what we are asking is, what is the volume between the 00:20:35.000 --> 00:20:40.000 paraboloid and the xy plane over the square of side one in the xy 00:20:40.000 --> 00:20:43.000 plane over the square of side one in the xy plane, 00:20:43.000 --> 00:20:49.000 x and y between zero and one. OK, so, what we'll do is we'll, 00:20:49.000 --> 00:20:54.000 so, see, here I try to represent the square. 00:20:54.000 --> 00:20:59.000 And, we'll just sum the areas of the slices as, 00:20:59.000 --> 00:21:03.000 say, x varies from zero to one. And here, of course, 00:21:03.000 --> 00:21:06.000 setting up the bounds will be easy because no matter what x I 00:21:06.000 --> 00:21:08.000 take, y still goes from zero to one. 00:21:08.000 --> 00:21:13.000 See, it's easiest to do double integrals what the region is 00:21:13.000 --> 00:21:18.000 just a rectangle on the xy plane because then you don't have to 00:21:18.000 --> 00:21:21.000 worry too much about what are the ranges. 00:21:21.000 --> 00:21:29.000 OK, so let's do it. Well, that would be the 00:21:29.000 --> 00:21:38.000 integral from zero to one of the integral from zero to one of 00:21:38.000 --> 00:21:42.000 1-x^2-y^2 dy dx. So, I'm dropping the 00:21:42.000 --> 00:21:44.000 parentheses. But, if you still want to see 00:21:44.000 --> 00:21:48.000 them, I'm going to put that in very thin so that you see what 00:21:48.000 --> 00:21:50.000 it means. But, actually, 00:21:50.000 --> 00:21:53.000 the convention is we won't put this parentheses in there 00:21:53.000 --> 00:21:57.000 anymore. OK, so what this means is first 00:21:57.000 --> 00:22:02.000 I will integrate 1-x^2-y^2 over y, ranging from zero to one with 00:22:02.000 --> 00:22:07.000 x held fixed. So, what that represents is the 00:22:07.000 --> 00:22:10.000 area in this slice. So, see here, 00:22:10.000 --> 00:22:13.000 I've drawn, well, what happens is actually the 00:22:13.000 --> 00:22:16.000 function takes positive and negative values. 00:22:16.000 --> 00:22:18.000 So, in fact, I will be counting positively 00:22:18.000 --> 00:22:21.000 this part of the area. And, I will be counting 00:22:21.000 --> 00:22:25.000 negatively this part of the area, I mean, 00:22:25.000 --> 00:22:32.000 as usual when I do an integral. OK, so what I will do to 00:22:32.000 --> 00:22:40.000 evaluate this, I will first do what's called 00:22:40.000 --> 00:22:48.000 the inner integral. So, to do the inner integral, 00:22:48.000 --> 00:22:55.000 well, it's pretty easy. How do I integrate this? 00:22:55.000 --> 00:22:58.000 Well, it becomes, so, what's the integral of one? 00:22:58.000 --> 00:23:01.000 It's y. Just anything to remember is we 00:23:01.000 --> 00:23:03.000 are integrating this with respect to y, 00:23:03.000 --> 00:23:08.000 not to x. The integral of x^2 is x^2 00:23:08.000 --> 00:23:14.000 times y. And, the integral of y^2 is y^3 00:23:14.000 --> 00:23:18.000 over 3. OK, and that we plug in the 00:23:18.000 --> 00:23:21.000 bounds, which are zero and one in this case. 00:23:21.000 --> 00:23:28.000 And so, when you plug y equals one, you will get one minus x^2 00:23:28.000 --> 00:23:34.000 minus one third minus, well, for y equals zero you get 00:23:34.000 --> 00:23:41.000 0,0, 0, so nothing changes. OK, so you are left with two 00:23:41.000 --> 00:23:45.000 thirds minus x^2. OK, and that's a function of x 00:23:45.000 --> 00:23:47.000 only. Here, you shouldn't see any y's 00:23:47.000 --> 00:23:49.000 anymore because y was your integration variable. 00:23:49.000 --> 00:23:55.000 But, you still have x. You still have x because the 00:23:55.000 --> 00:24:01.000 area of this shaded slice depends, of course, 00:24:01.000 --> 00:24:05.000 on the value of x. And, so now, 00:24:05.000 --> 00:24:11.000 the second thing to do is to do the outer integral. 00:24:11.000 --> 00:24:17.000 So, now we integrate from zero to one what we got, 00:24:17.000 --> 00:24:22.000 which is two thirds minus x^2 dx. 00:24:22.000 --> 00:24:26.000 OK, and we know how to compute that because that integrates to 00:24:26.000 --> 00:24:31.000 two thirds x minus one third x^3 between zero and one. 00:24:31.000 --> 00:24:33.000 And, I'll let you do the computation. 00:24:33.000 --> 00:24:43.000 You will find it's one third. OK, so that's the final answer. 00:24:43.000 --> 00:24:46.000 So, that's the general pattern. When we have a double integral 00:24:46.000 --> 00:24:49.000 to compute, first we want to set it up carefully. 00:24:49.000 --> 00:24:51.000 We want to find, what will be the bounds in x 00:24:51.000 --> 00:24:53.000 and y? And here, that was actually 00:24:53.000 --> 00:24:57.000 pretty easy because our equation was very simple. 00:24:57.000 --> 00:25:00.000 Then, we want to compute the inner integral, 00:25:00.000 --> 00:25:02.000 and then we compute the outer integral. 00:25:02.000 --> 00:25:11.000 And, that's it. OK, any questions at this point? 00:25:11.000 --> 00:25:16.000 No? OK, so, by the way, 00:25:16.000 --> 00:25:20.000 we started with dA in the notation, right? 00:25:20.000 --> 00:25:26.000 Here we had dA. And, that somehow became a dy 00:25:26.000 --> 00:25:28.000 dx. OK, 00:25:28.000 --> 00:25:35.000 so, dA became dy dx because when we 00:25:35.000 --> 00:25:39.000 do the iterated integral this way, 00:25:39.000 --> 00:25:45.000 what we're actually doing is that we are slicing our origin 00:25:45.000 --> 00:25:49.000 into small rectangles. OK, that was the area of this 00:25:49.000 --> 00:25:52.000 small rectangle here? Well, it's the product of its 00:25:52.000 --> 00:25:58.000 width times its height. So, that's delta x times delta 00:25:58.000 --> 00:26:03.000 y. OK, so, delta a equals delta x 00:26:03.000 --> 00:26:08.000 delta y becomes... So actually, 00:26:08.000 --> 00:26:18.000 it's not just becomes, it's really equal. 00:26:18.000 --> 00:26:28.000 So, the small rectangles for. Now, it became dy dx and not dx 00:26:28.000 --> 00:26:30.000 dy. Well, that's a question of, 00:26:30.000 --> 00:26:32.000 in which order we do the iterated integral? 00:26:32.000 --> 00:26:36.000 It's up to us to decide whether we want to integrate x first, 00:26:36.000 --> 00:26:38.000 then y, or y first, then x. 00:26:38.000 --> 00:26:41.000 But, as we'll see very soon, that is an important decision 00:26:41.000 --> 00:26:44.000 when it comes to setting up the bounds of integration. 00:26:44.000 --> 00:26:47.000 Here, it doesn't matter, but in general we have to be 00:26:47.000 --> 00:26:51.000 very careful about in which order we will do things. 00:26:51.000 --> 00:26:57.000 Yes? Well, in principle it always 00:26:57.000 --> 00:27:00.000 works both ways. Sometimes it will be that 00:27:00.000 --> 00:27:04.000 because the region has a strange shape, you can actually set it 00:27:04.000 --> 00:27:06.000 up more easily one way or the other. 00:27:06.000 --> 00:27:09.000 Sometimes it will also be that the function here, 00:27:09.000 --> 00:27:12.000 you actually know how to integrate in one way, 00:27:12.000 --> 00:27:15.000 but not the other. So, the theory is that it 00:27:15.000 --> 00:27:18.000 should work both ways. In practice, 00:27:18.000 --> 00:27:25.000 one of the two calculations may be much harder. 00:27:25.000 --> 00:27:33.000 OK. Let's do another example. 00:27:33.000 --> 00:27:39.000 Let's say that what I wanted to know was not actually what I 00:27:39.000 --> 00:27:42.000 computed, namely, the volume below the 00:27:42.000 --> 00:27:45.000 paraboloid, but also the negative of some 00:27:45.000 --> 00:27:48.000 part that's now in the corner towards me. 00:27:48.000 --> 00:27:51.000 But let's say really what I wanted was just the volume 00:27:51.000 --> 00:27:54.000 between the paraboloid and the xy plane, 00:27:54.000 --> 00:27:58.000 so looking only at the part of it that sits above the xy plane. 00:27:58.000 --> 00:28:01.000 So, that means, instead of integrating over the 00:28:01.000 --> 00:28:05.000 entire square of size one, I should just integrate over 00:28:05.000 --> 00:28:08.000 the quarter disk. I should stop integrating where 00:28:08.000 --> 00:28:18.000 my paraboloid hits the xy plane. So, let me draw another picture. 00:28:18.000 --> 00:28:28.000 So, let's say I wanted to integrate, actually -- So, 00:28:28.000 --> 00:28:36.000 let's call this example two. So, we are going to do the same 00:28:36.000 --> 00:28:39.000 function but over a different region. 00:28:39.000 --> 00:28:44.000 And, the region will just be, now, this quarter disk here. 00:28:44.000 --> 00:28:55.000 OK, so maybe I should draw a picture on the xy plane. 00:28:55.000 --> 00:29:14.000 That's your region, R. OK, so in principle, 00:29:14.000 --> 00:29:19.000 it will be the same integral. But what changes is the bounds. 00:29:19.000 --> 00:29:23.000 Why do the bounds change? Well, the bounds change because 00:29:23.000 --> 00:29:27.000 now if I set, if I fixed some value of x, 00:29:27.000 --> 00:29:31.000 then I want to integrate this part of the slice that's above 00:29:31.000 --> 00:29:36.000 the xy plane and I don't want to take this part that's actually 00:29:36.000 --> 00:29:39.000 outside of my disk. So, I should stop integrating 00:29:39.000 --> 00:29:42.000 over y when y reaches this value here. 00:29:42.000 --> 00:29:46.000 OK, on that picture here, on this picture, 00:29:46.000 --> 00:29:52.000 it tells me for a fixed value of x, the range of values for y 00:29:52.000 --> 00:29:55.000 should go only from here to here. 00:29:55.000 --> 00:30:00.000 So, that's from here to less than one. 00:30:00.000 --> 00:30:07.000 OK, so for a given, x, the range of y is, 00:30:07.000 --> 00:30:16.000 well, so what's the lowest value of y that we want to look 00:30:16.000 --> 00:30:20.000 at? It's still zero. 00:30:20.000 --> 00:30:24.000 From y equals zero to,what's the value of y here? 00:30:24.000 --> 00:30:28.000 Well, I have to solve in the equation of a circle, 00:30:28.000 --> 00:30:31.000 OK? So, if I'm here, 00:30:31.000 --> 00:30:39.000 this is x^2 y^2 equals one. That means y is square root of 00:30:39.000 --> 00:30:44.000 one minus x^2. OK, so I will integrate from y 00:30:44.000 --> 00:30:48.000 equals zero to y equals square root of one minus x^2. 00:30:48.000 --> 00:30:55.000 And, now you see how the bound from y will depend on the value 00:30:55.000 --> 00:30:59.000 of x. OK, so while I erase, 00:30:59.000 --> 00:31:05.000 I will let you think about, what is the bound for x now? 00:31:05.000 --> 00:31:06.000 It's a trick question. 00:31:35.000 --> 00:31:48.000 OK, so I claim that what we will do -- We write this as an 00:31:48.000 --> 00:31:57.000 iterated integral first dy then dx. 00:31:57.000 --> 00:32:04.000 And, we said for a fixed value of x, the range for y is from 00:32:04.000 --> 00:32:09.000 zero to square root of one minus x^2. 00:32:09.000 --> 00:32:13.000 What about the range for x? Well, the range for x should 00:32:13.000 --> 00:32:15.000 just be numbers. OK, remember, 00:32:15.000 --> 00:32:18.000 the question I have to ask now is if I look at all of these 00:32:18.000 --> 00:32:21.000 yellow slices, which one is the first one that 00:32:21.000 --> 00:32:24.000 I will consider? Which one is the last one that 00:32:24.000 --> 00:32:27.000 I want to consider? So, the smallest value of x 00:32:27.000 --> 00:32:30.000 that I want to consider is zero again. 00:32:30.000 --> 00:32:33.000 And then, I will have actually a pretty big slice. 00:32:33.000 --> 00:32:35.000 And I will get smaller, and smaller, 00:32:35.000 --> 00:32:36.000 and smaller slices. And, it stops. 00:32:36.000 --> 00:32:39.000 I have to stop when x equals one. 00:32:39.000 --> 00:32:42.000 Afterwards, there's nothing else to integrate. 00:32:42.000 --> 00:32:49.000 So, x goes from zero to one. OK, and now, 00:32:49.000 --> 00:32:52.000 see how in the inner integral, the bounds depend on in the 00:32:52.000 --> 00:32:55.000 inner integral, the bounds depend on x. 00:32:55.000 --> 00:32:58.000 In the outer one, you just get numbers because 00:32:58.000 --> 00:33:02.000 the questions that you have to ask to set up this one and set 00:33:02.000 --> 00:33:05.000 up that one are different. Here, the question is, 00:33:05.000 --> 00:33:07.000 if I fix a given, x, if I look at a given slice, 00:33:07.000 --> 00:33:10.000 what's the range for y? Here, the question is, 00:33:10.000 --> 00:33:16.000 what's the first slice? What is the last slice? 00:33:16.000 --> 00:33:21.000 Does that make sense? Everyone happy with that? 00:33:21.000 --> 00:33:28.000 OK, very good. So, now, how do we compute that? 00:33:28.000 --> 00:33:36.000 Well, we do the inner integral. So, that's an integral from 00:33:36.000 --> 00:33:42.000 zero to square root of one minus x^2 of one minus x^2 minus y^2 00:33:42.000 --> 00:33:48.000 dy. And, well, that integrates to 00:33:48.000 --> 00:34:00.000 y-x^2y-y^3 over three from zero to square root of one minus x^2. 00:34:00.000 --> 00:34:06.000 And then, that becomes, well, the root of one minus x^2 00:34:06.000 --> 00:34:13.000 minus x^2 root of one minus x^2 minus y minus x^2 to the three 00:34:13.000 --> 00:34:16.000 halves over three. And actually, 00:34:16.000 --> 00:34:20.000 if you look at it for long enough, see, this says one minus 00:34:20.000 --> 00:34:23.000 x^2 times square root of one minus x^2. 00:34:23.000 --> 00:34:26.000 So, actually, that's also, 00:34:26.000 --> 00:34:30.000 so, in fact, that simplifies to two thirds 00:34:30.000 --> 00:34:35.000 of one minus x^2 to the three halves. 00:34:35.000 --> 00:34:40.000 OK, let me redo that, maybe, slightly differently. 00:34:40.000 --> 00:34:50.000 This was one minus x^2 times y. So -- -- one minus x^2 times y 00:34:50.000 --> 00:35:02.000 becomes square root of one minus x^2 minus y^3 over three. 00:35:02.000 --> 00:35:04.000 And then, when I take y equals zero, I get zero. 00:35:04.000 --> 00:35:08.000 So, I don't subtract anything. OK, so now you see this is one 00:35:08.000 --> 00:35:11.000 minus x^2 to the three halves minus a third of it. 00:35:11.000 --> 00:35:21.000 So, you're left with two thirds. OK, so, that's the integral. 00:35:21.000 --> 00:35:27.000 The outer integral is the integral from zero to one of two 00:35:27.000 --> 00:35:31.000 thirds of one minus x^2 to the three halves dx. 00:35:31.000 --> 00:35:37.000 And, well, 00:35:37.000 --> 00:35:42.000 I let you see if you remember single variable integrals by 00:35:42.000 --> 00:35:47.000 trying to figure out what this actually comes out to be is it 00:35:47.000 --> 00:35:54.000 pi over two, or pi over eight, actually? 00:35:54.000 --> 00:36:12.000 I think it's pi over eight. OK, well I guess we have to do 00:36:12.000 --> 00:36:15.000 it then. I wrote something on my notes, 00:36:15.000 --> 00:36:16.000 but it's not very clear, OK? 00:36:16.000 --> 00:36:18.000 So, how do we compute this thing? 00:36:18.000 --> 00:36:20.000 Well, we have to do trig substitution. 00:36:20.000 --> 00:36:24.000 That's the only way I know to compute an integral like that, 00:36:24.000 --> 00:36:27.000 OK? So, we'll set x equal sine 00:36:27.000 --> 00:36:35.000 theta, and then square root of one minus x^2 will be cosine 00:36:35.000 --> 00:36:41.000 theta. We are using sine squared plus 00:36:41.000 --> 00:36:51.000 cosine squared equals one. And, so that will become -- -- 00:36:51.000 --> 00:36:59.000 so, two thirds remains two thirds. 00:36:59.000 --> 00:37:04.000 One minus x^2 to the three halves becomes cosine cubed 00:37:04.000 --> 00:37:08.000 theta. dx, well, if x is sine theta, 00:37:08.000 --> 00:37:13.000 then dx is cosine theta d theta. 00:37:13.000 --> 00:37:17.000 So, that's cosine theta d theta. And, 00:37:17.000 --> 00:37:19.000 well, if you do things with substitution, 00:37:19.000 --> 00:37:22.000 which is the way I do them, then you should worry about the 00:37:22.000 --> 00:37:25.000 bounds for theta which will be zero to pi over two. 00:37:25.000 --> 00:37:30.000 Or, you can also just plug in the bounds at the end. 00:37:30.000 --> 00:37:36.000 So, now you have the two thirds times the integral from zero to 00:37:36.000 --> 00:37:41.000 pi over two of cosine to the fourth theta d theta. 00:37:41.000 --> 00:37:48.000 And, how do you integrate that? Well, you have to use double 00:37:48.000 --> 00:37:55.000 angle formulas. OK, so cosine to the fourth, 00:37:55.000 --> 00:38:02.000 remember, cosine squared theta is one plus cosine two theta 00:38:02.000 --> 00:38:14.000 over two. And, we want the square of that. 00:38:14.000 --> 00:38:23.000 And, so that will give us -- -- of, 00:38:23.000 --> 00:38:29.000 well, we'll have, it's actually one quarter plus 00:38:29.000 --> 00:38:37.000 one half cosine to theta plus one quarter cosine square to 00:38:37.000 --> 00:38:41.000 theta d theta. And, how will you handle this 00:38:41.000 --> 00:38:42.000 guy? Well, using, 00:38:42.000 --> 00:38:45.000 again, the double angle formula. 00:38:45.000 --> 00:38:51.000 OK, so it's getting slightly nasty. 00:38:51.000 --> 00:38:54.000 So, but I don't know any simpler solution except for one 00:38:54.000 --> 00:38:56.000 simpler solution, which is you have a table of 00:38:56.000 --> 00:38:59.000 integrals of this form inside the notes. 00:38:59.000 --> 00:39:08.000 Yes? No, I don't think so because if 00:39:08.000 --> 00:39:12.000 you take one half times cosine half times two, 00:39:12.000 --> 00:39:15.000 you will still have half, OK? 00:39:15.000 --> 00:39:18.000 So, if you do, again, the double angle 00:39:18.000 --> 00:39:22.000 formula, I think I'm not going to bother to do it. 00:39:22.000 --> 00:39:37.000 I claim you will get, at the end, pi over eight 00:39:37.000 --> 00:39:46.000 because I say so. OK, so exercise, 00:39:46.000 --> 00:39:54.000 continue calculating and get pi over eight. 00:39:54.000 --> 00:39:56.000 OK, now what does the show us? Well, this shows us, 00:39:56.000 --> 00:40:00.000 actually, that this is probably not the right way to do this. 00:40:00.000 --> 00:40:02.000 OK, the right way to do this will be to integrate it in polar 00:40:02.000 --> 00:40:05.000 coordinates. And, that's what we will learn 00:40:05.000 --> 00:40:09.000 how to do tomorrow. So, we will actually see how to 00:40:09.000 --> 00:40:11.000 do it with much less trig. 00:40:36.000 --> 00:40:46.000 So, that will be easier in polar coordinates. 00:40:46.000 --> 00:40:51.000 So, we will see that tomorrow. OK, so we are almost there. 00:40:51.000 --> 00:40:53.000 I mean, here you just use a double angle again and then you 00:40:53.000 --> 00:40:55.000 can get it. And, it's pretty 00:40:55.000 --> 00:41:00.000 straightforward. OK, so one thing that's kind of 00:41:00.000 --> 00:41:07.000 interesting to know is we can exchange the order of 00:41:07.000 --> 00:41:11.000 integration. Say we have an integral given 00:41:11.000 --> 00:41:14.000 to us in the order dy dx, we can switch it to dx dy. 00:41:14.000 --> 00:41:18.000 But, we have to be extremely careful with the bounds. 00:41:18.000 --> 00:41:22.000 So, you certainly cannot just swap the bounds of the inner and 00:41:22.000 --> 00:41:26.000 outer because there you would end up having this square root 00:41:26.000 --> 00:41:30.000 of one minus x^2 on the outside, and you would never get a 00:41:30.000 --> 00:41:33.000 number out of that. So, that cannot work. 00:41:33.000 --> 00:41:37.000 It's more complicated than that. OK, so, well, 00:41:37.000 --> 00:41:42.000 here's a first baby example. Certainly, if I do integral 00:41:42.000 --> 00:41:46.000 from zero to one, integral from zero to two dx 00:41:46.000 --> 00:41:51.000 dy, there, I can certainly switch the bounds without 00:41:51.000 --> 00:41:54.000 thinking too much. What's the reason for that? 00:41:54.000 --> 00:41:59.000 Well, the reason for that is this corresponds in both cases 00:41:59.000 --> 00:42:04.000 to integrating x from zero to two, and y from zero to one. 00:42:04.000 --> 00:42:08.000 It's a rectangle. So, if I slice it this way, 00:42:08.000 --> 00:42:11.000 you see that y goes from zero to one for any x between zero 00:42:11.000 --> 00:42:13.000 and two. It's this guy. 00:42:13.000 --> 00:42:18.000 If I slice it that way, then x goes from zero to two 00:42:18.000 --> 00:42:22.000 for any value of y between zero and one. 00:42:22.000 --> 00:42:25.000 And, it's this one. So, here it works. 00:42:25.000 --> 00:42:28.000 But in general, I have to draw picture of my 00:42:28.000 --> 00:42:32.000 region, and see how the slices look like both ways. 00:42:32.000 --> 00:42:41.000 OK, so let's do a more interesting one. 00:42:41.000 --> 00:42:48.000 Let's say that I want to compute an integral from zero to 00:42:48.000 --> 00:42:55.000 one of integral from x to square root of x of e^y over y dy dx. 00:42:55.000 --> 00:42:59.000 So, why did I choose this guy? Which is the guy because as far 00:42:59.000 --> 00:43:03.000 as I can tell, there's no way to integrate e^y 00:43:03.000 --> 00:43:06.000 over y. So, this is an integral that 00:43:06.000 --> 00:43:10.000 you cannot compute this way. So, it's a good example for why 00:43:10.000 --> 00:43:13.000 this can be useful. So, if you do it this way, 00:43:13.000 --> 00:43:14.000 you are stuck immediately. So, instead, 00:43:14.000 --> 00:43:15.000 we will try to switch the order. 00:43:15.000 --> 00:43:20.000 But, to switch the order, we have to understand, 00:43:20.000 --> 00:43:25.000 what do these bounds mean? OK, so let's draw a picture of 00:43:25.000 --> 00:43:29.000 the region. Well what I am saying is y 00:43:29.000 --> 00:43:33.000 equals x to y equals square root of x. 00:43:33.000 --> 00:43:48.000 Well, let's draw y equals x, and y equals square root of x. 00:43:48.000 --> 00:43:52.000 Well, maybe I should actually put this here, 00:43:52.000 --> 00:43:56.000 y equals x to y equals square root of x. 00:43:56.000 --> 00:44:01.000 OK, and so I will go, for each value of x I will go 00:44:01.000 --> 00:44:07.000 from y equals xo to y equals square root of x. 00:44:07.000 --> 00:44:11.000 And then, we'll do that for values of x that go from x 00:44:11.000 --> 00:44:15.000 equals zero to x equals one, which happens to be exactly 00:44:15.000 --> 00:44:22.000 where these things intersect. So, my region will consist of 00:44:22.000 --> 00:44:26.000 all this, OK? So now, if I want to do it the 00:44:26.000 --> 00:44:29.000 other way around, I have to decompose my region. 00:44:29.000 --> 00:44:32.000 The other way around, I have to, so my goal, 00:44:32.000 --> 00:44:35.000 now, is to rewrite this as an integral. 00:44:35.000 --> 00:44:36.000 Well, it's still the same function. 00:44:36.000 --> 00:44:41.000 It's still e to the y over y. But now, I want to integrate dx 00:44:41.000 --> 00:44:45.000 dy. So, how do I integrate over x? 00:44:45.000 --> 00:44:50.000 Well, I fix a value of y. And, for that value of y, 00:44:50.000 --> 00:44:54.000 what's the range of x? Well, the range for x is from 00:44:54.000 --> 00:44:57.000 here to here. OK, what's the value of x here? 00:44:57.000 --> 00:45:02.000 Let's start with an easy one. This is x equals y. 00:45:02.000 --> 00:45:08.000 What about this one? It's x equals y^2. 00:45:08.000 --> 00:45:16.000 OK, so, x goes from y2 to y, and then what about y? 00:45:16.000 --> 00:45:19.000 Well, I have to start at the bottom of my region. 00:45:19.000 --> 00:45:24.000 That's y equals zero to the top, which is at y equals one. 00:45:24.000 --> 00:45:31.000 So, y goes from zero to one. So, switching the bounds is not 00:45:31.000 --> 00:45:36.000 completely obvious. That took a little bit of work. 00:45:36.000 --> 00:45:40.000 But now that we've done that, well, 00:45:40.000 --> 00:45:44.000 just to see how it goes, it's actually going to be much 00:45:44.000 --> 00:45:48.000 easier to integrate because the inner integral, 00:45:48.000 --> 00:45:52.000 well, what's the integral of e^y over y with respect to x? 00:45:52.000 --> 00:46:01.000 It's just that times x, right, from x equals y^2 to y. 00:46:01.000 --> 00:46:04.000 So, that will be, well, if I plug x equals y, 00:46:04.000 --> 00:46:09.000 I will get e to the y minus, if I plug x equals y^2, 00:46:09.000 --> 00:46:16.000 I will get e to the y over y times y^2 into the y times y, 00:46:16.000 --> 00:46:20.000 OK? So, now, if I do the outer 00:46:20.000 --> 00:46:27.000 integral, I will have the integral from zero to one of e 00:46:27.000 --> 00:46:33.000 to the y minus y^e to the y dy. And, that one actually is a 00:46:33.000 --> 00:46:37.000 little bit easier. So, we know how to integrate 00:46:37.000 --> 00:46:39.000 e^y. We don't quite know how to 00:46:39.000 --> 00:46:43.000 integrate ye^y. But, let's try. 00:46:43.000 --> 00:46:49.000 So, let's see, what's the derivative of ye^y? 00:46:49.000 --> 00:46:55.000 Well, there's a product rule that's one times e^y plus y 00:46:55.000 --> 00:46:59.000 times the derivative of e^y is ye^y. 00:46:59.000 --> 00:47:03.000 So, if we do, OK, let's put a minus sign in 00:47:03.000 --> 00:47:06.000 front. Well, that's almost what we 00:47:06.000 --> 00:47:09.000 want, except we have a minus e^y instead of a plus e^y. 00:47:09.000 --> 00:47:15.000 So, we need to add 2e^y. And, I claim that's the 00:47:15.000 --> 00:47:18.000 antiderivative. OK, if you got lost, 00:47:18.000 --> 00:47:21.000 you can also integrate by integrating by parts, 00:47:21.000 --> 00:47:25.000 by taking the derivative of y and integrating these guys. 00:47:25.000 --> 00:47:29.000 Or, but, you know, that works. Just, your first guess would 00:47:29.000 --> 00:47:32.000 be, maybe, let's try minus y^e to the y. 00:47:32.000 --> 00:47:35.000 Take the derivative of that, compare, see what you need to 00:47:35.000 --> 00:47:39.000 do to fix. And so, if you take that 00:47:39.000 --> 00:47:45.000 between zero and one, you'll actually get e minus 00:47:45.000 --> 00:47:47.000 two. OK, 00:47:47.000 --> 00:47:50.000 so, tomorrow we are going to see 00:47:50.000 --> 00:47:53.000 how to do double integrals in polar coordinates, 00:47:53.000 --> 00:47:55.000 and also applications of double integrals, 00:47:55.000 --> 00:47:58.000 how to use them for interesting things.