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So basically, for the last few weeks
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we've been doing derivatives, now
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we're going to do integrals.
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OK, so more precisely, we're going
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to be talking about double integrals.
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OK, so just to motivate the notion,
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let me just remind you that when you have a function of one
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variable, say f of x.
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And if you take its integral from say a to b of f of x d x,
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well, that corresponds to the area below the graph of f
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over the interval from a to b.
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So the picture is something like, at a, you have b,
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you have the graph of f.
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And then what the integral measures
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is the area of this region.
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And when we say the area of this region,
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of course, if f is positive, that's what happens,
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if f is negative, then we can negatively
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the area below the x-axis.
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OK, so now when you have a function of two variables,
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then you can try to do the same thing.
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Namely, you can plot its graph.
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Its graph will be a surface in space.
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And then we can try to look for the volume below the graph.
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And that's what we will call the double integral of a function
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over a certain region.
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OK, so let's take a look at the graph.
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Let's say that we have a function of two variables x and y.
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Then we look at the volume that's below the graph, z
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equals f of xy.
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So a picture for what this means.
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I have a function of x and y.
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I can draw its graph.
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The graph will be the surface with equation z equals f of x
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and y.
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And while I have to decide where I will integrate the function,
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so far that I will choose some region in the xy plane.
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And I will integrate the function on that region.
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So it's over a region r in the xy plane.
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So I have this region r.
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And I look at the piece of a graph that is above this region.
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And we'll try to compute the volume of this solid here.
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OK, that's what the double integral will measure.
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So we'll call that the double integral of a region r of f of xy,
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dA.
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And I will have to explain what the notation means.
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So dA here stands for a piece of area, a stands for area.
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And well, it's a double integral.
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So that's why we have two integral signs.
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And we'll have to indicate somehow the region of a region
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where we're integrating.
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We'll come up with more concrete notations
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when we see how to actually compute these things.
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That's the basic definition.
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So actually, how do, well, I mean, that's not
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really much of a definition yet.
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How do we actually define this rigorously?
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Well, remember the integral in one variable,
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you probably saw a definition where you take your interval
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from a to b and you cut it into little pieces.
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And then for each little piece, you take a value of a function
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and you multiply by the weight for the piece
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that gives you a rectangular slice.
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And then you sum all of these rectangular slices together.
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So here we'll do the same thing.
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So well, let me put a picture up and explain what it does.
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So we are going to cut our region into little pieces,
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say, little rectangles, or actually anything we want.
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And then for each piece, with a small area delta A,
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we'll take the area delta A times the value of a function
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in there that will give us the volume of a small box
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that sits under the graph.
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And then we'll add all these boxes together
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that gives us an estimate of a volume.
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And then to get, actually, the integral
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will be defined as a limit as we subdivide
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into smaller and smaller boxes and with some more and more pieces.
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So actually, what we do, oh, I still have a board here.
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So the actual definition involves cutting R into small pieces
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of area, let's call delta A, maybe delta A, I, the area of the I
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piece, and then, so maybe in the xy plane, we have our region
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and we'll cut it maybe using some grid.
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And then we'll have each small piece.
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Each small piece will have a real delta A, or delta A, I.
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And it will be at some point, let's call it xy, yi, yi, xi.
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And then we'll consider the sum over all the pieces of f
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at that point, xy, yi, times the area of a small piece.
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So what that corresponds to in the three-dimensional picture
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is just I sum the volumes of all of these little columns
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that set under the graph.
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And then, so what I do is, actually, I take the limit,
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as the size of the pieces tends to 0,
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so I have more and more smaller and smaller pieces.
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And that gives me the double integral.
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That's not a very good sentence, but whatever.
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OK, so that's the definition.
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Of course, we'll have to see how to compute it.
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We don't actually compute it.
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When you compute an integral in single variable calculus,
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you don't do that.
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You don't cut into little pieces and sum the pieces together.
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You've learned how to integrate functions using various formulas.
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And similarly, here, we'll learn how
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to actually compute these things without doing
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that cutting into small pieces.
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Any questions first about the concept
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or what the definition is?
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Yes?
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Well, so we'll have to learn which tricks work and how exactly.
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But what we'll do, actually, is we'll
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reduce the calculation of a double integral
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to two calculations of single integrals.
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And so for these, certainly, all the tricks you've learned
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in single variable calculus will come in handy.
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OK, so yes, that's a strong signal, by the way,
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that if you've forgotten everything
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about single variable calculus, now
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might be a good time to actually brush up on integrals.
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The usual integrals and the usual substitution
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of the integrals are very useful.
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So, yes, so how do we compute these things?
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That's what we have to come up with.
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And, well, going back to what we did with derivatives,
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to understand variations of functions and derivatives,
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what we did was really we took slices parallel
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to an excess or another one.
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So, in fact, here the key is also the same.
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So what we're going to do is we're going to do
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a lot of small boxes like that, and something completely
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at random, will actually somehow scan our region,
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scan through our region by parallel planes.
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OK, so let me put up actually a slightly different picture
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up here.
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So what I'm going to do is I'm going to take planes,
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say, in this picture, parallel to the waisie plane,
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I'll take a moving plane, and I'll take a plane
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parallel to the waisie plane, I'll take a moving plane
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that scans from the back to the front or from the front
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to the back.
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So that means, you know, I set a value of x0 of x,
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and I look at the slice x equals x0,
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and then I will do that for all values of x0.
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So now in each slice, well, I get what looks
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a lot like a single variable integral.
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OK, and that integral will tell me what is the area
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in this, well, I guess it was supposed to be green,
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but it all comes as black.
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So let's say the black shaded slice,
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and then when I add all of these areas together
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as the value of x changes, I will get the volume.
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OK, let me try to explain that again.
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So to compute this integral,
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what we do is actually we take slices.
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So let's consider, let's call s of x,
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the area of a slice, well, by a plane parallel
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to the y-z plane.
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OK, so on the picture, s of x is just the area of this thing
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in the vertical wall.
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Now, if you sum all of these, well, let's see,
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why does that work?
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So if you take the region between two parallel slices,
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but are very close to each other,
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what's the volume between these two things?
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Essentially s of x times the thickness of this very thin slice,
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and the thickness would be delta xO dx,
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if you take a limit with more and more slices.
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OK, so that the volume will be the integral of s of x dx.
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Well, what should be the range for x?
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Well, we would have to start at the very lowest value of x
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that ever happens in our region, and we'd have to go all the way
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to the very largest value of x, OK, from the very far back
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to the very far front.
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So on this picture, we'd probably start over here at the back,
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and with end over here at the front.
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So let me just say from the minimum x to the maximum x.
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And now, how do we find s of x?
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Well, s of x will be actually again an integral,
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but now it's an integral of a variable y
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because when we look at this slice, what changes from left to right is y.
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So let me actually find that done.
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For a given x, the area s of x, you can compute as an integral
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of f of xy dy.
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OK, well now x is a constant, and y will be the variable of integration.
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What's the range for y?
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Well, it's from the leftmost point here to the rightmost point here
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on the given slice.
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So there's a big catch here.
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That's a very important thing to remember.
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What is the range of integration for y depends actually on x.
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See, if I take this slice that's pictured on that diagram,
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then the range for y goes all the way from the very left to the very right.
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But if I take a slice that's, say, near the very front,
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then in fact, only a very small segment of it will be in my region.
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So the range of values for y will be much less.
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Let me actually draw a 2D picture for that.
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So remember, we fix x, and we fix a value of x.
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And for a given value of x, what we'll do is we'll slice our graph
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by this plane parallel to the y-zip plane.
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So now we imagine the graph is sitting above that.
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And that's the region r.
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I have the region r.
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And I have the graph of a function above this region r.
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And I'm trying to find the area between this segment and the graph above it
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in this vertical plane.
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Well, to do that, I have to integrate from y going from here to here.
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I want the area of a piece that sits above this red segment.
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And so in particular, the end point, the extreme values for y depend on x.
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Because see, if I slice here instead, well, my bounds for y will be smaller.
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So now if I put the two things together, what I will get is actually from you
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where I have to integrate over x and integral over y.
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And this is called an iterated integral because we iterate twice the process of taking an integral.
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So again, what's important to realize here, I'm going to say that several times
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over the next few days, but that's because it's the single most important thing to remember about the whole integral.
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The bounds here are just going to be numbers.
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Because the question that I'm asking myself here is what is the first value of x by which I want to slice.
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And what is the last value of x?
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Which range of x do I want to look at to take my red slices?
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And the answer is I would go all the way from here, that's my first slice, to somewhere here, that's my last slice.
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For any value in between these, I will have some red segment and I will want to integrate over that.
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On the other hand, here, the bounds will depend on the outer variable x.
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Because if I fix a value of x, what the values of y will be depends on x in general.
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So I think probably we should do lots of examples to convince ourselves and see how it works.
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It's called an integrated integral because first we integrate over y and then we integrate again over x.
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So we can do that.
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I mean, y depends on x or x depends. No, actually x and y vary independently of each other inside here.
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What is more complicated is how the bounds on y depend on x.
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But actually, you could also do it the other way around.
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First integrate over x and then over y.
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And then the bounds for x would depend on y. We'll see that on an example.
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For your y values, are you using the starting point?
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So for y, I'm using the range of values for y that corresponds to the given value of x.
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Remember, this is just like a plot in the x-y plane, above that we have the graph.
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Maybe I should have a picture here instead. For a given value of x, so that's a given slice,
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I have a range of values for y that is from, you know, this picture is the leftmost point on that slice to the rightmost point on that slice.
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So where I start and where I stop depends on the value of x.
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Does that make sense?
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Okay. Okay. No more questions? No? Okay.
245
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So let's do a first example.
246
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So let's say that we want to integrate the function 1-x square minus y squared over the region defined by x between 0 and 1 and y between 0 and 1.
247
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So what does that mean geometrically?
248
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Well, z equals 1-x square minus y squared.
249
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And it's a variation on, you know, something, I think actually we've plotted that one, right?
250
00:20:17.640 --> 00:20:22.640
When we've, that was our first example of a function of two variables, possibly.
251
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And so we saw that the graph is this paraboloid pointing downwards.
252
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Okay. It's what you get by taking a parabola, parabola, and rotating it.
253
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And now what we're asking is what is the volume between the paraboloid and the x-y plane over the square of side 1 in the x-y plane?
254
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x and y between 0 and 1. Okay.
255
00:20:45.640 --> 00:21:01.640
So what we'll do is we'll, so see, here I've tried to represent the square and we'll just sum the areas of the slices as, say, x varies from 0 to 1.
256
00:21:01.640 --> 00:21:09.640
And here, of course, setting up the bounds will be easy because no matter what x I take, why still goes from 0 to 1?
257
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See, it's easiest to do double integrals when the region is just a rectangle in the x-y plane.
258
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Because then you don't have to worry too much about what are the ranges.
259
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Okay, so let's do it. Well, that would be the integral from 0 to 1 of the integral from 0 to 1 of 1 minus x square minus y squared,
260
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d y dx. Okay, so I'm dropping the parentheses, but if you still want to see them, I'm going to put them in very thin.
261
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So as you see what it means, but actually the convention is we won't put these parentheses in there anymore.
262
00:21:54.640 --> 00:22:03.640
Okay, so what this means is first I will integrate 1 minus x square minus y squared over y ranging from 0 to 1 with x health fixed.
263
00:22:03.640 --> 00:22:10.640
So what that represents is the area in this slice. So see here actually I've drawn.
264
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Well, what happens is actually the function takes positive and negative values.
265
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So in fact, I will be counting positively this part of the area. And I will be counting negatively this part of the area.
266
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I mean, as usual when I do an integral.
267
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So what I will do to evaluate this, I will first do what's called the inner integral.
268
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So to do the inner integral, well, it's pretty easy.
269
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How do I integrate this? Well, it becomes, so what's the integral of 1?
270
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It's y. Just the only thing to remember is we're integrating respect to y, not to x.
271
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The integral of x squared is x squared times y.
272
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And the integral of y squared is y cubed over 3.
273
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And then we plug in the bounds, which are 0 and 1 in this case.
274
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And so when you plug y equals 1, you will get 1 minus x squared minus 1 third.
275
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Minus, well, from y equals 0, you get 0, 0, 0. So nothing changes.
276
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So you're left with 2 thirds minus x squared. And that's a function of x only.
277
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Here, you shouldn't see any y's anymore because y was your integration variable.
278
00:23:49.640 --> 00:24:03.640
But you still have x. You still have x because the area of this shaded slice depends, of course, on the value of x.
279
00:24:03.640 --> 00:24:11.640
And so now the second thing to do is to do the outer integral.
280
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So now we integrate from 0 to 1, what we got, which is 2 thirds minus x squared, dx.
281
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And we know how to compute that because that integrates to 2 thirds x minus 1 third x cubed
282
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between 0 and 1. And I let you do the computation. You will find it's 1 third.
283
00:24:38.640 --> 00:24:42.640
OK, so that's the final answer.
284
00:24:42.640 --> 00:24:48.640
So that's the general pattern. See, when we have a double integral to compute, first we want to set it up carefully.
285
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We want to find what will be the bounds in x and y. And here that was actually pretty easy because our region was very simple.
286
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Then we want to compute the inner integral and then we compute the outer integral. And that's it.
287
00:25:05.640 --> 00:25:14.640
Any questions at this point?
288
00:25:14.640 --> 00:25:27.640
OK, so by the way, we started with a dA in the notation. Here we had dA. And that somehow became a dy dx.
289
00:25:27.640 --> 00:25:42.640
So dA became dy dx because when we do the outer integral this way, what we are actually doing is that we are slicing our region
290
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into small rectangles. Now what's the area of a small rectangle here?
291
00:25:49.640 --> 00:25:58.640
Well, it's the product of its width times its height. So that's delta x times delta y.
292
00:25:58.640 --> 00:26:25.640
So delta x equals delta x delta y became, so actually it's not just becomes it's really equal for the small rectangles for.
293
00:26:25.640 --> 00:26:32.640
Now it became dy dx and not dx dy. Well, that's a question of in which order we do the iterative integral.
294
00:26:32.640 --> 00:26:43.640
It's up to us to decide whether we want to integrate x first, then y or y first, then x. But as we'll see very soon, that is an important decision when it comes to setting up the bounds of integration.
295
00:26:43.640 --> 00:26:50.640
Here it doesn't matter, but in general we have to be very careful about in which order we'll do things.
296
00:26:50.640 --> 00:27:06.640
Yes? Well, always works. Yeah, it's well, in principle it always works both ways. Sometimes it will be that you know because the region has a strange shape, you can actually set it up more easily one way than the other.
297
00:27:06.640 --> 00:27:12.640
Sometimes it will also be that the function here, you actually know how to integrate in one way but not the other.
298
00:27:12.640 --> 00:27:25.640
So the theory is that it should work both ways. In practice, one of the two calculations may be much harder.
299
00:27:25.640 --> 00:27:32.640
Okay. Let's do another example.
300
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Let's say that what I wanted to know was not actually what I computed, namely the volume below the paraboloid but also the negative of some part that's in the corner towards me.
301
00:27:48.640 --> 00:27:57.640
But let's say really what I wanted was just the volume between the paraboloid and the xy plane. So looking only at the part of it that sits above the xy plane.
302
00:27:57.640 --> 00:28:05.640
So that means instead of integrating over the entire square of size one, I should just integrate over the quarter disk.
303
00:28:05.640 --> 00:28:17.640
I should stop integrating where my paraboloid hits the xy plane. So let me draw another picture.
304
00:28:17.640 --> 00:28:28.640
So let's say I wanted to integrate actually.
305
00:28:28.640 --> 00:28:33.640
So let's call this example two.
306
00:28:33.640 --> 00:28:41.640
So it's the same. We're going to do the same function but over a different region and the region will just be now.
307
00:28:41.640 --> 00:28:55.640
This quarter disk here. So maybe I should draw a picture in the xy plane.
308
00:28:55.640 --> 00:29:13.640
So let's see what I'm going to do.
309
00:29:13.640 --> 00:29:37.640
So in principle it will be the same integral but what changes is the bounds. Why do the bounds change? Well the bounds change because now if I fix some value of x then I will want to integrate this part of the slice that's above the xy plane and I don't want to take this part that's actually outside of my disk.
310
00:29:37.640 --> 00:29:46.640
So I should stop integrating over y when y reaches this value here. On that picture here.
311
00:29:46.640 --> 00:29:55.640
On this picture tells me for a fixed value of x the range of values for y should go only from here to here.
312
00:29:55.640 --> 00:30:13.640
So that's from here to less than one. So for given x the range of y is well.
313
00:30:13.640 --> 00:30:28.640
So what's the lowest value of y that I want to look at? It's still zero. From y equals zero to what's the value of y here? Well I have to solve in the equation of a circle.
314
00:30:28.640 --> 00:30:41.640
So if I'm here this is x squared plus y squared equals one. That means y is square root of one minus x squared.
315
00:30:41.640 --> 00:30:56.640
So I will integrate from y equals zero to y equals square root of one minus x squared. And now you see how the bound for y will depend on the value of x.
316
00:30:56.640 --> 00:31:05.640
So one of your ways I will let you think about what is the bound for x now.
317
00:31:05.640 --> 00:31:34.640
So this is a very good question.
318
00:31:34.640 --> 00:31:49.640
So I claim that what we'll do.
319
00:31:49.640 --> 00:32:09.640
So we'll write this as an iterative integral. First dy than the x. And we said for a fixed value of x the range for y is from zero to square root of one minus x squared.
320
00:32:09.640 --> 00:32:25.640
So what about the range for x? Well the range for x should just be numbers. Remember the question I have to ask now is if I look at all of these yellow slices which one is the first one that I want to consider which one is the last one that I want to consider.
321
00:32:25.640 --> 00:32:39.640
And then I will have there I will actually have a pretty big slice and I will get smaller and smaller and smaller slices. And it stops I have to stop when x equals one.
322
00:32:39.640 --> 00:32:48.640
Afterwards there's nothing else to integrate. So x goes from zero to one.
323
00:32:48.640 --> 00:33:03.640
Okay and now see how in the inner integral the bounds depend on x in the outer one you just get numbers because the questions that you have to ask to set up this one set up that one are different.
324
00:33:03.640 --> 00:33:09.640
Here the question is if I fix a given x if I look at a given slice what's the range for y here.
325
00:33:09.640 --> 00:33:23.640
The question is what's the first slice what is the last slice. Does that make sense? Everyone happy with that? Okay very good.
326
00:33:23.640 --> 00:33:43.640
So now how do we compute that. Well we do the inner integral so that's integral from zero to square root of one minus x squared of one minus x squared minus y squared the y.
327
00:33:43.640 --> 00:34:00.640
And well that integrates to y minus x squared y minus y cubed over three from zero to square root of one minus x squared.
328
00:34:00.640 --> 00:34:15.640
And then that becomes well root of one minus x squared minus x squared root of one minus x squared minus one minus x squared to the three halves of a free.
329
00:34:15.640 --> 00:34:23.640
And actually if you look at it for long enough see this is one minus x squared times square root of one minus x squared.
330
00:34:23.640 --> 00:34:35.640
And also so in fact that simplifies to two thirds of one minus x squared to the three halves.
331
00:34:35.640 --> 00:34:51.640
Okay let me redo that maybe slightly differently this was one minus x squared times y so.
332
00:34:51.640 --> 00:35:05.640
And then when I take y equals zero I get zero so I don't subtract anything.
333
00:35:05.640 --> 00:35:14.640
Okay so now you see this is one minus x squared to the three halves minus a third of it so you're left with two thirds.
334
00:35:14.640 --> 00:35:31.640
So that's the inner integral the outer integral is integral from zero to one of two thirds of one minus x squared to the three halves the x.
335
00:35:31.640 --> 00:35:47.640
And well I let you see if you remember single variable integral also by trying to figure out how this actually comes out to be.
336
00:35:47.640 --> 00:36:08.640
Is it by over two or by over eight truly. I think it's by over eight.
337
00:36:08.640 --> 00:36:24.640
Okay well I guess we have to do it then. I've also been on my notes but it's not very clear okay so how do we compute this thing well we have to do to take substitution that's the only way I know to compute anything or like that.
338
00:36:24.640 --> 00:36:45.640
So we'll set x equal sine theta and then square root of one minus x squared will be cosine theta we're using sine square plus cosine square equals one.
339
00:36:45.640 --> 00:36:55.640
And so that will become.
340
00:36:55.640 --> 00:37:12.640
So two thirds remains two thirds one minus x squared to the three halves becomes cosine cubed theta dx well sine if x is sine theta then dx is cosine theta.
341
00:37:12.640 --> 00:37:29.640
So that's cosine theta theta. And well if you do things to the way I do them then you should worry about the bounds for theta which will be zero to pi over two or you can you know also just plug into the bounds at the end.
342
00:37:29.640 --> 00:37:51.640
So now you have the two thirds times the integral from zero to pi over two of cosine to the four theta d theta. And how do you integrate that well you have to use double angle formulas.
343
00:37:51.640 --> 00:38:14.640
Okay so cosine to the four remember cosine square theta is one plus cosine two theta over two and we want the square of that.
344
00:38:14.640 --> 00:38:24.640
And so that will give us. Of well we'll have.
345
00:38:24.640 --> 00:38:32.640
So actually. One quarter plus one half cosine to theta.
346
00:38:32.640 --> 00:38:45.640
And how will you handle this guy while using again for the ball angle formula.
347
00:38:45.640 --> 00:38:50.640
Okay so it's getting slightly nasty.
348
00:38:50.640 --> 00:39:07.640
So but I don't know any simpler solution except for one simpler solution which is you have a table of integrals of this form inside the notes.
349
00:39:07.640 --> 00:39:15.640
No I don't think so because if you take one half times cosine half times two you will still have a half.
350
00:39:15.640 --> 00:39:21.640
So if you do again for the one-angle formula I think I'm not going to bother to do it.
351
00:39:21.640 --> 00:39:26.640
I claim you will get at the end.
352
00:39:26.640 --> 00:39:46.640
So I say so. So okay so exercise.
353
00:39:46.640 --> 00:39:53.640
Continue calculating and get by over eight.
354
00:39:53.640 --> 00:39:59.640
Okay now what does this show us well this shows us actually that this is probably not the right way to do this.
355
00:39:59.640 --> 00:40:05.640
Okay the right way to do this will be to integrate it in polar coordinates and that's what we'll learn how to do tomorrow.
356
00:40:05.640 --> 00:40:16.640
Okay so we'll actually see how to do it with much less trig.
357
00:40:16.640 --> 00:40:45.640
So that will be easier in polar coordinates.
358
00:40:45.640 --> 00:40:48.640
Okay so we'll see that tomorrow.
359
00:40:48.640 --> 00:40:51.640
Okay so but you know we're almost there.
360
00:40:51.640 --> 00:40:56.640
I mean here you just use the double angle again and then you can get it and it's it's pretty straightforward.
361
00:40:56.640 --> 00:41:08.640
Okay so one thing that's kind of interesting to know is we can exchange the order of integration.
362
00:41:08.640 --> 00:41:18.640
So if we have an integral given to us in the order dy dx we can switch it to dx dy but we have to be extremely careful with the bounds.
363
00:41:18.640 --> 00:41:30.640
So you certainly cannot just swap the bounds of the inner and outer because where you would end up you know having this square root of one minus x square on the outside and you would never get a number out of that.
364
00:41:30.640 --> 00:41:40.640
So that cannot work it's more complicated than that. Okay so well here's a first baby example.
365
00:41:40.640 --> 00:41:52.640
Certainly if I do integral from 0 to 1 integral from 0 to 2 dx dy I can very can certainly switch the bounds without thinking too much.
366
00:41:52.640 --> 00:42:05.640
What's the reason for that? Well the reason for that is you know this corresponds in both cases to integrating x from 0 to 2 and y from 0 to 1 it's a rectangle.
367
00:42:05.640 --> 00:42:13.640
So if I slice it this way you see that y goes from 0 to 1 for any x between 0 and 2 it's this guy.
368
00:42:13.640 --> 00:42:23.640
If I slice it that way then x goes from 0 to 2 for any value of y between 0 and 1 and it's this one.
369
00:42:23.640 --> 00:42:32.640
So here it works but in general I have to draw a picture of my region and see how the slices look like both ways.
370
00:42:32.640 --> 00:42:55.640
Okay so let's do a more interesting one. Let's say that I want to compute integral from 0 to 1 of integral from x to square root of x of e to the y of y and dy dx.
371
00:42:55.640 --> 00:43:03.640
So why did I choose this guy? I chose this guy because as far as I can tell there's no way to integrate e to the y of a y.
372
00:43:03.640 --> 00:43:08.640
So this is integral but you cannot compute this way.
373
00:43:08.640 --> 00:43:11.640
So it's a good example for y this can be useful.
374
00:43:11.640 --> 00:43:22.640
So if you do it this way you're stuck immediately. So instead we'll try to switch the order. But to switch the order we have to understand what do these bounds mean.
375
00:43:22.640 --> 00:43:32.640
So let's draw a picture of the region. Well what we're saying is y goes from y equals x to y equals square root of x.
376
00:43:32.640 --> 00:43:55.640
Well let's draw y equals x and y equals square root of x. Well maybe I should actually put this here y equals x to y equals square root of x.
377
00:43:55.640 --> 00:44:06.640
Okay and so I will go for each value of x. I will go from y equals x to y equals square root of x.
378
00:44:06.640 --> 00:44:17.640
And I will do that for values of x that go from x equals 0 to x equals 1 which happens to be exactly where this thing intersects.
379
00:44:17.640 --> 00:44:32.640
So my region will consist of all this. Okay. So now if I want to do it the other way around I have to decompose my region the other way around I have to.
380
00:44:32.640 --> 00:44:38.640
So my goal now is to verify this as an integral well it's still the same thing it's still e to the y over y.
381
00:44:38.640 --> 00:44:48.640
But now I want to integrate the x to y. So how do I integrate over x? Well I fix a value of y.
382
00:44:48.640 --> 00:44:55.640
And for that value of y what's the range for x? Well the range for x is from here to here.
383
00:44:55.640 --> 00:45:07.640
Okay what's the value of x here? Let's start with the easy one. This is x equals y. What about this one? It's x equals y squared.
384
00:45:07.640 --> 00:45:23.640
Okay so x goes from y squared to y. And then what about y? Well I have to start at the bottom of my region that's y equals 0 to the top which is at y equals 1.
385
00:45:23.640 --> 00:45:35.640
So y goes from 0 to 1. So switching the bounds is not completely obvious. That took a little bit of work.
386
00:45:35.640 --> 00:45:51.640
But now that we've done that well just to see how it goes it's actually going to be much easier to integrate because in our integral well what's the integral of e to the y over y with respect to x?
387
00:45:51.640 --> 00:46:13.640
Just that times x. From x equals y squared to y. So that will be well if I plug x equals y I will get e to the y minus if I plug x equals y squared I will get e to the y over y times y squared.
388
00:46:13.640 --> 00:46:28.640
Okay so now if I do the outer integral. I will have the integral from 0 to 1 of e to the y minus y e to the y.
389
00:46:28.640 --> 00:46:43.640
And that one actually is a little bit easier. So we know how to integrate e to the y. We don't quite know how to integrate y e to the y but let's try.
390
00:46:43.640 --> 00:46:59.640
So that's the derivative of y e to the y. Well by the product pool that's 1 times e to the y plus y times the derivative of e to the y is y e to the y.
391
00:46:59.640 --> 00:47:11.640
So if we do let's put a minus sign in front. Well that's almost what we want except we have a minus e to the y instead of a plus e to the y. So we need to add 2 e to the y.
392
00:47:11.640 --> 00:47:24.640
And I claim that the anti derivative. Okay if you got lost you can also do it by integrating by parts by taking the derivative of y and integrating this guy.
393
00:47:24.640 --> 00:47:35.640
Or that you know that works just you know your first guess would be maybe let's try minus y e to the y. Take the derivative of that compare see what you need to do to fix.
394
00:47:35.640 --> 00:47:45.640
So if you take that between 0 and 1 you will eventually get e minus 2.
395
00:47:45.640 --> 00:47:58.640
Okay so tomorrow we are going to see how to do double integrals in polar coordinates and also applications of double integrals. How to use them for interesting things.