Calculus: Integrating with Respect to y: Part II

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Applications of Integration
Area Between Two Curves Not in x
Finding Areas by Integrating with Respect to y: Part Two Page [1 of 3]
Now we see that, in fact, there’s a possibility of finding areas bounded by curves that aren’t even functions, and, in fact, maybe the solution is to instead of stacking our rectangles in this manner – sort of a manner where we have a small change in x for the base and the h8 is the function f(x) – maybe we should be stacking like this, one on top of each other, in which case the small change will be a change in the y-value, and thus we should be stacking and integrating with respect to y.
Let me try to actually illustrate this with a quick little picture and then write down exactly how this would work, and then take a quick look at an example. So suppose that I have two things that are relations. So one maybe looks sort of like this, really funky that’s not a function. Let me call this f(y) = x. It’s not a function of x, it’s a function of y. You’ve got to tilt your head. If you take your head and just tilt it 90 degrees or – tilt it in the clockwise direction so what used to be positive x is now positive y – then you see it is a function that way. So I write x = f(y).
What if I have another relation, and maybe this one is x = g(y). Then if I want to find the area bounded by let’s say here, that’s y = a, and let’s say here, y = b, then that region, which you can see right here is this region, doesn’t lend itself very easily to stacking in a vertical way because here I go from the orange to the orange curve inside here, and then all of the sudden I sort of am hitting those edges there. Then all of the sudden I’m going from the green to the green. So it doesn’t seem to lend itself naturally to be integrating or be putting the rectangles in this manner.
But notice what happens when you put the rectangles in this way. Well, it’s very beautiful. I start, say, over here; and as I go it’s always a very simple way that I can describe the rectangles. They always go from the orange down to the green. Do you see that? So, in fact, this is an example – an illustration – where we should stack rectangles in this manner. When the second rectangle is like this, one on top of each other, it is a good idea. I refer this as y-easy, and the thinking being that, in fact, it’s easy to stack on top of each other in a y-direction rather than next to each other in an x-direction. So I call this y-easy; and when you can stack like this, I call that x-easy.
Okay, so what would the area be? Well, it would be a sum. Let’s just think about what we need to do. We need to sum up the areas of the rectangles, and we’re going to sum from where to where? We’re going to start…where? We’re going to start from way down here, and then we’re going to sort of move our way up to here. So we’re going to start from b and go up to a – so from here to here – and then what are we doing? We want to now sum up the areas of the rectangles – the rectangle’s area is base times h8.
Now what’s the h8? Let’s do the h8 first. The h8, you notice, is a very, very tiny amount, isn’t it? It’s a very tiny change. Now is it a tiny change in x? No, it’s not a tiny change in x because this is a change in the y-direction. So, in fact, this is a change in y. We used to write dx when it was a tiny change in x, but now we write dy.
Now what’s the base? Well, the base is now really long, but how do I find that? It’s actually this length minus that length, so it’s actually f(y) - g(y). But for thinking purposes, I wouldn’t think this way. If I were you, what I would think of is I integrate from the bottom to the top. So, in fact, I would like to write this now in a slightly different way. Let me write this for you as I really see it. Here’s how I see it. The way I see it is the following – that’s correct, by the way, but here’s how I see it – I see it as we integrate from the bottom value to the top. And what do I subtract? I take the function on the right and subtract the function on the left. So I take right value minus left with respect to y.
If you were looking at this with respect to x, you see, if you were stacking this way, then what would you do? Then you would integrate from left to right, and you would take the top function minus the bottom function. Do you see how the roles are all sort of reversed here? Then I would go from left to right, and I’d take the top function and subtract off the bottom function. When I have this shape, I integrate from the bottom to the top, and I take the right thing and subtract off the left thing.
Okay, let’s actually look at this in practice. I think memorizing these is not a great idea. It’s better to think of the picture, draw in an arbitrary rectangle, and figure out the area. Let’s get an actual example to try. Let’s find the area bounded by the region given by y = x - 2 and x = y2. So the first thing we have to do is to sketch a reasonably accurate picture of this. So I’ll do that right now for you. Let’s see, the x = y2, that sort of looks like a parabola – a happy-face parabola, in fact – but it’s not quite because the roles of x and y have been switched. Usually we write y = x2, which looks like this. But since the roles have been switched, what we need to do is we need to again tilt our head 90 degrees, or , clockwise so that what used to be positive x is now positive y, and then you can see the happy-face parabola. So that actually helps us in placing this parabola. This goes in here. Look at that! That actually is, I think, my best parabola to date. That is incredible. And this other thing is a line. It has slope 1, intercept -2, so it’s down here somewhere, and it looks something like this. So there’s the line. And you can see the region that we’re cutting out, it’s sort of another almond-like region. So this is the region that we want here, this yellow region. We want to find the area. Notice that it doesn’t lend itself very naturally to putting in rectangles of this sort because over here, I admit, they have the general shape of going from the upper part of the parabola down to the line; but then once you get way to the left, what happens? I start to go from the parabola to itself. So there’s no general form that captures the true spirit of the rectangle. But if I stack this way – horizontally – this looks really good because I can stack way down here; and if the rectangles are horizontal, stack one on top of each other. They always go from the line down to the parabola. Always. So I should be stacking with respect to y here.
Anyway, if you get stuck on this, as I always do, once I decide on how I’m going to stack, I draw a generic rectangle in, and I think about them being stacked on top of each other just like that. So what’s the area? Well, the area is going to be an integral. And what do I need? I need these points of intersection. Now where am I integrating from? Well, if I’m stacking like this, the low point is where I’m starting, so that is actually going to be a y-value. So notice that the roles of x and y here have changed. I’m going to start from this h8, and I’m going to stack them all the way up until I get to this h8. So I’ve got to find the y-values where these 2 curves intersect, which means I have to find out what happens when these x’s are equal. I’ve got to find the y.
So what I do here is I have to solve both of these for x, and then set the x’s equal. So if I actually solve this for x, I see that x = y + 2, and I know this x equals y2, so I set y2 equal to y + 2, and I want to find out which y-values produce the solution. So I bring everything over to this side, and I attempt to factor. Will it factor? Who knows? It’s always a great mystery. The signs will be different, and it looks like putting 2 and a 1 here actually produce a good solution. So I have y = -1, y = 2. Let’s see how accurate my graph is. Look how accurate my graph is! I just freehanded this. This certainly could be -1; and if that’s -1, look, 1, 2…that actually could be 2. So my graph is actually drawn almost to scale. Good for me! Cool!
Now, so what do we have here? Now I’m going to start down here, and I’m going to build my way up, and I’m going to sum up. So I’m going to sum from -1 to 2. Then I’ve got to put in here the area of the generic rectangle. That’s base times h8. Now what’s the base? Well, that base is going to be this length, which is going to be this value, which is going to be whatever this is here minus this value. So what is this value? Well, if I’m at a particular x…if I’m at a particular y-value here, what is this x? Well, this x is going to be the x-value on the line. So if I put a y here, for example, then this we just saw is going to be…if I’m at a particular y-value, then what is x? x is going to be y + 2. I already solved this for x: x = y + 2. So I’m at a y; I know this is x + 2. So I have y + 2, and then I have to subtract off the parabola, and what’s the parabola if I’m at y? I see the parabola is x, which is y2. So that actually is this length right here. It’s this value – this is y2 – which is y + 2, and I have to subtract off this value, which is y2. Notice that I’m all in terms of y here, not in terms of x’s. So, in fact, if I’m at a particular level, this x-value is given by y + 2; this x-value is given by y2. So I have to subtract these 2 to give me y + 2 - y2.
That’s the base. What’s the h8? Well, the h8 is a very, very tiny change in what? It’s a tiny change in the y-direction, so that’s dy. Once we set that up, we have an integral that we can do; it’s all in terms of y’s. Notice now there are no x’s, just y’s. If we evaluate this, what do we get? This gives me a , and I evaluate this for -1 to 2. So first I plug in the 2. If I plug in the 2, I see , which is just 2. If I plug in the 2 here, I see plus 4. If I plug in the 2 here, I see -8/3; and I subtract off what I get when I plug in -1. When I plug in -1 here I see just 1/2. Here I see -2; and here I see a minus, but another minus is a plus 1/3. So that’s the answer, but we could probably simplify that a little bit if we’re careful. Let’s see. So here I see a 2 plus 4, which is 6, and then a minus a minus is a plus another 2, so that’s 8. So we have 8. And then what do we have here? This gives me 3 and 2 is 5/6 here…so I have - , and that’s going to be 8 - 1, Iguess that’s going to be 21/6. So if you take 8 times 6, which is - , and we get . So the answer is , which you can reduce a little bit. So the area of this is units squared, but the important thing to notice is how we got there. We got there by stacking in this direction rather than stacking in this direction and just carefully writing everything out.
Let’s try one last example to really get a sense of this. Let’s again look at x = y2, and now look at the region bounded by that curve and the curve x = (4y - 1) 2. Well, what do these graphs look like? Let me try to show you these. This is the good, old-fashioned parabola. Let me draw that in in my good, old-fashioned blue. If I do that, it will look like this. What does this parabola look like? Well, this actually looks like sort of a good, old-fashioned parabola, but first of all I’ve got to shift it up by one because notice that when y = 1, this whole thing is zero, and then this makes it tighter – this 4 makes it tighter, so it’s going to be a tighter thing. So if we tighten that up here and sketch the graph of this, what would it look like? It would look something like this. It comes out very tight like this, and comes out very tight like this. So you can see the little teeny region there that we’re forming. It’s right inside, and that’s the area I want to set up the integral for. Notice that doesn’t lend itself naturally to being x-easy because if it were x-easy, then the rectangles would always have the same form when placed this way, but you can see they don’t. Here I’m touching the red to the blue, but then later in life I’m going from red to red. So there’s no nice way of stacking in this direction. However, if I stack in this direction, I always go red on the left and then blue on the right. So this is going to be one where I think we want to do a y-easy. Here’s that arbitrary rectangle right in there, and you can see we’re going to stack them. So the first thing I need to do is know this lower point and then stack up to the upper point. So how do I find those particular y-values? Well, I set the x’s equal to each other. So let’s set the x’s equal to each other really fast, and we see y2 = (4y - 1) 2. Let’s solve this. How do we do that? Well, this is sort of a little complicated thing here, so I guess what I would do is I would first just square that out and see 4y2 - 8y + 4. So I just squared that out. I got y2 - 2y + 1, and multiplied everything through by the 4. Bring this over to that side, and I would see 0 = 3y2 - 8y + 4, and let’s see if this can factor. Who knows? I’ll try 3y and y. Both signs have to be the same, and they both have to be negative. So we put in 2 things here that give us 4 when we multiply them and then produce an 8 in here. So let’s see, 4 times I think 2 and 2 because that would give me a 6 here, and another 2 gives me an 8. Great! So therefore either this is zero, so y = 2/3, or this is zero, y = 2. Is my diagram drawn even close to scale? Well, yeah, this looks like 2/3 over here; and way up there, that could be 2. So we see roughly speaking those points of intersection.
And then if we set up the integral, what would it look like? Well, we’ve got to think now. I’m summing this way, so I’m starting from my low point, so I’m starting at 2/3, and I’m going to sum all the way up to 2. Then what am I summing up? Well, the areas of these rectangles that I’ve drawn in here. Well, that’s base times h8. But what’s the base? The base at a particular y-h8 is going to be the x-value off of the blue parabola, which is this one. So I have to take that x-value, which is y2 – at h8 y, I’m at y2 – and I subtract off what I get on the red one. What’s the red one? Well, I come down here and see that’s this one, so I have to subtract off the x-value at y, which would be exactly (4y - 1) 2. That’s the base. What’s the h8? The h8 is a tiny change in the up-and-down direction. It’s a tiny change in y, so it’s dy. Now we have to evaluate that integral, which you can do actually by just squaring all of that out and distributing and producing the usual integral and evaluating at these two points. If you do that, I believe you get , and I’ll let you try that and see if I’m right, and hopefully I am. That’s not the point here, that’s pretty easy to do. The point is to actually set up this integral, that’s the hard part.
All right, one time a student said to me, “Well, gee, if we see that all of these things are going this way, couldn’t we just swap the roles of x and y? So write this as y = x2, and write this as y = (4x - 1) 2, so then in some sense the picture would be just exactly like that. Then could we just do everything like we normally would and just get everything in terms of x’s?” Answer? Yes. You could certainly do that if you wanted to. Sometimes, though, you might make a mistake in swapping and whatnot. Sometimes the function’s a little bit too complicated to actually get an accurate picture of what’s going on here, but if you wanted to you could certainly do that. However, I think it’s valuable to understand exactly what’s going on, and this, I think, really emphasizes the notion of stacking these rectangles either this way or this way, and then you get to know exactly where the points of integration are and subtracting which from which. So either way is completely fine with me, and I hope fine with your instructor, but the idea is that when you have relations, you need to integrate with respect to y quite often to avoid the ambiguity of the rectangles colliding against sort of each other. This is sort of y-easy; if you can stack this way, then you’re x-easy.
Okay, so much for relations. I’ll see you at the next section. Bye.