Type I regions in three dimensions
Definition of Type 1 regions. Visualizing what is and isn't a Type I region. Created by Sal Khan.
Want to join the conversation?
- So, basically, a Type I region passes the vertical line test for each of it's bounding functions...right??(8 votes)
- Yes; functions by definition pass the vertical line test. In the case of 3-variable functions: for any combination of x and y, there is a unique z value assigned to it. In other words, a function is defined so that there is only one height (z) for every point in 2-D space (x,y). Therefore a vertical line will only pass through the surface once. Type I regions are bounded by functions of (x,y), so each boundary passes the vertical line test.
N.B. It follows that any vertical line passing through a Type I region will cross its boundary surface exactly twice (once in & once out).(7 votes)
- I don't get is why the dubmbell is not a type I region.(4 votes)
- Great question. I'm also unsure of why that is the case, but here is hopefully a good enough explanation.
We begin with the definition of a type I region;
(x, y, z)means that it is a collection of points in 3D space, or simply a 3D figure with volume.
(x, y) ϵ Dsimply means that x and y are part of a domain, or that the region is not infinite.
Now comes the hard part to understand.
f1(x, y) ≤ z ≤ f2(x, y)put into words is simply that for all z, there is a surface (because f(x, y) is a surface) f1 that is the lower bound and f2 that is the upper bound.
Using the examples, we see that for the sphere, the upper bound is the green hemisphere, whereas the lower bound is the purple hemisphere, which are both surfaces that bound all z. Likewise, for the cylinder, the green circle and purple circle are the bounds.
For the dumbbell, there cannot be a lower or upper bound that can completely bound or "wrap around" the 3D figure.
Another thing to note is that because the bounds are surfaces, for any (x, y) (which can be imagined as a vertical line parallel to the z-axis), there can be at most 2 values of z where it crosses the boundary surfaces. This can be considered the 3D version of the vertical line test of sorts.
I'm unsure if this is correct, but hopefully I helped.(5 votes)
- Is a surface like the plane z = 3x + 5y + 2 a Type I reigon(5 votes)
- No, because a plane is a single surface. A Type I region is a region between two surfaces.(5 votes)
- Could you please explain how exactly to break up the dumbbell region to make it type 1 region? At7:28Sal says that the UPPER part becomes BOTTOM region, and vice versa... I don't get that and how it makes the dumbbell shape into two separate type 1 regions?(3 votes)
- Good Call.
I don't think Sal was as clear as he could have been on that.
Since the shape is symmetrical above and below the y axis, he was trying to show that the bottom part could be flipped over so that the narrow part, which is the top of the bottom part, is the same as the bottom of the top part. In essence then, you only need to integrate the top part and multiply the result by 2, since the volume of the top half and bottom half are the same.(1 vote)
- For the dumbbell example, you could just knock the dumbbell on its side (rotate it 90 degrees about either the x or the y axis). Then you would have two surfaces meeting the criteria. but that is just a matter of what you choose to label the coordinates. Is the type of the region dependent on the choice of coordinate orientation?(3 votes)
- I believe it does matter on your coordinate choice (otherwise everything would be rather arbitrary).(1 vote)
- I'm using his coloring scheme to understand the lower bounds and upper bounds - at7:10, is there is an accidental switch in the colors (purple was the lower bound and green was the upper bound before)?(2 votes)
- Yes, he makes careless mistakes like that all the time.(3 votes)
- Ok so he is saying that dumbbell wouldn't fit into type 1 but how come sphere applies to that? Y values also changes when you go from top the bottom(for sphere)(3 votes)
- For the sphere, if you fix an (x, y) inside its D, there are two points on the sphere's boundary, with those (x, y) coordinates: the one above it, and the one below it. For the dumbbell, that is not the case. There are places (x, y) inside its D where anywhere from two to four points share the same (x, y) coordinates. So it fails to be a Type I region. Hope that helps!(1 vote)
- Could this be considered to be topology?(2 votes)
- After doing a little bit of research, I guess that you can call this topology, although it is much closer to mathematical analysis (i.e. calculus), but regions are related to topics like topological spaces. Hope that I helped.(1 vote)
- what does he mean in7:20that z cant take any value in between those two flate plane boundaries? of course it can? z is in between the same way as why the cylinder works?(1 vote)
- No, for certain x and y within the domain, z cannot take on every value in between. For instance, for most x and y in the domain, z cannot be zero, even though z=0 is in between the two planes.(3 votes)
- How to do this problem
If S is a unit sphere x^2+y^2+z^2=1 then what is the value of surface integral [(2x^2+3)-y^2+5z^2]ds is ?(1 vote)
In this and the next few videos, I hope to explore different types of regions in three dimensions. And these will be useful for thinking about how to evaluate different double and triple integrals and also some interesting proofs in multivariable calculus. So the first type of region, and it's appropriately named, we will call a type 1 region. At first, I'll give a formal definition. And hopefully, the formal definition makes some intuitive sense. But then I'll draw a couple of type 1 regions, and then I'll show you what would not be a type 1 region because sometimes that's the more important question. So type 1 region, maybe a type one region R, is the set-- and these little curly brackets means set-- is the set of all x, y's, and z's. It's the set of all points in three dimensions such that the x and y's are part of some domain, are a member-- that's what this little symbol represents-- are a member of some domain. And z can-- essentially varies between two functions of x and y. So let me write it over here. So f1 of x, y is kind of the lower bound on z. So this is going to be less than or equal to z, which is less than or equal to another function of x and y, which is going to be less than or equal to f2 of x and y. And let me close the curly brackets to show that this was all a set. This is a set of x, y's, and z's. And right here, we are defining that set. So what would be a reasonable type 1 region? Well, a very simple type one region is a sphere. So let me draw a sphere right over here. So in a sphere, where it intersects the x, y plane-- that's essentially this domain D right over here. So I'll do it in blue. So let me draw my best attempt at drawing that domain so that this is the domain D right over here for a sphere. This is a sphere centered at 0, but you could make the same argument for a sphere anywhere else. So that is my domain. And then f1 of x, y, which is a lower bound of z, will be the bottom half of the sphere. So you really can't see it well right over here, but it would be-- these contours right over here would be on the bottom half. And I can even color in this part right over here. The bottom surface of our sphere would be f1 of x, y, and f2 of x, y as you could imagine, will be the top half of the sphere, the top hemisphere. So it'll look something like that. This thing that I'm drawing right over here is definitely a type 1 region. As we'll see, this could be a type 1 type 2, or a type 3 region. But it's definitely a type one region. Another example of a type 1 region-- and actually this might even be more obvious. So let me draw some axes again, and let me draw some type of a cylinder. Just to make it clear that our domain, where the x, y plane does not have to be inside of our region-- let's imagine a cylinder that is below-- well, actually I'll draw it above-- that is above the x, y plane. So this is the bottom of the cylinder. It's right over here. And once again, it doesn't have to be centered around the z-axis. But I'll do it that way just for this video. Actually, I could draw it a little bit better than that. So this is the bottom surface of our cylinder, and then the top surface of our cylinder might be right over here. And these things actually don't even have to be flat. They could actually be curvy in some way. And in this situation, so in this cylinder-- let me draw it a little bit neater. In this cylinder right over here, our domain are all of the values that the x and y's can take on. So our domain is going to be this region right over here in the x, y plane. And then for each of that, those x, y pairs, f1 of x, y defines the bottom boundary of our region. So f1 of x, y is going to be this right over here. So you give me any of these x, y's in this domain D, and then you evaluate the function at those points, and it will correspond to this surface right over here. And then f2 of x, y, once again, give me any one of those x, y points in our domain, and you evaluate f2 at those points, and it will give you this surface up here. And we're saying that z will take on all the values in between, and so it is really this whole solid-- it's really this entire solid area. Likewise, over here, z could take on any value between this magenta surface and this green surface. So it would essentially fill up our entire volume so it would become a solid region. Now, you might be wondering what would not be a type 1 region? So let's think about that. So it would essentially be something that we could not define in this way, and I'll try my best to draw it. But you could imagine a shape that does something funky like this. So there's like one big-- I guess you could imagine a sideways dumbbell. So a sideways dumbbell-- and I'll maybe curve it out a little bit. So this is the kind of the top of the dumbbell-- or an hour glass, I guess you could say, or a dumbbell. It would look something like that. So I'm trying my best to draw it. It would look something like that. And the reason why this is not definable in this way-- it becomes obvious if you kind of look at a cross section of it. There's no way to define only two functions that's a lower bound and an upper bound in terms of z. So even if you say, hey, maybe my domain will be all of the x, y values that can be taken on-- let me see how well I can draw this. So you say my x, y values-- so let me try to draw this whole thing a little bit better, a better attempt. So you might say, OK, for something like a dumbbell-- let me clear out that part as well. For something like a dumbbell-- so let me erase that. So for something like a dumbbell, maybe my domain is right over here. So these are all the x, y values that you can take on. But in order to have a dumbbell shape, for any one x, y, z is going to take on-- there's not just an upper and a lower bound, and z doesn't take on all values in between. Well, let me just draw it a more clearly. So our dumbbell-- maybe it's centered on the z-axis. This is the middle of our dumbbell, and then it comes out like that. And then up here, the z-axis-- so it looks like that. And then it goes below the x, y plane, and it does kind of a similar thing. It goes below the x, y plane and looks something like that. So notice, for any given x, y, what would be-- if you attempted to make it a type 1 region, you would say, well, maybe this is the top surface. And maybe you would say down here is the bottom surface. But notice, z can't take on every value in between. You kind of have to break this up if you wanted to be able to do something like that. You would have to break this up into two separate regions where this would be the bottom region, and then this right over here would be another top region. So this dumbbell shape itself is not a type 1 region, but you could actually break it up into two, separate type 1 regions. So, hopefully, that helps out. And actually, another way to think about, this might be an easier way-- if we were to look at it from this direction, and if we were to just think about the z, y, if we were just thinking about what's happening on the z, y plane-- so that's a z, and this is y right over here-- our dumbbell shape would look something like this, my best attempt to draw our dumbbell shape. And so if you get a given x or y, maybe x is even 0, and you're sitting right here on the y-axis, notice z is not, even up here, cannot be a function of just y. On this top part, there's two possible z-values that we need to take on for that given y-- two possible z-values for that given y. So you can't define it simply in terms of just one lower bound function and one upper bound function.