3D divergence theorem intuition
Intuition behind the Divergence Theorem in three dimensions. Created by Sal Khan.
Want to join the conversation?
- Is there such a thing a 4D or even 5D?(5 votes)
- Sure, a lot of mathematics has been generalised into higher dimensions (often n dimensions, meaning you can have as many as you want/need).
We live in three spatial dimensions (space) and one temporal dimension (time): spacetime, so it's also 4D in a sense (except we can't really control our movement in the time dimension).(6 votes)
- So is divergence theorem the same as Gauss' theorem? Also, we have been taught in my multivariable class that Gauss' theorem only relates the Flux over a surface to the divergence over the volume it bounds and if you had for example a path in three dimensions you would apply Green's theorem and the line integral would be equivalent to the Curl of the vector field integrated over the surface it bounds - not the divergence - so I guess I don't understand how its being "generalized to more dimensions"?(2 votes)
- Gauss Theorem is just another name for the divergence theorem.
It relates the flux of a vector field through a surface to the divergence of vector field inside that volume. So the surface has to be closed! Otherwise the surface would not include a volume.
So you can rewrite a surface integral to a volume integral and the other way round.
On the other hand the Stokes Theorem relates a line integral along a closed loop through a vector field to the the surface integral of the curl of the vector field. Here the closed lopp has to beed the boundary of the surface over which you take to surface integral.
In other words you can turn a line integral to an surface integral and the other way round.
The Greens theorem is just a 2D version of the Stokes Theorem. Just remember Stokes theorem and set the z demension to zero and you can forget about Greens theorem :-)
So in general Stokes and Gauss are not related to each other. They are NOT the same thing in an other dimenson.(5 votes)
- i dont understand how the normal vector is inward facing and outward facing(2 votes)
- Imagine a piece of paper with a pencil stuck halfway through it. Both sides of the pencil are perpendicular to the paper. One side points to the inside, if you start from the center of the pencil. The other points to the outside. Since both sides are perpendicular to the paper, they both model a normal vector to the paper's surface.
Vectors have components that describe which way they face from the origin. So a vector with components (2, 5) would point up and to the right, into the first quadrant, while a vector with components (-2, -5), even though it has exactly the same length, would point down and to the left, into the third quadrant. If you imagine a vector that is normal to this surface, on the outside top right part of the surface like where Sal first drew the normal vector, then the first vector would be pointing out of the surface, and the second vector would be pointing into it. Both could be normal to the surface (perpendicular to the surface's tangent line).(2 votes)
- Is this considered Calculus 3?(1 vote)
- Yep. Multivariable calculus.
Pretty much every playlist after the practice AP Calc tutorials are Calc III,(8 votes)
- I AM GONE CRAZY. SOMEBODY HELP ME! ok my question is Why is that normal unit vector ñ perpendicular to do curve, not the vector field F, since ñ is multiplied by F, the unit normal vector ñ should be perpendicular to the F vectors, so why ?(2 votes)
- N is perpendicular to the curve because it is the cross product of the tangential functions of the curve. The vector field F is not multiplied by n but dotted by n which shows how much of the vector field at each point is in the direction of n. This is necessary to determine the magnitude of the flux traveling out (or into) the curve/surface at every point along it.(2 votes)
- Can you calculate the field of g?(2 votes)
- is F.n dr as stated in this video the same as F.n dS?
I remember in past videos on 2-D Divergence Theorem it is stated as F.n dS.(2 votes)
- Thank you for clear explanation!!
And I have a question about application of div theorem.
How about this case?
When vector field is offered for 3 dimension(x,y,z),
I'd like to calculate 'out amount(out let)' in 2 dimension(x,y).
Namely, If given vector field is 3D, and given domain is 2D,
can I use the 2D or 3D div theorem??
I'd like to extend appreciate!!(2 votes)
- There is a very similar operation, just using curl instead of divergence. What is the difference between the two operations? What is the one giving us versus the other?(2 votes)
- Such an intuitive video, thank you so much Sal! I have a question though, at0:55, shouldn't it be a unit normal vector? Since if not, the magnitude of the resulting dot product would vary depending on the magnitude of the normal vector? (Don't we purely want the component of the vector field going in the perpendicular direction to the boundary, cos(theta) * |F||1| (|unit normal vector| = 1))? Thanks.(2 votes)
- yeah, i was thinking same. pro'lly a typo - since at6:30he has used correct symbol i.e. hat or cap instead of arrow. Though still didnt mention the term "unit" but yeah - error, rather than intentional i'd assume.(1 vote)
We've already explored a two-dimensional version of the divergence theorem. If I have some region-- so this is my region right over here. We'll call it R. And let's call the boundary of my region, let's call that C. And if I have some vector field in this region, so let me draw a vector field like this. If I draw a vector field just like that, our two-dimensional divergence theorem, which we really derived from Green's theorem, told us that the flux across our boundary of this region-- so let me write that out. The flux across the boundary, so the flux is essentially going to be the vector field. It's going to be our vector field F dotted with the normal outward-facing vector. So the normal vector at any point is this outward-facing vector, So our vector field, dotted with the normal-facing vector at our boundary times our little chunk of the boundary. If we were to sum them all up over the entire boundary-- let me write that a little bit neater-- that's the same thing as summing up over the entire region. So let's sum up over the entire region. So summing up over this entire region, each little chunk of area, dA-- we could call that dx dy if we wanted to if we're dealing in the xy domain right over here, but each little chunk of area times the divergence of F, which is really saying, how much is that vector field pulling apart? So it's times the divergence of F. And hopefully, it made intuitive sense. The way that I drew this vector field right over here, you see everything's kind of coming out. You could almost call this a source right here, where the vector field seems like it's popping out of there. This has positive divergence right over here. And so because of this, you actually see that the vector field at the boundary is actually going in the direction of the normal vector, pretty close to the direction of the normal vector, so it makes sense. You have positive divergence, and this is going to be a positive value. The vector field is going, for the most part, in the direction of the normal vector. So the larger this is, the larger that is. So hopefully, some intuitive sense. If you had another vector field-- so let me draw another region-- that looked like this, so I could draw a couple situations. So one where there's very limited divergence, maybe it's just a constant. The vector field doesn't really change as you go in any given direction. Over here you'll get positive fluxes. I don't know what the plural of flux is. You'll get positive fluxes, because the vector field seems to be going in roughly the same direction as our normal vector. But here, you'll get a negative flux. So stuff is coming in here. If you imagine your vector field is essentially some type of mass density times volume, and we've thought about that before, this is showing how much stuff is coming in, and then stuff is coming out. So your net flux will be close to zero. Stuff is coming in, and stuff is coming out. Here, you're just saying, hey, stuff is constantly coming out of this surface. So hopefully, this gives you a sense that here you have very low divergence, and you would have a low flux, total aggregate flux, going through your boundary. Here you have a high divergence, and you would have a high aggregate flux. I could draw another situation. So this is my region R. And let's say that we have negative divergence, or we could even call it convergence. Convergence isn't an actual technical term, but you could imagine if the vector field is converging within R, well, the divergence is going to be negative in this situation. It's actually converging, which is the opposite of diverging. So the divergence is negative in this situation. And also the flux across the boundary is going to be negative. Because as we see here, the way I drew it, across most of this boundary, the vector field is going in the opposite direction. It's going in the opposite direction as our normal vector at any point. So hopefully, this gives you a sense of why there's this connection between the divergence over the region and the flux across the boundary. Well, now we're just going to extend this to three dimensions, and it's the exact same reasoning. If we have a-- and I'll define it a little bit more precisely in future videos-- a simple, solid region. So let me just draw it. And I'm going to try to draw it in three dimensions. So let's say it looks something like that. And one way to think about it is this is going to be a region that doesn't bend back on itself. And if you have a region that bends back on itself-- well, we'll think about it in multiple ways. But out of all the volumes of three dimensions that you can imagine, these are the ones that don't bend back on themselves. And there are some that you might not be able to imagine that would also not make the case. But even if you had ones that bent back on themselves, you could separate them out into other ones that don't. So here is just a simple solid region over here. I'll make it look three dimensional. So maybe if it was transparent, you would see it like that. And then you see the front of it like that. So it's this kind of elliptical, circular, blob-looking thing. So that could be the back of it. And then if you go to the front, it could look something like that. So this is our simple solid region. I'll call it-- well, I'll call it R still. But we're dealing with a three dimensional. We are now dealing with a three-dimensional region. And now the boundary of this is no longer a line. We're now in three dimensions. The boundary is a surface. So I'll call that S. S is the boundary of R. And now let's throw on a vector field here. Now, this is a vector field in three dimensions. And now let's imagine that we actually have positive divergence of our vector field within this region right over here. So we have positive divergence. So you can imagine that it's kind of-- the vector field within the region, it's a source of the vector field, or the vector field is diverging out. That's just the case I drew right over here. And the other thing we want to say about vector field S, it's oriented in a way that its normal vector is outward facing, so outward normal vector. So the normal, it's oriented so that the surface-- the normal vector is like that. The other option is that you have an inward-facing normal vector. But we're assuming it's an outward-facing N. Well, then we just extrapolate this to three dimensions. We essentially say the flux across the surface. So the flux across the surface, you would take your vector field, dot it with the normal vector at the surface, and then multiply that times a little chunk of surface, so multiply that times a little chunk of surface, and then sum it up along the whole surface, so sum it up. So it's going to be a surface integral. So this is flux across the surface. It's going to be equal to-- if we were to sum up the divergence, if we were to sum up across the whole volume, so now if we're summing up things on every little chunk of volume over here in three dimensions, we're going to have to take integrals along each dimension. So it's going to be a triple integral over the region of the divergence of F. So we're going to say, how much is F? What is the divergence at F at each point? And then multiply it times the volume of that little chunk to sense of how much is it totally diverging in that volume. And then you sum it up. That should be equal to the flux. It's completely analogous to what's here. Here we had a flux across the line. We had essentially a two-dimensional-- or I guess we could say it's a one-dimensional boundary, so flux across the curve. And here we have the flux across a surface. Here we were summing the divergence in the region. Here we're summing it in the volume. But it's the exact same logic. If you had a vector field like this that was fairly constant going through the surface, on one side you would have a negative flux. On the other side, you would have a positive flux, and they would roughly cancel out. And that makes sense, because there would be no diverging going on. If you had a converging vector field, where it's coming in, the flux would be negative, because it's going in the opposite direction of the normal vector. And so the divergence would be negative as well, because essentially the vector field would be converging. So hopefully this gives you an intuition of what the divergence theorem is actually saying something very, very, very, very-- almost common sense or intuitive. And now in the next few videos, we can do some worked examples, just so you feel comfortable computing or manipulating these integrals. And then we'll do a couple of proof videos, where we actually prove the divergence theorem.