Example of calculating the flux across a surface by using the Divergence Theorem. Created by Sal Khan.
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- At2:55isn't the height (z) of the region not always z=1-x^2 ? sometimes it is z=1-x^2 and sometimes it is the plane y=2-z?(8 votes)
- Look at the xz plane, then you should notice that the region of xz plane (which is a region limited by -1<x<1 and 0<z<1-x^2) goes up in the y axis direction limited by y=2-z .
I hope this solves your answer because i don't know to write english properly, :P
- At the beginning of the video, shouldn't the surface integral also be dotted with the normal vector of the surface in order to get the flux?(3 votes)
- That is true, but the field is dotted with the vector-differential ds. The surface normal is incorporated in there.
He derives the form used in this video in the following video:
- Because the function (a*x^2 + b*x^4 + c*x^6) is even, shouldn't we multiply it by 2 and change the limit of integration of x to be from 0 to 1? That way we would get a different result from zero... 5/3.(3 votes)
- (a*x^2 + b*x^4 + c*x^6) is not the function that is integrated. It is a function of the form (a*x + b*x^3 + c*x^5) which is odd. Only after the integration we get the even function.(2 votes)
- For triple integrals, how do you know which variable to integrate with respect to first?(2 votes)
- Most of the time, it won't matter, just make sure the last one doesn't end up with variables(4 votes)
- isn't d(xy)/dx=y ? so shouldn't div F be x+y?(1 vote)
- Remember that when you are taking the divergence you are taking the partial of each term with respect to that term's "direction". So you are taking the partial with respect to y of the second term. That gives:
d(xy)/dy = x(3 votes)
- Does that triple integral in the divergence theorem count as a volume integral?(1 vote)
- yes. but you are suppose to insert some function inside the integral which is the result of the divergence of F.(2 votes)
- what if the surface is not closed. for instance a hemisphere?(1 vote)
- Then I think Stoke's theorem would be more appropriate, relating the surface integral over an arbitrary shape with the line integral over its "edge" (in lack of the proper word)(1 vote)
Let's see if we might be able to make some use of the divergence theorem. So I have this region, this simple solid right over here. x can go between negative 1 and 1. z, this kind of arch part right over here, is going to be a function of x. That's the upper bound on z. The lower bound on z is just 0. And then y could go anywhere between 0, and then it's bounded here by this plane, where we can express y as a function of z. y is 2 minus z along this plane right over here. And we're given this crazy vector field. It has natural logs and tangents in it. And we're asked to evaluate the surface integral, or actually, I should say the flux of our vector field across the boundary of this region, across the surface of this region right over here. And surface integrals are messy as is, especially when you have a crazy vector field like this. But you could imagine that there might be a way to simplify this, perhaps using the divergence theorem. The divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it, or I'll just call it over the region, of the divergence of F dv, where dv is some combination of dx, dy, dz. The divergence times each little cubic volume, infinitesimal cubic volume, so times dv. So let's see if this simplifies things a bit. So let's calculate the divergence of F first. So the divergence of F is going to be the partial of the x component, or the partial of the-- you could say the i component or the x component with respect to x. Well, the derivative of this with respect to x is just x. The derivative of this with respect to x, luckily, is just 0. This is a constant in terms of x. Now let's go over here, the partial of this with respect to y. The partial of this with respect to y is just x. And then this is just a constant in terms of y, so it's just going to be 0 when you take the derivative with respect to y. And then, finally, the partial of this with respect to z, well, this is just a constant in terms of z. Doesn't change when z changes. So the partial with respect to z is just going to be 0 here. And so taking the divergence really, really, really simplified things. The divergence of F simplified down to 2x. And so now we can restate the flux across the surface as a triple integral of 2x. So let me just write 2x here. And let's think about the ordering. So y can go between 0 and this plane that is a function of z. So let's write that down. So y is bounded below by 0 and above by this plane 2 minus z. z is bounded below by 0 and above by-- you could call them these parabolas of 1 minus x squared. And then x is bounded below by negative 1 and bounded above by 1. So negative 1 is less than or equal to x is less than or equal to 1. And so this is probably a good order of integration. We can integrate with respect to y first, and then we'll get a function of z. Then we can integrate with respect to z, and we'll get a function of x. And then we can integrate with respect to x. So let's do it in that order. So first we'll integrate with respect to y, so we have dy. y is bounded below at 0 and above by the plane 2 minus z. So this right over here is a plane y is equal to 0. And this up over here is the plane y is equal to 2 minus z. Then we can integrate with respect to z. And z, once again, is bounded below by 0 and bounded above by these parabolas, 1 minus x squared. And then, finally, we can integrate with respect to x. And x is bounded below by negative 1 and bounded above by 1. So let's do some integration here. So the first thing, when we're integrating with respect to x-- sorry, when we're integrating with respect to y, 2x is just a constant. So this expression right over here is just going to be 2x times y, and then we're going to evaluate it from 0 to 2 minus z. So it's going to be 2x times 2 minus z minus 2x times 0. Well, that second part's just going to be 0. So this is going to simplify as-- I'll write it this way-- 2x times 2 minus z. And actually, I'll just leave it like that. And then we're going to integrate this with respect to z. And that's going to go from 0 to 1 minus x squared, and then we have our dz there. And then after that, we're going to integrate with respect to x, negative 1 to 1 dx. So let's take the antiderivative here with respect to z. This you really can just view as a constant. We can actually even bring it out front, but I'll leave it there. So this piece right over here-- I'll do it in z's color-- this piece right over here, see, we can leave the 2x out front. Actually, I'll leave the 2x out front of the whole thing. It's going to be 2x times-- so the antiderivative of this with respect to z is going to be 2z. Antiderivative of this is negative z squared over 2, and we are going to evaluate this from 0 to 1 minus x squared. When we evaluate them at 0, we're just going to get 0 right over here. And so we really just have to worry about when z is equal to 1 minus x squared. Did I do that right? Yep. 2z, and then minus z squared over 2. You take the derivative, you get negative z. Take the derivative here, you just get 2. So that's right. So this is going to be equal to 2x-- let me do that same color-- it's going to be equal to 2x times-- let me get this right, let me go into that pink color-- 2x times 2z. Well, z is going to be 1 minus x squared, so it's going to be 2 minus 2x squared. That was just 2 times that. And then minus-- I'll just write 1/2 times this quantity squared. So this quantity squared is going to be 1 minus 2x squared plus x to the fourth. That's just some basic algebra right over there. And then from that, you're going to subtract this thing evaluated at 0, which is just going to be 0. So [? y, you ?] just won't even think about that. And now we need to simplify this a little bit. And we are going to get, if we simplify this, we get 2 minus 2x squared minus 1/2, and then plus-- so this is negative 1/2 times negative 2x squared. So it's going to be positive x squared minus 1/2 x to the fourth. Now, let's see, can we simplify this part? Let me just make sure we know what we're doing here. So we have this 2x right over there. I want to make sure I got the signs right. Yep, looks like I did. And now let's look at this. So let's see, can I simplify a little bit? I have 2 minus 1/2, which is 3/2. So I have 3/2. That's that term and that term take into account. And then I have negative 2x squared plus x squared. So that's just going to result in negative x squared, if I take that term and that term. And then I have negative 1/2 x to the fourth, and I'm multiplying this whole thing by 2x. And so that's going to give us-- we have, let's see, 2x times 3/2. And I want to make sure. I'm doing this slowly, so I don't make any careless mistakes. The 2's cancel out. You get 3x, and then 2x times negative x squared is negative 2x to the third. And then 2x times this right over here. The 2 cancels out with the negative 1/2, you have negative x to the fifth. So all of this simplifies to this right over here. So our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth, and then we have dx. And now we just take the antiderivative with respect to x, which is going to be 3/2 x squared minus-- let's see, x to the fourth power-- minus 1/2, because it's going to be 2/4, x to the fourth. Is that right? Because if you multiply it, you're going to 2. Yep, x to the third, and then minus x to the sixth over 6. And it's going to go from 1 to negative 1 or negative 1 to 1. So when you evaluate it at 1-- I'll just write it out real fast. So first, when you evaluate it at 1, you get 3/2 minus 1/2 minus 1/6. And then from that, we are going to subtract 3/2 minus 1/2 plus 1/6. Or actually, no, they're actually all going to cancel out. Is that right? Are they all going to cancel out? Yep, I think that's right. They all cancel out. So it's actually going to be plus, or I should say minus 1/6 right over here. And then all of these cancel out. That cancels with that. That cancels with that because we're subtracting the negative 1/2. And that cancels with that. And so we are actually left with 0. So after doing all of that work, this whole thing evaluates to 0, which was actually kind of a neat simplification. So this whole thing right over here evaluated, very conveniently, evaluated to be equal to 0.