And, then I have a single variable integral.
然后就变成了一个单变量积分。
How do I compute the line integral along the curve that goes all around here?
应该怎样沿着围绕这个区域的曲线，做线积分呢？
And so I want to compute for the line integral along that curve.
那我想计算那条曲线上的线积分。
And, I still want to compute the line integral along a closed curve.
但仍然想要沿着封闭曲线的线积分计算。
The other way around, I have to, so my goal, now, is to rewrite this as an integral.
重新分解积分域。。。,我的目标是，用另外一种形式重写这个积分。
Then I can actually -- --replace the line integral for flux by a double integral over R of some function.
那么我就能名正言顺地,用R上的某个函数的二重积分来替代通量的线积分。
OK, so if I give you a curve that's not closed, and I tell you, well, compute the line integral, then you have to do it by hand.
如果给你们一条非封闭曲线，然后让你们计算线积分，你们必须动手一点点来计算。
Any questions about how I set up this flux integral?
关于建立通量积分，又问题吗？
Then I don't have actually to compute the flux integral.
我不需要真的计算通量积分。
So, now, if I compare my double integral and, sorry, my triple integral and my flux integral, I get that they are, indeed, the same.
比较这个二重积分的话，抱歉。。。，比较这个三重积分和通量积分，就可以看到，它们是一样的。
And this is finally where I have left the world of surface integrals to go back to a usual double integral.
也就是最终要摆脱曲面积分，回到常规的二重积分。
So, in both cases, we need the vector field to be defined not only, I mean, the left hand side makes sense if a vector field is just defined on the curve because it's just a line integral on c.
了解这两种表述后，我们不仅需要向量场，就是左边这里，这是曲线c上的线积分，向量场在曲线上有定义。
I mean if that is your favorite path then that is fine, but it is not very easy to compute the line integral along this, especially since I didn't tell you what the definition is.
如果你喜欢也可以，但是沿着这条路径积分却不好计算，尤其是还没告诉你曲线的定义。
If there are no other questions then I guess we will need to figure out how to compute this guy and how to actually do this line integral.
如果没有问题的话，我们就来算算它吧，如何计算这个线积分的值呢？
I. research concentrated on what seemed to be difficult intellectual tasks, such as playing grandmaster level chess, or proving theorems in integral calculus.
早期的人工智能研究集中在了一些看起来很困难的智力活动上，如高水平国际象棋，或者积分定理的证明。
Well, let's say I give you a curve and I ask you to compute this integral.
给出一条曲线，计算沿着这条曲线的积分。
And, so the total line integral for this thing is equal to the integral along C prime, I guess the outer one.
那么，它的总的线积分,就等于沿着C’，我是指外面这个的线积分。
So, using Green's theorem, the way we'll do it is I will, instead, compute a double integral.
那么，使用格林公式，我们去计算二重积分。
Here I didn't set this up as a iterated integral yet.
这里还没有变成累次积分。
Well, if I have a gradient field, then if I try to compute this line integral, I know it will be the value of the function at the end point minus the value at the starting point.
如果有一个梯度场，然后想计算线积分，我知道积分值是,起点的函数值减去终点的函数值。
But, what I want to point out is if you have to compute the two sides separately, well, this is just, you know, your standard flux integral.
需要指出的是，如果要分别计算两边的积分，这就是标准的通量积分。
And instead I will want to reduce things to a surface integral.
相反，我想要把它变成面积分。
Next, I should try to look at my double integral and see if I can make it equal to that.
然后观察二重积分,看看能不能使两式相等。
The first one that I will mention is actually something you thought maybe you could do with a single integral, but it is useful very often to do it as a double integral.
第一点就是，有些你以为是用一重积分来做的，但却通常是用二重积分来完成的。
Now, if I want to compute this integral, so let's first do the inner integral.
如果想计算这个积分，先做内积分。
On c1, how do I evaluate my integral?
那么怎么在c1上计算积分？
OK, so the claim is if I have a line integral to compute, that it doesn't matter which path I take b as long as it goes from point a to point b.
也就是，如果我需要计算一个线积分，无论我取怎样的路径,只要路径是从点a到点。
OK, so vertically simple means it will be something which I can setup an integral over the z variable first easily.
垂直简单的意思是,在其中建立的对z积分会比较简单。
I get that the line integral on c1 — Well, a lot of stuff goes away.
得到c1上的线积分，大部分就消去了。
So, we'll call that the double integral of our region, R, of f of xy dA and I will have to explain what the notation means.
称之为区域R上fdA的二重积分,会向大家解释这些符号的含义的。
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