We have a line integral along a curve.
对于沿曲线的线积分。
A line integral for flux just becomes this.
求通量的线积分就变成这样了。
So that will be the line integral of Pdx plus Qdy.
这就变成了Pd x +Qdy的线积分。
So, what it does actually is it computes a line integral.
它实际上做的是计算线积分。
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 this we can compute using the definition of the line integral.
而且我们能用线积分的定义计算出来。
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。
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。
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？
Remember, you have to know how to set up and evaluate a line integral of this form.
注意，大家需要知道，如何建立和计算这种形式的线积分。
This relates a line integral for one field to a surface integral from another field.
这把一个向量场的线积分，和另外一个向量场的曲面积分联系起来。
And then we had to evaluate the line integral for the work done along this path.
然后计算，沿此路径所做的功。
Then, we just have to, well, the line integral is just the change in value of a potential.
那么我们只需要知道，线积分正是势函数值的变化。
So, you should remember, what is this line integral, and what's the divergence of a field?
你们需要记住，什么是线积分，什么是场的散度？
If you cannot parameterize the curve then it is really, really hard to evaluate the line integral.
如果无法对曲线参数化，那么就很难计算线积分了。
Just as we do work, when we compute this line integral, usually we don't do it geometrically like this.
就像做功一样，当计算这线积分时，通常不这样用几何方法来做。
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。
If we have to compute a line integral, we have to do it by finding a parameter and setting up everything.
如果我们必须计算线积分，就必须通过寻找一个参数，并建立起一切。
So, that's a really strange statement if you think about it because the left-hand side is a line integral.
那么，如果你仔细想，会有一种很奇特的结论,因为，左面的是一个线积分。
If we want to compute the line integral along this guy then we have to break it into a sum of three terms.
如果要计算沿这条曲线的线积分，我们不得不把它分解成3部分。
It tells you, if you take the line integral of the gradient of a function, what you get back is the function.
它说，如果你对一个函数的梯度做线积分，就能得到原函数。
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, to say that a vector field with conservative means 0 that the line integral is zero along any closed curve.
一个保守的向量场就是说,沿任意闭曲线的线积分的结果是。
And so that's why you have this line integral that makes perfect sense, but you can't apply Green's theorem to it.
这就是为什么，这个线积分，有着完美的定义，但却不能对它使用格林公式的原因。
OK, so the proof, so just going to prove that the line integral is path independent; the others work the same way.
那么这个证明，只需证出线积分与路径无关的，其它的也是用同样的方法。
In the line integral in the plane, you had two variables that you reduced to one by figuring out what the curve was.
在平面上的线积分中，有两个变量，可以通过了解曲线的形成规律，从而去掉一个变量。
And let's take my favorite curve and compute the line integral of that field, you know, the work done along the curve.
对我喜欢的曲线，计算其上的线积分，在这条线上所做的功。
But, you know, it gives you an example where you can turn are really hard line integral into an easier double integral.
但是你知道，它给了你一个例证，其中你可以，把复杂的线积分化成简单些的二重积分。
But it is still a line integral so it is still going to turn into a single integral when you plug in the correct values.
但是这仍然是一个线积分，代入计算就会发现，这仍然会变为单变量积分。
This one is a line integral. So, you use the method to explain here, namely, you express x and y in terms of a single variable.
这一个是线积分，可以用这里的办法来做，也就是用一个变量来表示出来。
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