# Closed curve line integrals of conservative vector fields | Multivariable Calculus | Khan Academy

In the last video, we saw that if a vector field can be
written as the gradient of a scalar field-- or another way
we could say it: this would be equal to the partial of our big
f with respect to x times i plus the partial of big f, our
scalar field with respect to y times j; and I'm just writing
it in multiple ways just so you remember what the gradient is
--but we saw that if our vector field is the gradient of
a scalar field then we call it conservative.
So that tells us that f is a conservative vector field.
And it also tells us, and this was the big take away from the
last video, that the line integral of f between two
points-- let me draw two points here; so let me draw my
coordinates just so we know we're on the xy plane.
My axes: x-axis, y-axis.
Let's say I have the point, I have that point and that point,
and I have two different paths between those two points.
So I have path 1, that goes something like that, so
I'll call that c1 and it goes in that direction.
And then I have, maybe in a different shades of
green, c2 goes like that.
They both start here and go to there.
We learned in the last video that the line integral
is path independent between any two points.
So in this case the line integral along c1 of f dot dr
is going to be equal to the line integral of c2, over the
path c2, of f dot dr. The line, if we have a potential in
a region, and we may be everywhere, then the line
integral between any two points is independent of the path.
That's the neat thing about a conservative field.
Now what I want to do in this video is do a little bit of
an extension of the take away of the last video.
It's actually a pretty important extension; it might
I've already written this here; I could rearrange
this equation a little bit.
So let me do it.
So let me a rearrange this.
I'll just rewrite this in orange.
So the line integral on path c1 dot dr minus-- I'll just go
subtract this from both sides --minus the line integral c2 of
f dot dr is going to be equal to 0.
All I did is I took this take away from the last video and
I subtracted this from both sides.
Now we learned several videos ago that if we're dealing with
a line integral of a vector field-- not a scalar field
--with a vector field, the direction of the
path is important.
We learned that the line integral over, say, c2 of f dot
dr, is equal to the negative of the line integral of minus c2
of f dot dr where we denoted minus c2 is the same path as
c2, but just in the opposite direction.
So for example, minus c2 I would write like this-- so let
me do it in a different color --so let's say this is minus
c2, it'd be a path just like c2-- I'm going to call this
minus c2 --but instead of going in that direction, I'm now
going to go in that direction.
So ignore the old c2 arrows.
We're now starting from there and coming back here.
So this is minus c2.
Or we could write, we could put, the minus on the other
side and we could say that the negative of the c2 line
integral along the path of c2 of f dot dr is equal to the
line integral over the reverse path of f dot dr. All I did is
I switched the negative on the other side; multiplied
both sides by negative 1.
So let's replace-- in this equation we have the minus of
the c2 path; we have that right there, and we have that right
there --so we could just replace this with
this right there.
So let me do that.
So I'll write this first part first.
So the integral along the curve c1 of f dot dr, instead of
minus the line integral along c2, I'm going to say plus the
integral along minus c2.
This-- let me switch to the green --this we've established
is the same thing as this.
The negative of this curve, or the line integral along this
path, is the same thing as the line integral, the positive of
the line integral along the reverse path.
So we'll say plus the line integral of minus c2 of
f dot dr is equal to 0.
Now there's something interesting.
Let's look at what the combination of the path
of c1 and minus c2 is.
c1 starts over here.
Let me get a nice, vibrant color.
c1 starts over here at this point.
It moves from this point along this curve c1 and
ends up at this point.
And then we do the minus c2.
Minus c2 starts at this point and just goes and comes back
to the original point; it completes a loop.
So this is a closed line integral.
So if you combine this, we could rewrite this.
Remember, this is just a loop.
By reversing this, instead of having two guys starting here
and going there, I now can start here, go all the way
there, and then come all the way back on this
reverse path of c2.
So this is equivalent to a closed line integral.
So that is the same thing as an integral along a closed path.
I mean, we could call the closed path, maybe, c1 plus
minus c2, if we wanted to be particular about
the closed path.
But this could be, I drew c1 and c2 or minus c2 arbitrarily;
this could be any closed path where our vector field f has a
potential, or where it is the gradient of a scalar field,
or where it is conservative.
And so this can be written as a closed path of c1 plus the
reverse of c2 of f dot dr. That's just a rewriting
of that, and so that's going to be equal to 0.
And this is our take away for this video.
This is, you can view it as a corollary.
It's kind of a low-hanging conclusion that you can make
after this conclusion.
So now we know that if we have a vector field that's the
gradient of a scalar field in some region, or maybe over the
entire xy plane-- and this is called the potential of f;
this is a potential function.
Oftentimes it will be the negative of it, but it's easy
to mess with negatives --but if we have a vector field that is
the gradient of a scalar field, we call that vector
field conservative.
That tells us that at any point in the region where this is
valid, the line integral from one point to another is
independent of the path; that's what we got from
the last video.
And because of that, a closed loop line integral, or a closed
line integral, so if we take some other place, if we take
any other closed line integral or we take the line integral of
the vector field on any closed loop, it will become 0 because
it is path independent.
So that's the neat take away here, that if you know that
this is conservative, if you ever see something like this:
if you see this f dot dr and someone asks you to evaluate
this given that f is conservative, or given that f
is the gradient of another function, or given that f is
path independent, you can now immediately say, that is going
to be equal to 0, which simplifies the math a good bit.