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## Multivariable calculus

### Course: Multivariable calculus > Unit 4

Lesson 3: Line integrals in vector fields- Line integrals and vector fields
- Using a line integral to find work
- Line integrals in vector fields
- Parametrization of a reverse path
- Scalar field line integral independent of path direction
- Vector field line integrals dependent on path direction
- Path independence for line integrals
- Closed curve line integrals of conservative vector fields
- Line integrals in conservative vector fields
- Example of closed line integral of conservative field
- Second example of line integral of conservative vector field
- Distinguishing conservative vector fields
- Potential functions

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# Closed curve line integrals of conservative vector fields

Showing that the line integral along closed curves of conservative vector fields is zero. Created by Sal Khan.

## Want to join the conversation?

- I'm having a hard time with this: Why would a vector field NOT be conservative. The constraint of being conservative seems like a big one, but can't think of a time where a field is not conservative.(6 votes)
- I couldn't explain this mathematically , but I could using a simple physics example.

Imagine we're on a planet with no atmosphere, meaning no air drag, and let's throw a ball upwards. As the ball goes upward the velocity we gave it is being converted in to height. Eventually the ball will reach its max height and start coming back downwards, converting its height in to velocity. The mechanical energy (sum of potential and kinetic energy) stays constant, or it's energy is conserved.

Now let's imagine we're back on Earth (with an atmosphere) and we throw that same ball up, some of its velocity is still converted in to potential energy, but it's also bumping in to air particles and transferring some of its kinetic energy to these particles. The ball has now lost some of its mechanical energy, meaning it must have been acted on by non-conservative forces.

If mechanical energy is being transferred out of an object, it's typically being acted on by a conservative field (I say typically because I've heard different definitions of mechanical energy).

Sorry if the example is too simplistic.(17 votes)

- A bit confusing because my book (Multivariable Calculus by Stewart) uses F to refer to the vector function and f to refer to the scalar function. So the gradient of f is F.

Not sure what the standard is, just a heads up(3 votes)- Don't think it matters really.(4 votes)

- If doing a line integral gives you the surface area above a curve, then how can it be 0 for a closed line integral that's path independent?(2 votes)
- You can think of it like this: there are 3 types of line integrals:

1) line integrals with respect to arc length (dS)

2) line integrals with respect to x, and/or y (surface area dxdy)

3) line integrals of vector fields

That is to say, a line integral can be over a scalar field or a vector field.

You are confusing integration over a scalar field (surface area) versus over a vector field (and in this case, a vector field that is conservative).(4 votes)

- Can you relate this to the work done on the particle by the vector field f?(2 votes)
- Assume that you drag something along any path by applying some force and eventually come back to the exact point that you started (closed curve). Since the distance you took is zero, the work done equals to zero.(3 votes)

- If the line integral on a vector field is equal to the work done, then won't the line integral over a closed curve always be equal to zero, since displacement is zero, whether or not the vector field is conservative?(3 votes)
- vector field being conservative is very important since otherwise the line integral wont be zero.. for a physical example consider a block travelling a closed path on a surface having friction.. the work done will not be zero because friction is not a conservative force.. so there will be energy loss due to friction(1 vote)

- So how do we know when the vector field can be expressed as a gradient?(2 votes)
- At7:02Sal says that in order to represent the vector field as the gradient of a scalar field, the vector field must be conservative. That a vector field is conservative can be tested by obtaining the curl (
`𝛁⃗⨉F⃗`

) of the vector field; if it's`0`

, then the field is conservative.

Here are the videos on divergence and curl: https://www.khanacademy.org/math/multivariable-calculus/partial_derivatives_topic(2 votes)

- Is the curl of a conservative vector field always zero? From the picture it seems like it should be, but that's just a wild guess. And would that mean that all vector fields with 0 curl are conservative?

Edit: I looked on Wikipedia, and it says that the curl of the gradient of a scalar field is always 0, which means that the curl of a conservative vector field is always zero.

But then can you go the other way and say that a vector field is conservative if it has a curl of 0?

http://en.wikipedia.org/wiki/Vector_calculus_identities#Curl_of_the_gradient

Further Edit: Yay Stokes' Theorem!(2 votes)- Yes that is exactly correct. The other way to prove that a vector field is conservative is to show that the dy of the i component = the dx of the j component, the dx of the k component = dz of the i component, and the dz of the j component = the dy of the k component. Essentially, this is the same thing as finding the curl of the vector field and making sure that the curl gives you 0 vector. If the curl does not give you 0 vector, it is not conservative.(2 votes)

- I have a question. How can the area be zero?(2 votes)
- There are not only positive values that areas can take, but also negative ones. If the vector field is conservative, then positive and negative areas cancel each other, making the total area equal to 0.

Note that this is a special property for conservative fields.(1 vote)

- Is this line integral anyway related to stokes theorem?

If so do we have to consider the "conservative"ness?(1 vote) - You mention that "the line integral is path independent between any two points". This statement is only true if Rot(f)=0 right? Or a field can only be conservative if Rot(f)=0?(1 vote)
- You are correct. Path independence is only valid for conservative vector fields (as the title suggests). This concept is very important in physics when one starts to analyze work and energy transitions. The work done to move an object from one height to another is independent of whatever crazy winding path you took between the two points. Why? Because the vector field representing the force of gravity is non-curling, it can be written as the gradient of some potential scalar function.(1 vote)

## Video transcript

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 already be obvious to you. 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.