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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.

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• 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.
• 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.
• 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
• Don't think it matters really.
• 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?
• 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).
• Can you relate this to the work done on the particle by the vector field f?
• 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.
• 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?
• 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
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• So how do we know when the vector field can be expressed as a gradient?
• At Sal 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
• 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?

Further Edit: Yay Stokes' Theorem!
• 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.
• I have a question. How can the area be zero?
• 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.
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• Is this line integral anyway related to stokes theorem?
If so do we have to consider the "conservative"ness?
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• 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?
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• 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.
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