If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Example of closed line integral of conservative field

Example of taking a closed line integral of a conservative field. Created by Sal Khan.

## Want to join the conversation?

• At couldn't g(y) be equal to any constant, not just zero? Surely g(y)=k will give the same result when you take the partial derivatives? •  Yes, the overall answer should have a constant I think. If you take the gradient of F(x,y) = x^3/3 + xy^2 + C, you get the correct vector field.
• Let us assume that I did not check if f = grad F, does that mean that when I continue with the problem I will eventually get zero for an answer? • I'm getting a little lost in the intuition for this example. If a normal integral is the area under the curve, then what does this example represent? • As a random aside, I don't think there's really such a thing as a "normal" integral, at least not up to where the calculus playlist is so far. All the integrals (so far) have been fundamentally the same. For example, the 'normal' integrals you are talking about are really a special case of line integral through a one-dimensional scalar field, special in the sense that the parametric curve through the field is a straight line (as another aside you can parameterise your 'normal' integral w.r.t. dx to one w.r.t dt and integrate 'backwards' like how Khan showed a few videos ago using the a + b - t trick. The point is, normally our x(t) = t and so we tend to 'de-parameterise' it to the familiar dx integral). Whether or not we can interpret it as area depends on the units of our variables and the context in which we are applying the integration. That is, integration is a mathematical tool that does what it does rather abstractly, and it is up to the user to endow the results with a physically meaningful interpretation.

I might have been taking an off-hand comment too far, but I hope my comment was mildly interesting.
• Is there any obvious reason for the omission of the x and y unit vectors from integral form? • At the video says that normally you can treat differentials (such as dt) as numbers and multiply them. What are some examples where you can't? • just a bit confused, my lecturer told us to check if a force is conservative or not by calculating it its curl. He said if the curl = 0 that means the force is conservative, the curl of this function doesn't = 0 but the force is conservative, how come?? • Why don't we have a +C (constant) in F(x,y) in the end (on )? • Strictly speaking Sal made a mistake not including the constant in the funtion F(x,y). It's supposed to + C. This is true because you can add a constant to F, and finding the gradient will yield the same result regardless of the value of that constant. Also because my Physics math lecturer has been shouting us in the face multiple times not to forget that particular constant xD
• At , the vector field must be equal to a scalar field. I don't quite understand the connection there. A scalar field is defined by magnitudes only, so how can it be that it equals a field that is defined by magnitude and direction on any point? • To determine the f vector is a conservative, it has to equal the gradient of a scalar field. I though the scalar field has to be pre-determined before you can solve this problem, otherwise it can be any scalar field you pick. Here Sal just take the anti-derivative of the f vector to obtain a scalar field, then of course the gradient of this derived scalar field will be equal to the f vector. If he put this scalar vector in the beginning of the question, I can understand this path is path independent according to this scalar field. But otherwise it can be any scalar field, and f vector is not conservative to just any scalar field. Am I getting this concept right? Thanks for any correction! • If you are give vector field f and can find ANY scalar potential F where grad(F) = f, that is sufficient.

Actually, if such an F exists, then it is unique up to a constant, so is, in a sense, pre-determined by f even if you haven't been told what it is.

Note that there are many fields f for which no such F exists and which are therefore not conservative.

E.g. wind velocity in a cyclone / tornado / typhoon - integrate round any path round the "eye" in a direction "with the wind" and all the components of the integral will be positive - and the result definitely not zero. In general wind velocity is not a conservative field. 