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# Second example of line integral of conservative vector field

Using path independence of a conservative vector field to solve a line integral. Created by Sal Khan.

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• At , it looks like you could solve this by:
1) substituting `x = cos(t)`, `y = sin(t)`, `dx = -sin(t)•dt`, `dy = cos(t)•dt`
2) integrate from t=0 to t=pi

Anyone know of a reason why that wouldn't be valid?
• Yes, that is correct. It's the same method used in earlier videos. I suggest starting with the "Line Integrals and vector fields" video here and viewing all of them. He is really just showing an alternate method of solving the integral here, a method which in some cases may actually be quicker.
• When he says the only solution for the potential is x^3/3+xy^2, shouldn´t he put on a constant, c?
• You could put a constant there. It would not be wrong, but since the constat will cancel itself out, we can set it to 0.
• In checking wether there exists that function F .. can't we use the method we learnt in ODE exact equations .. I mean can't we get Mx and Ny and if they are equal we can say that this function F ( or psi in ODE part ) exists ? Am I right or the situation is different ?
• i am confused here. try and explain this clearer.
• im just thinking couldnt we just simply define our own path from 0 to pi like a straight line to bypass dealing with sine's and cosine, i know finding capital F and evaluating at endpoints is more simple but just wondering if defining a simpler path is also a method for solving
• You could define your own path as long as you know the vector field is conservative. Conservative vector fields are path independent meaning you can take any path from A to B and will always get the same result.
Showing that capital F exists is the way you find out if the vector field is conservative. So it is a necessary step. If there is no capital F that exists for that vector field, then your vector field is not conservative so you can't define your own path. You will get the wrong answer.

Hope I helped.
p.s. There are other ways to show the vector field is conservative. For example if curl(f) = 0
• Why is it that f being written as del F makes the path conservative?
• What is conservative is not the path, but the vector field. And actually that is the definition "A vector field that can be written as the gradient of a scalar field is called 'conservative'".

As you are currently studying, conservative fields have some very nice mathematical shortcuts, and that is why it's very nice to work with conservative fields, this is because, as they can be represented by a scalar field, you don't have to deal with vector calculus, which removes a dimension of work.

It also means that conservative fields are "simpler" than non-conservative ones, (again because they can be represented as scalar fields).
• Around . Isn't the nabla an operator, so the vector arrow should be over F, right?
• Here F is not a vector field because it outputs just one number, so we would not write an arrow over it. It's not standard notation to write an arrow over the nabla operator when it is a gradient, but some people do to highlight that the result will be a vector. Hope this helps :)
• What is the significance of the answer being negative, if any? Thanks for replies.
• If you try to plot the field, you see that it opposes the path at the positive y-axis.
• Just wondering, would it be possible to evaluate this like if it were a normal line integral and not a vector line integral?
(1 vote)
• You could, but it would be very long and tedious.
(1 vote)
• Another method to solve this in polar coordinates:
(say i^= i-hat unit vector and p = phi)
f = r^2*i^ + 2*cos(p)*sin(p)*j^
dr_vector = r*d_p*p^ (p^ = unit vector in phi direction (azimutal angle)

you cant write the phi unit vector as: -sin(p)*i^+cos(p)*j^
now dot product f with dr gives you:

-sin(p)*d_p+2*cos^2(p)*sin(p)d_p
integrate from 0 to pi with respect to p (phi):

[cos(p)] + [-2/3*cos^3(p)] all from 0 to pi

this is: (-1 -1) - 2/3(-1-1) = -2 + 4/3 = -2/3
(1 vote)
• If a vector field has components Q(x,y) and P(x,y) in the i,j direction correspondingly, then if Q(x,y) and P(x,y) are continuous functions in space R, can't we tell for sure that the specific vector field is the gradient of a scalar valued function F(x,y), so it is conservative?
(1 vote)