Path Integral Setup for Given Initial and Final Points

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The discussion focuses on setting up a path integral from the point (0,0,0) to (1,1,1) for the vector field <x^2, 2yz, y^2>. The initial attempt incorrectly substitutes y and z as zero, leading to an erroneous simplification of the integral. It is clarified that y and z vary along the path and cannot be treated as constants. The concept of exact differentials is introduced, emphasizing that the integral can only be separated if it meets specific criteria. The importance of parameterizing the path for accurate integration is also highlighted, indicating that practice is necessary to gain comfort with path integrals.
Safder Aree
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Homework Statement


The path integral from (0,0,0) to (1,1,1) of $$<x^2,2yz,y^2>$$.
I am a little confused about the setup.

Homework Equations


$$\int_{a}^{b} v.dl$$

The Attempt at a Solution



Here is how I set it up.
$$\int_{0}^{1}x^2 dx + \int_{0}^{1}2yz dy + \int_{0}^{1}y^2 dz$$

Since the initial values of all of them are 0 can I not substitute y=0 and z=0?
So the equation looks like:
$$\int_{0}^{1}x^2 dx + \int_{0}^{1}2y(0) dy + \int_{0}^{1}(0)^2 dz$$
$$=1/3 + 0 + 0$$

I know this is wrong but where am I making the error? Thank you so much.
 
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You can only break the integral up into three parts like that because ##x^2\:dx+2xy\:dy+y^2\:dz## is what is called an exact differential. In general, you cannot do this.

You cannot set y and z equal to zero because, as you move along the path, they vary from 0 to 1. So their "average" value is not zero.
 
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Gene Naden said:
You can only break the integral up into three parts like that because ##x^2\:dx+2xy\:dy+y^2\:dz## is what is called an exact differential. In general, you cannot do this.

You cannot set y and z equal to zero because, as you move along the path, they vary from 0 to 1. So their "average" value is not zero.

That actually makes perfect sense now. If say (0,0,0) to (1,0,0) then I could right?
 
You could go from (0,0,0) to (0,0,1) and from there to (0,1,1) and from there to (1,1,1)
 
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Gene Naden said:
You could go from (0,0,0) to (0,0,1) and from there to (0,1,1) and from there to (1,1,1)
Right sorry, that's what i meant. Thank you.
 
@Safder Aree : You do understand though, that generally you have to be given a particular path and work the integral out by using a parameterization of that path, right? Your problem as stated couldn't be solved if the integral hadn't been independent of path.
 
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LCKurtz said:
@Safder Aree : You do understand though, that generally you have to be given a particular path and work the integral out by using a parameterization of that path, right? Your problem as stated couldn't be solved if the integral hadn't been independent of path.

Yes I do understand that, I'm not quite comfortable with path integrals quite yet but hopefully it'll come with practice.
 

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