Path Integral Setup for Given Initial and Final Points

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Homework Help Overview

The discussion revolves around evaluating a path integral from the point (0,0,0) to (1,1,1) for the vector field . Participants express confusion regarding the setup of the integral and the implications of treating variables as constants during integration.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to break down the integral into separate components but questions the validity of substituting variables to zero. Other participants clarify that the variables y and z cannot be treated as constants and emphasize the importance of the path taken in the integral.

Discussion Status

Participants are exploring the implications of treating the integral as an exact differential and discussing the necessity of a specific path for evaluation. Some guidance has been provided regarding the nature of path integrals and the importance of parameterization, though the original poster expresses a lack of comfort with the concept.

Contextual Notes

There is an acknowledgment that the problem cannot be solved without a specified path, and the discussion reflects uncertainty about the assumptions made in the setup of the integral.

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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