Path integral question

In summary, the student was attempting to solve the path integral from (0,0,0) to (1,1,1) of <x^2,2yz,y^2> and was confused about the setup. After receiving an explanation, they realized that the integral could not be broken up into three parts and that setting y and z equal to zero was incorrect. The student also clarified that the integral could only be solved if a specific path was given. f
  • #1

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.
 
  • #2
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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  • #3
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?
 
  • #4
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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  • #5
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.
 
  • #6
@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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  • #7
@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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