# Path integral question

• Safder Aree
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

## 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.

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.

Safder Aree
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)

Safder Aree
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.

Safder Aree
@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.