# Path integral question

## 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.

Related Calculus and Beyond Homework Help News on Phys.org
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.

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)

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.

LCKurtz