MHB Finding potential function for field F

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To find the potential function for the vector field \(\vec{F} = (y+z)\hat{i} + (x+z)\hat{j} + (x+y)\hat{k}\), it is essential to first confirm that the field is irrotational, which allows for the existence of a scalar function \(\phi\) such that \(\nabla \phi = \vec{F}\). The integration of the components leads to the potential function \(\phi = xy + xz + yz + C\), which aligns with the answer provided in the textbook. The initial misunderstanding stemmed from incorrectly handling the integration process and the relationships between the components. Clarification on these steps resolved the confusion regarding the potential function. Understanding the proper method for finding potential functions is crucial in vector calculus.
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In doing a seemingly simple homework problem, I came across a snag...
It says to find the potential function \(f\) for the vector field \(\vec{F}\). The problem states
$$\vec{F}=(y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}$$
So I figured that just the simple integral of each section respective to its component direction would give the potential function
$$f=x(y+z)+y(x+z)+z(x+y)+C$$
However, the answer in the back of the book says other wise. It claims that
$$f=x(y+z)+zy+C$$
Now I can kind of recognize that distributing out my original answer would give something similar to this, but wouldn't it really be
$$f=2x(y+z)+2zy+C$$
Which is ultimately
$$f=2xy+2xz+2zy+C$$
Does anyone have any idea how they got to this answer? I'm a little stuck here.
 
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skatenerd said:
In doing a seemingly simple homework problem, I came across a snag...
It says to find the potential function \(f\) for the vector field \(\vec{F}\). The problem states
$$\vec{F}=(y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}$$
So I figured that just the simple integral of each section respective to its component direction would give the potential function
$$f=x(y+z)+y(x+z)+z(x+y)+C$$
However, the answer in the back of the book says other wise. It claims that
$$f=x(y+z)+zy+C$$
Now I can kind of recognize that distributing out my original answer would give something similar to this, but wouldn't it really be
$$f=2x(y+z)+2zy+C$$
Which is ultimately
$$f=2xy+2xz+2zy+C$$
Does anyone have any idea how they got to this answer? I'm a little stuck here.

I assume you have first shown that the field is irrotational, i.e. that [math]\displaystyle \begin{align*} \nabla \times \mathbf{F} = \mathbf{0} \end{align*}[/math]. Now, if it is irrotational, then there exists a scalar function [math]\displaystyle \begin{align*} \phi \left( x, y, z \right) \end{align*}[/math] so that [math]\displaystyle \begin{align*} \nabla \phi = \mathbf{F} \end{align*}[/math]. This would mean that [math]\displaystyle \begin{align*} \left( \frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} , \frac{\partial \phi}{\partial z } \right) = \left( y + z, x + z, x + y \right) \end{align*}[/math]. Equating each of these components gives

[math]\displaystyle \begin{align*} \frac{\partial \phi}{\partial x} &= y + z \\ \phi &= \int{ \left( y + z \right) \, \partial x } \\ \phi &= x\,y + x\,z + f(y, z) \\ \\ \frac{\partial \phi}{\partial y} &= x + z \\ \phi &= \int{ \left( x + z \right) \, \partial y } \\ \phi &= x\,y + y\,z + g(x, z) \\ \\ \frac{\partial \phi}{\partial z} &= x + y \\ \phi &= \int{ \left( x + y\right) \, \partial z } \\ \phi &= x\,z + y\,z + h(x, y) \end{align*}[/math]

When we compare all the different components to [math]\displaystyle \begin{align*} \phi \end{align*}[/math] we find that [math]\displaystyle \begin{align*} \phi = x\,y + x\,z + y\,z + C \end{align*}[/math].
 
Sorry to anybody who may have been trying to answer this at the moment, but I just realized how I was doing it wrong. I guess I didn't exactly understand the full process of finding a potential function. Nevermind!

- - - Updated - - -

Thanks ProveIt. That makes things pretty simple. Got it now
 
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