Finding potential function for field F

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SUMMARY

The discussion focuses on finding the potential function \(f\) for the vector field \(\vec{F}=(y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}\). The initial attempt to derive \(f\) resulted in \(f=x(y+z)+y(x+z)+z(x+y)+C\), which was incorrect. The correct potential function is \(f=x(y+z)+zy+C\), derived by recognizing that the field is irrotational, allowing the use of the gradient to find \(f\). The final form of \(f\) is confirmed as \(f=x\,y + x\,z + y\,z + C\).

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  • Knowledge of irrotational fields and the curl operator
  • Familiarity with partial derivatives and integration
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  • Study the properties of irrotational vector fields and their implications
  • Learn how to compute the curl of a vector field using vector calculus
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skate_nerd
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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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