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Finding potential function for field F

skatenerd

Active member
Oct 3, 2012
114
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.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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 \(\displaystyle \displaystyle \begin{align*} \nabla \times \mathbf{F} = \mathbf{0} \end{align*}\). Now, if it is irrotational, then there exists a scalar function \(\displaystyle \displaystyle \begin{align*} \phi \left( x, y, z \right) \end{align*}\) so that \(\displaystyle \displaystyle \begin{align*} \nabla \phi = \mathbf{F} \end{align*}\). This would mean that \(\displaystyle \displaystyle \begin{align*} \left( \frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} , \frac{\partial \phi}{\partial z } \right) = \left( y + z, x + z, x + y \right) \end{align*}\). Equating each of these components gives

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

When we compare all the different components to \(\displaystyle \displaystyle \begin{align*} \phi \end{align*}\) we find that \(\displaystyle \displaystyle \begin{align*} \phi = x\,y + x\,z + y\,z + C \end{align*}\).
 

skatenerd

Active member
Oct 3, 2012
114
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!

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