How do you find a potential function of the field:
F= (ysinz)i + (xsinz)j + (xycosz)k
I'm lost on questions like this...
Let $\displaystyle F = \nabla \phi$. Then:
$\displaystyle \frac{\partial \phi}{\partial x} = y \sin z$ .... (1)
$\displaystyle \frac{\partial \phi}{\partial y} = x \sin z$ .... (2)
$\displaystyle \frac{\partial \phi}{\partial z} = xy \cos z$ .... (3)
These pde's have to be solved simultaneously.
From (1): $\displaystyle \phi = x y \sin z + g(y, z)$ where $\displaystyle g(y, z)$ is an arbitrary function of y and z.
Substitute this into (2): $\displaystyle x \sin z + \frac{\partial g}{\partial y} = x \sin z \Rightarrow \frac{\partial g}{\partial y} = 0 \Rightarrow g(y, z) = C$ where C is an arbitrary constant.
So you can say that $\displaystyle \phi = x y \sin z$. Check that this function does this job.
If we have a vector field described as…
$\displaystyle \overrightarrow{f} =f_{x} (x,y,z)\cdot $$\displaystyle \overrightarrow{i} + f_{y} (x,y,z)\cdot$$\displaystyle \overrightarrow{j} + f_{z} (x,y,z)\cdot$$\displaystyle \overrightarrow{k}$
… and it is conservative the following relation holds…
$\displaystyle \overrightarrow{f}= \nabla {U(x,y,z)} $$\displaystyle = \frac{\partial U}{\partial x}\cdot$$\displaystyle \overrightarrow{i} $$\displaystyle +\frac{\partial U}{\partial y}\cdot $$\displaystyle \overrightarrow{j} + $$\displaystyle \frac{\partial U}{\partial z}\cdot $$\displaystyle \overrightarrow{k}$
… where $\displaystyle U(*,*,*)$ is a scalar function which is called potential. In the case you have proposed is…
$\displaystyle f_{x}= y\cdot \sin z$
$\displaystyle f_{y}= x\cdot \sin z$
$\displaystyle f_{z}= x\cdot y\cdot \cos z$
… so that you can verify that the scalar function…
$\displaystyle U(x,y,z)= x\cdot y\cdot \sin z$
… satisfies the conditions to be the potential of the field given by you…
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
P.S. Again i didn't note that fantastic had posted before me... i apologize for that...