For $\displaystyle \Phi (x,y,z) = x^{xy} \sin(x+z) + y \ln (z) $

find the flux vector $\displaystyle \vec{J} = \triangledown \Phi$

and prove that curl $\displaystyle \vec{J} = 0$

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- Sep 22nd 2009, 05:42 AMJimmy_WFind Flux Vector
For $\displaystyle \Phi (x,y,z) = x^{xy} \sin(x+z) + y \ln (z) $

find the flux vector $\displaystyle \vec{J} = \triangledown \Phi$

and prove that curl $\displaystyle \vec{J} = 0$ - Oct 4th 2009, 02:08 AMJimmy_W
Is this correct?

$\displaystyle \triangledown \Phi = \frac{\partial \Phi}{\partial x} i + \frac{\partial \Phi}{\partial y} j + \frac{\partial \Phi}{\partial z} k$

where

$\displaystyle \frac{\partial \Phi}{\partial x} = e^{xy} \cos(x+z) + \sin(x+z) y e^{xy}$

$\displaystyle \frac{\partial \Phi}{\partial y} = \sin(x+z) x e^{xy} + \ln(z)$

and

$\displaystyle \frac{\partial \Phi}{\partial z} = e^{xy} \cos(x+z) + \frac{y}{x} $

and

$\displaystyle \triangledown \Phi = (e^{xy} \cos(x+z) + \sin(x+z) y e^{xy}) i + (\sin(x+z) x e^{xy} + \ln(z) ) j + (e^{xy} \cos(x+z) + \frac{y}{x})k$

Is there anything more I need to do with this? - Oct 4th 2009, 02:26 AMMarkW
In the k component it should be y/z, not y/x.

Other than that, all you need to do is calculate the curl and it will cancel out nicely.

Do you know how to find the general flux vector in the second part of question 1?