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 October 27th, 2009, 12:13 PM #1 Newbie   Joined: Oct 2009 Posts: 27 Thanks: 0 Vector Field 3 Find the divergence and the curl of the vector field $\LARGE \b{F}(x,y,z)\,=\,(yz,\,xz,\,xy)$ Please show the calculations.
 October 27th, 2009, 01:22 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 How far have you got? Do you know the definition of divergence and curl?
 October 27th, 2009, 06:41 PM #3 Newbie   Joined: Oct 2009 Posts: 27 Thanks: 0 Re: Vector Field 3 i know none of this, we just started this stuff, honestly my professor is horrible at explaining things, so i'm looking for help anywhere i can find it. It usually helps me to see the steps someone would take to work this out.
 October 28th, 2009, 06:21 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 For a 3-dimensional vector field $\bf {F},$ the divergence is given by $\nabla\cdot\bf{F}$ and the curl is given by $\nabla\times\bf {F},$ where the differential operator $\nabla$ can be thought of as $\nabla=(\partial/\partial x,\partial/\partial y,\partial/\partial z).$ Example: $\bf{F}=(x,xy,xyz)$ $\nabla\cdot\bf{F}=\frac{\partial}{\partial x}x+\frac{\partial}{\partial y}(xy)+\frac{\partial}{\partial z}(xyz)=1+x+xy$ \begin{align}\nabla\times\bf{F}&=\left(\frac{\part ial}{\partial y}(xyz)-\frac{\partial}{\partial z}(xy)\,,\,\frac{\partial}{\partial z}x-\frac{\partial}{\partial x}(xyz)\,,\,\frac{\partial}{\partial x}(xy)-\frac{\partial}{\partial y}x\right)\\ &=\left(xz\,,\,-yz\,,\,y\right)\end{align}
 October 28th, 2009, 05:38 PM #5 Newbie   Joined: Oct 2009 Posts: 27 Thanks: 0 Re: Vector Field 3 but wouldn't: $\frac{\partial}{\partial x}\,yz\,=\,0$ $\frac{\partial}{\partial y}\,xz\,=\,0$ $\frac{\partial}{\partial z}\,xy\,=\,0$
October 28th, 2009, 08:40 PM   #6
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Re: Vector Field 3

Quote:
 Originally Posted by OSearcy4 but wouldn't: $\frac{\partial}{\partial x}\,yz\,=\,0$ $\frac{\partial}{\partial y}\,xz\,=\,0$ $\frac{\partial}{\partial z}\,xy\,=\,0$

Yes.

He was just using another example, F(x,xy,xyz). You would be correct for the divergence of your function.

 October 28th, 2009, 08:41 PM #7 Newbie   Joined: Oct 2009 Posts: 27 Thanks: 0 Re: Vector Field 3 so where would i go from there?
 October 28th, 2009, 08:56 PM #8 Newbie   Joined: Oct 2009 Posts: 27 Thanks: 0 Re: Vector Field 3 $\nabla \times \b{F}\,=\,(\frac{\partial}{\partial y}xy\,-\,\frac{\partial}{\partial z}xz,\,\frac{\partial}{\partial z}yz\,-\,\frac{\partial}{\partial x}xy,\,\frac{\partial}{\partial x}xz\,-\,\frac{\partial}{\partial y}yz)$ $=\,(y\,-\,x,\,y\,-\,y,\,z\,-\,z)$ $=\,(y\,-\,x,\,0,\,0)$ would this be the correct answer? is there a better way to write this or work this out?
 October 29th, 2009, 05:47 AM #9 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Check your calculations for the first element of $\nabla\times\bf{F}.$.
 October 29th, 2009, 06:09 AM #10 Newbie   Joined: Oct 2009 Posts: 27 Thanks: 0 Re: Vector Field 3 oops it would be (0,0,0) and that's it??

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