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July 31st, 2012, 04:21 AM   #1
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3-dimensional scalar field

Hi!

A 3-d scalar field h(x,y,z)=x^2+y^2+z^2-zx

I need to find an expression for the scalar field h in cylindrical polar coordinates.

So far I have

cyclindrical polar coordinates are related to the cartesian coordinates (x,y,z) by
x=p cos O, y=p sin O, z=z

we see that x^2+y^2+z^2= p^2(cos^2 O + sin^2 O)+z^2 and - zx = -z(p cos O)

hence
h(p,O,z)=p^2(cos^2 O+sin^2 O) +z^2 -z(p cos O)
(p>0, z>1)

Not sure about the last bit.

Can anyone help

Arron
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July 31st, 2012, 05:57 AM   #2
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Re: 3-dimensional scalar field

Quote:
Originally Posted by arron1990
Not sure about the last bit.
I don't see any faults in your approach. though I am not very knowledgeable on this part of mathematics (Polar Coordinate).
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July 31st, 2012, 06:30 AM   #3
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Re: 3-dimensional scalar field

Quote:
Originally Posted by arron1990
hence
h(p,O,z)=p^2(cos^2 O+sin^2 O) +z^2 -z(p cos O)
(p>0, z>1)

Not sure about the last bit.

Can anyone help

Arron
You know, I hope, that so you can simplify that a bit. Yes, (which looks like a p), although I would have used "r", must be greater than 0-. But what reason do you have to require that z> 1? There is no such condition in your original post.
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August 1st, 2012, 03:31 AM   #4
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Re: 3-dimensional scalar field

Hi!

Sorry not sure about that simplification.

Could you go over that.

Thanks
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August 2nd, 2012, 08:29 AM   #5
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Re: 3-dimensional scalar field

I'm not sure what "simplification" you are talking about. Do you mean "? That comes from the Pythagorean theorem: if a right triangle has legs of length a and b, hypotenuse of length c, then . Dividing on both sides by , . In such right triangle, "sine" is defined as "opposite side divided by near side" so is sine and is cosine (or vice-versa, it doesn't matter). Then .

To "simplify" what you have replace your "" with "1".


But, again, why are you requiring that z> 1?
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