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 July 31st, 2012, 04:21 AM #1 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 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
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).

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 $cos^2 O+ sin^2 O= 1$ so you can simplify that a bit. Yes, $\rho$ (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.

 August 1st, 2012, 03:31 AM #4 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Re: 3-dimensional scalar field Hi! Sorry not sure about that simplification. Could you go over that. Thanks
 August 2nd, 2012, 08:29 AM #5 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: 3-dimensional scalar field I'm not sure what "simplification" you are talking about. Do you mean "$sin^2(x)+ cos^2(x)= 1$? That comes from the Pythagorean theorem: if a right triangle has legs of length a and b, hypotenuse of length c, then $a^2+ b^2= c^2$. Dividing on both sides by $c^2$, $$$\frac{a}{c}$$^2+ $$\frac{b}{c}$$^2= 1$ . In such right triangle, "sine" is defined as "opposite side divided by near side" so $\frac{a}{c}$ is sine and $\frac{b}{c}$ is cosine (or vice-versa, it doesn't matter). Then $$$\frac{a}{c}$$^2+ $$\frac{b}{c}$$^2= sin^2(O)+ cos^2(O)= 1$. To "simplify" what you have replace your "$sin^2(O)+ cos^2(O)= 1$" with "1". But, again, why are you requiring that z> 1?

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