July 31st, 2012, 04:21 AM  #1 
Member Joined: Jan 2012 Posts: 82 Thanks: 0  3dimensional scalar field
Hi! A 3d scalar field h(x,y,z)=x^2+y^2+z^2zx 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  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: 3dimensional scalar field Quote:
 
July 31st, 2012, 06:30 AM  #3  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: 3dimensional scalar field Quote:
 
August 1st, 2012, 03:31 AM  #4 
Member Joined: Jan 2012 Posts: 82 Thanks: 0  Re: 3dimensional 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: 3dimensional 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 viceversa, 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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