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October 25th, 2017, 02:13 PM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  Switching scalar field to polar coordinates
I am trying to take this scalar field $\phi$(r)=$\exp[x^2+y^2] * sin(x^2  y^2)$ And switch to polar coordinates. But every time I do so I get trig functions sandwiched together. The $x^2 + y^2$ in the exp is simple as it's just r^2. However, the $x^2  y^2$ in the sin function becomes a mess when I substitute in the $r^2sin^2(\theta)*r^2cos^2(\theta)$ in there. Is there a way to do this and get a more simple answer? Thanks! 
October 25th, 2017, 02:20 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics 
I'm not sure what the notation $\Phi(r)$ denotes since you haven't written it in polar yet and even so, $\Phi: \mathbb{R}^2 \to \mathbb{R}$ regardless of coordinates. Otherwise, I'm not sure why you expect this to be "nice" in polar coordinates. Its not nice in rectangular either. I don't see why $\sin(x^2  y^2)$ is any nicer than $\sin(r^2 \sin^2 \theta  r^2 \cos^2 \theta)$. 
October 25th, 2017, 02:45 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
$\displaystyle sin(x^2 y^2)= sin(r^2cos^2(\theta) r^2sin^2(\theta))= sin(r^2(cos^2(\theta) sin^2(\theta))= sin(r^2cos(2\theta))$. That is about as simple as it gets.

October 25th, 2017, 02:52 PM  #4  
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  Quote:
Thank you sir  

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coordinates, field, polar, scalar, switching 
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