My Math Forum Switching scalar field to polar coordinates

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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: 635 Thanks: 401 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. Thanks from SenatorArmstrong
October 25th, 2017, 02:52 PM   #4
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Quote:
 Originally Posted by Country Boy $\displaystyle sin(x^2- y^2)= sin(r^2cos^2(\theta)- r^2sin^2(\theta))= sin(r^2(cos^2(\theta)- sin^2(\theta))= sin(r^2cos(2\theta))$. That is about as simple as it gets.
Okay thanks. That's what I had on my exam, but kept second guessing myself. The cos tangled in the sin function looks bizarre to me. Not used to that.

Thank you sir

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