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October 25th, 2017, 02:13 PM   #1
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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!
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October 25th, 2017, 02:20 PM   #2
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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 $\sin(r^2 \sin^2 \theta - r^2 \cos^2 \theta)$.
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October 25th, 2017, 02:45 PM   #3
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$\displaystyle sin(x^2- y^2)= sin(r^2cos^2(\theta)- r^2sin^2(\theta))= sin(r^2(cos^2(\theta)- sin^2(\theta))= sin(r^2cos(2\theta))$. That is about as simple as it gets.
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October 25th, 2017, 02:52 PM   #4
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Quote:
Originally Posted by Country Boy View Post
$\displaystyle sin(x^2- y^2)= sin(r^2cos^2(\theta)- r^2sin^2(\theta))= sin(r^2(cos^2(\theta)- sin^2(\theta))= sin(r^2cos(2\theta))$. 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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