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Consider the Cartesian coordinate system with a vector $$f=(a(x^2+y^2)^{N/2}\cos(N\theta),a(x^2+y^2)^{N/2}\sin(N\theta),bz)$$where $a,b\in\mathbb{R}$ fixed and $N$ is an integer.
For $f'=f/|f|$, I want to determine the integral $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f'\cdot\left(\frac{\partial f'}{\partial x}\times\frac{\partial f'}{\partial y}\right) dxdy.$$

To do that, we transform to "almost spherical" coordinates
$r=\sqrt{a^2(x^2+y^2)^N+b^2z^2}$
$\tan\theta=\frac{y}{x}$
$\tan\phi=\frac{a(x^2+y^2)^{N/2}}{bz}$
with $\theta\in[0,2\pi]$, $\phi\in[0,\pi]$.

How do I rewrite the integral to these new coordinates?

I find that $f=(r\sin\phi\cos(N\theta),r\sin\phi\sin(N\theta),r\cos\phi)$ and $f'=(\sin\phi\cos(N\theta),\sin\phi\sin(N\theta),\cos\phi)$, but how do I rewrite $\frac{\partial f'}{\partial x}\times\frac{\partial f'}{\partial y}$?

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  • $\begingroup$ Recall the chain rule: $\frac{\partial f'}{\partial x} = \frac{\partial f'}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f'}{\partial \theta}\frac{\partial \theta}{\partial x}+\frac{\partial f'}{\partial \phi}\frac{\partial \phi}{\partial x}$, and similarly for $\frac{\partial f'}{\partial y}$. $\endgroup$ – user3482749 Dec 27 '18 at 16:49
  • $\begingroup$ @user3482749 And how do the integral variables change? We integrate over $x$ and $y$ from $-\infty$ to $\infty$, what does that become in the new coordinates? $\endgroup$ – Pierre LeFèvre Dec 27 '18 at 17:33
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First, to answer your question, you need to find the Jacobian Matrix $\mathcal{J}$ between the 2 set of coodinates. And then $dxdy = |\mathcal{J}|drd\theta$, where $|\mathcal{J}|$ is the determinant of the Jacobian Matrix.

There are a couple more problems in your thinking

  • Your original integral is 2-D, I feel it is better to just think $z$ as a constant and ignore the third coordinate $z$ (or $\phi$) in the transformation
  • $f$ is a vector already, then $f'$ is a very confusing notation. It can be a matrix (gradient) $$ f' = [\partial_x f, \partial_y f, \partial_z f] $$ It can be a vector (curl) $$ f' = \nabla \times f $$ It can also be a scalar (divergence) $$ f' = \nabla \cdot f $$

not sure which one you are referring to , but I guess it is the third one

  • I noticed that $\theta$ is not a new variable you defined, it is already in the definition of your $f$ function. Is your transformation definition of $\theta$ consistent with the meaning of $\theta$ in $f$ function?

So if you define the transformation between $x,y$ and $r,\theta$ as $$ r = a(x^2 + y^2) ^{N/2} \quad \tan\theta = \frac{y}{x} $$

Your Jacobian matrix is $$ \mathcal{J} = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \\ \end{bmatrix} = \begin{bmatrix} \frac{2}{Nr}(\frac{r}{a})^{2/N}\cos\theta & -(\frac{r}{a})^{2/N}\sin\theta \\ \frac{2}{Nr}(\frac{r}{a})^{2/N}\sin\theta & (\frac{r}{a})^{2/N}\cos\theta \\ \end{bmatrix} $$

so

$$ dxdy = |\mathcal{J}|drd\theta = \frac{2}{Nr}\left(\frac{r}{a}\right)^{4/N}drd\theta $$

Just a sanity check, when $N=2, a=1$, we can observer that it falls back to the form of polar cooridnate $$ dxdy = rdrd\theta $$

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  • $\begingroup$ Thanks for the reply! Just to clarify on the points you made: $f'$ is my (indeed poor) notation for the unit vector in the direction of $f$, i.e. $f':=f/|f|$. Further $\theta$ is indeed consistent with the $\theta$ in $f$. Nice username btw :) $\endgroup$ – Pierre LeFèvre Dec 27 '18 at 21:47
  • $\begingroup$ So to find $dx$, $dy$ in the integral we basically take $\theta=\pi/2$. Must we also fill that in in our expression for $f’\cdot (\partial_x f’\times \partial_y f’)$? $\endgroup$ – Pierre LeFèvre Dec 28 '18 at 9:25
  • $\begingroup$ Oh sorry I missed your definition of $f'$, now it makes sense. I will update the solution later (probably tonight). btw. I don't understand your comment about $\theta=\pi/2$. There is no dependence of $\theta$ in the conversion of $dxdy$ because all $\theta$ cancels out in the $\cos^2\theta + \sin^2\theta=1$ calculation. Maybe you misunderstand how to calculate the determinant of a matrix. $\endgroup$ – MoonKnight Dec 28 '18 at 18:36

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