Vrftsjtryk

Consider the Cartesian coordinate system with a vector $$f=(a(x^2+y^2)^{N/2}cos(Ntheta),a(x^2+y^2)^{N/2}sin(Ntheta),bz)$$where $a,binmathbb{R}$ fixed and $N$ is an integer.

For $f'=f/|f|$, I want to determine the integral
$$int_{-infty}^{infty}int_{-infty}^{infty}f'cdotleft(frac{partial f'}{partial x}timesfrac{partial f'}{partial y}right) dxdy.$$

To do that, we transform to "almost spherical" coordinates
$r=sqrt{a^2(x^2+y^2)^N+b^2z^2}$
$tantheta=frac{y}{x}$
$tanphi=frac{a(x^2+y^2)^{N/2}}{bz}$

with $thetain[0,2pi]$, $phiin[0,pi]$.

How do I rewrite the integral to these new coordinates?

I find that $f=(rsinphicos(Ntheta),rsinphisin(Ntheta),rcosphi)$ and $f'=(sinphicos(Ntheta),sinphisin(Ntheta),cosphi)$, but how do I rewrite $frac{partial f'}{partial x}timesfrac{partial f'}{partial y}$?

First, to answer your question, you need to find the Jacobian Matrix $mathcal{J}$ between the 2 set of coodinates. And then $dxdy = |mathcal{J}|drdtheta$, 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
$$
r = a(x^2 + y^2) ^{N/2} quad tantheta = frac{y}{x}
$$

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

so

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

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