How to transform integral after coordinate transformation 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 $\frac{\partial f'}{\partial x}\times\frac{\partial f'}{\partial y}$?
 A: 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
$$
dxdy = rdrd\theta
$$
