# 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 $$\frac{\partial f'}{\partial x}\times\frac{\partial f'}{\partial y}$$?

• 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 $\frac{\partial f'}{\partial y}$. – user3482749 Dec 27 '18 at 16:49
• @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? – Pierre LeFèvre Dec 27 '18 at 17:33

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 $$dxdy = rdrd\theta$$

• 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 :) – Pierre LeFèvre Dec 27 '18 at 21:47
• So to find $dx$, $dy$ in the integral we basically take $\theta=\pi/2$. Must we also fill that in in our expression for $f’\cdot (\partial_x f’\times \partial_y f’)$? – Pierre LeFèvre Dec 28 '18 at 9:25
• 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. – MoonKnight Dec 28 '18 at 18:36