Problem. Consider the following maps from $\Bbb{R^2}$ to $\Bbb{R^2}$:
(i) the map $(x,y) \mapsto(2x+5y+1, x +3y),$
(ii) the map $(x,y)\mapsto (x+y^2, y+x^2)$
(iii) the map given in polar coordinates as $(r, \theta)\mapsto (r, \theta+r^3)$ for $r \ne 0$, with the origin mapping to the origin.
Which of maps in the above list preserves areas?
My Attempt.
$\phi:\Bbb{R^2}\to\Bbb{R^2}$ is area preserving if $$m(\phi(A))=m(A)$$ for any $A \subset \Bbb{R}^2$. (Here, $m$ denotes the Lebesgue Measure on $\Bbb{R^2}$.)
We know the change of variables for double integral is:
If $S$ and $T$ be two regions in the plane and $\phi:\Bbb{R^2}\to \Bbb{R^2}$ such that $\phi$ is continuously differentiable, one-one and $\phi(T)=S$. Then for any $f:\Bbb{R^2}\to \Bbb{R}$ $$\iint_{S}f(x,y)dxdy=\iint_T f(\phi(u,v))|J_\phi(u,v)|dudv$$
So for a map $\phi:\Bbb{R^2}\to\Bbb{R^2}$ we have:
$$m(\phi(A))=\iint_{\phi(A)}dxdy=\iint_A|J_\phi(u,v)|dudv\dots\dots(*)$$
So if $|J_\phi(u,v)|=1$ then from $(*)$ we have $m(\phi(A))=m(A)$
So I check the Jcobian determinant of the given functions in the problem:
For (i) modulus of the Jacobian determinant of $(x,y) \mapsto(2x+5y+1, x +3y)$ is $1$. Therefore this map preserves area
(ii) Now the modulus of the Jacobian determinat of $(x,y)\mapsto (x+y^2, y+x^2)$ is $|1-4xy|$. So the area may not be preserved for all regions (as the area will be scaled by the factor $|J_\phi|$). So I think this is not area preserving map.
(iii) For the function $(r, \theta)\mapsto (r, \theta+r^3)$ I think I need the corresponding formula of change of variables in polar coordinates. I try to replace $dxdy$ in $(*)$ by $rdrd\theta$ and find the jacobian($=3r^2$) but it was not so satisfactory to me...Please help in this function to proceed.
Is everything correct till now? Please mention whether I've made any mistake there. Thank you.
PS: (iii) I use the change of variable like the following:
$$\iint_{S}f(r,\theta)rdrd\theta=\iint_T f(\phi(r,\theta))|J_\phi(r,\theta)|rdrd\theta$$ Here, $\phi(T)=S$, and $S,T$ are two regions in $r\theta$ plane. Then for the map in (iii), $J_\phi(r,\theta)=3r^2$. So it doesn't preserve area. Is the formula I use here right?
It is unsatisfactory because the answer key to this particular problem indicates that: "there are 2 maps in the list that preserve area".
EDIT. I think the problem here is to determine the Jacobian matrix of the function given in (iii). Now for any differentiable function $f:\Bbb{R^2}\to \Bbb{R^2}$ where $f(x,y)=(f_1(x,y), f_2(x,y))$ the Jacobian matrix is $J = (\nabla f_1(x,y), \nabla f_2(x,y))$.
What if the function is given in terms of $r\theta$ as in (iii). How to find the Jacobian then?
Your third map $f$ is area-preserving as well. This can be seen intuitively as follows: Partition the plane ${\mathbb R}^2$ into very very thin concentric annuli. Then $f$ takes each annulus and rotates it somewhat about the origin, whereby the rotation angle ($=r^3$) depends on the radius $r$ of the annulus. When $r\ll1$ this angle is very small, and for large $r$ it can be anything in the interval $[0,2\pi[\>$.
For a proof we imbed $f$ into the following product of three maps: $$\eqalign{{\rm polar}:\quad &(x,y)\mapsto(r,\phi):=\bigl(\sqrt{x^2+y^2}, \>{\rm arg}(x,y)\bigr)\>,\cr f:\quad &(r,\phi)\mapsto (\rho,\theta):=\bigl(r,\>\phi+r^3\bigr)\>,\cr {\rm rect}:\quad&(\rho,\theta)\mapsto(u,v):=\bigl(\rho\cos\theta,\rho\sin\theta\bigr)\ .\cr}$$ For area considerations we have to compute the Jacobian of the composition $${\rm rect}\circ f\circ{\rm polar}:\quad(x,y)\mapsto(u,v)\ .$$ Now $J_f(r,\phi)\equiv1$, $J_{\rm rect}(\rho,\theta)=\rho$, and $$J_{\rm polar}(x,y)={1\over J_{\rm rect}(r,\phi)}={1\over r}\ .$$ As $\rho=r$ (by definition of $f$) the product of the three Jacobians is $\equiv1$, which proves that your $f$ is area preserving with respect to euclidean area in the plane..