A conceptual reason for why the Jacobian of a rotation by a changing angle is $1$? Consider
$$f(x,y)=(x\cos r^2+y\sin r^2, y\cos r^2-x\sin r^2)\qquad\text{with }r=\sqrt{x^2+y^2},$$
as a map $\mathbb{R}^2 \to \mathbb{R}^2$. Geometrically, $f(x,y)$ is obtained from the vector $(x,y)$ by rotating it by angle $r^2$. (we rotate the more we move away from the origin).

Wolframe Alpha claim that $\det (df)=1$ holds identically. Is there an elegant rigorous way of seeing this, without too much computations?

Should this be "obvious" in retrospect? I was a bit surprised by this result... and direct computation is not so nice to do by hand (although tractable).
I thought using the chain rule (treating the angle as a function of $x,y$) but got nowhere.
Edit:
I agree that roughly speaking, near a point $p$, this function is like a standard rotation by a fixed angle $|p|^2$. However, I do not consider this a rigorous explanation. Indeed, this vague intuition is still with us when we replace $r^2=x^2+y^2$ by $x^4+y^4$, but then the Jacobian is non-constant. 

So, this property doesn't even hold for smooth radially symmetric angle function $\theta(x,y)$.

Is this just a coincidence then? Can we charavterize the angle functions (or at least radially symmetric ones) which satisfy this?
(Wolframe says the Jacobian remains $1$ when we replace $r^2$ by $r$ or $r^4$. Perhaps this remains true for any power of $r$?)

Partition the plane into very thin annular sets $A_k: \>r_{k-1}\leq\sqrt{x^2+y^2}< r_k$ $(k\in {\mathbb N}_{\geq1})$. The map $f$ (approximately) rotates each $A_k$ by an angle $\phi_k$ around the origin, hence is (approximately) an isometry on $A_k$. This implies $\mu\bigl(f(X)\bigr)=\mu(X)$ for  any $X\subset A_k$. For an arbitrary set $B\subset{\mathbb R}^2$ consider the annular subsets $B_k:=B\cap A_k\subset A_k$ and obtain
$$\mu\bigl(f(B)\bigr)=\sum_{k=1}^\infty \mu\bigl(f(B_k)\bigr)=\sum_{k=1}^\infty \mu(B_k)=\mu(B)\ .$$
This shows that $f$ is globally area preserving. (But I'd say that using the chain rule on $df$ is simpler.)
