At the risk of asking an extremely stupid question, suppose that $P\subset\mathbb{R}^2$ is a convex polygon with area $1$ that contains the origin, and let $r$ denote the farthest distance between the origin and a point in $P$, i.e. $r = \max_{x\in P} \x\$. Let $S$ denote a sector of a circle with radius $r$ that is centered at the origin such that $S$ also has area $1$ (i.e. the angle of the sector must therefore be $2/r^2$ radians). My question: is there an obvious areapreserving map $f:P\to S$ such that, for any point $x\in P$, we have $\x\\leq \f(X)\$?

1$\begingroup$ Let $C_\rho$ denote the centered circle of radius $\rho$. Given $\rho$ and $\rho'$, you could map the intersection $C_\rho\cap P$ to an arc $C_{\rho'}\cap S$ in some canonical way. Now you only have to figure out the unique monotonic map $\rho\mapsto\rho'$ that would make this area preserving when applied to all of $P$. That map only depends on the radial profiles of $P$ and $S$. $\endgroup$ – Yoav Kallus Dec 10 '13 at 22:18

1$\begingroup$ P.S. The map will have $\rho'\ge\rho$ because a uniform random point in $P$ has a smaller distance from the origin than a uniform random point in $S$ (which has the same distribution of distances from the origin as a random point in the disk of radius $r$). $\endgroup$ – Yoav Kallus Dec 10 '13 at 22:24

1$\begingroup$ @Yoav: The origin stays fixed, then a small circle surrounding it is mapped onto a simple closed curve surrounding the small circle and not identical to it. Then the map cannot be areapreserving. $\endgroup$ – Wlodek Kuperberg Dec 11 '13 at 0:08

2$\begingroup$ Does the map have to be continuous? $\endgroup$ – Lev Borisov Dec 11 '13 at 0:38

1$\begingroup$ $f$ can be piecewise continuous and still areapreserving in the sense that for any $X\subset P$, $f(X)=X$. $\endgroup$ – Yoav Kallus Dec 11 '13 at 3:07
Begin with a lemma: Let $\ T=\Delta oab$ be a triangle contained in a circle centered at the origin $o$ and let the arc $\alpha$ of the circle be such that the circle's sector $S$ based on $\alpha$ has the same area as $T$. Let the map $f_T$ from $T$ onto $S$ be defined so that $f_T(o)=o$, $f_T$ maps the segment $ab$ linearly (with respect to the arc length) onto $\alpha$, and $f_T$ maps every segment $ox$ ($x\in ab$) linearly onto the radius $of(x)$. Then $f_T$ is areapreserving and, obviously, satisfies the condition $\ x\\le \ f_T(x)\$.
That $f_T$ is area preserving is intuitively clear and can be easily verified by the Jacobian of $f_T$.
Now, to answer the question in the affirmative, cut $P$ into triangles $T_1, T_2,\ldots, T_k$ along segments connecting $o$ with all vertices of $P$. In case $o$ is a boundary point of $P$, the triangles form a fan adjacent to, but not surrounding $o$, and if $o$ is an interior point of $P$, they form a rosette (cycle) surrounding $o$. Split your sector $S$ into subsectors $S_1, S_2,\ldots, S_k$ so that the area of $S_i$ equals the area of $T_i$, and so that their neighbororder is the same as the neighbororder of the triangles $T_1, T_2,\ldots, T_k$. This last condition makes sense only in case $o$ is a boundary point of $P$, otherwise the order of $S_1, S_2,\ldots, S_k$ does not matter. In either case, $f$ is obtained by piecing together the functions $f_{T_i}$ mapping $T_i$ to $S_i$ as described at the beginning; in the first case $f$ will be continuous; and in the second one  only piecewise continuous. Also, in the second case $f$ remains undefined on one edge of one of the triangles. By the way, polygon $P$ need not be convex, only starshaped with $o$ as a star center.
It remains to decide whether the construction given in this answer is obvious or not, as the question reads. This, I suppose, is a matter of opinion.