Let $E\subset \Bbb{R}^3$ be a measurable set (i.e. $\int_{\Bbb{R}^n}1_{E}$ exists) and let $v(E)\ne 0$. Let $f$ be a linear function $f:\Bbb{R}^3\to \Bbb{R}$, and let $(x_0,y_0,z_0)$ be the center of mass of $f$,which in the question is also presented, as a remainder, as $f(x_0,y_0,z_0)={1\over v(E)}\int_{E}f$. Show that if there is $\theta\in (0,2\pi)$ such that $(x,y,z)\in E \iff (x\cos\theta-y\sin\theta,x\sin \theta+y\cos\theta,z)\in E$, then $x_0=y_0=0$.
I posted a similar question a few days ago, but here I am pretty confused. I really can't seem to understand how the ranges of $x$ and $y$ look in the integral itself. I also don't really see how it matters that these are trigonometric figure and how I can use it(no change of variables seems useful or relevant here.).
Is there any way to determine the bounds for $x$ and $y$ in the integral, say, $\int\int\int_{E}f(x,y,z)xdxdydz$?