Assume that $f$ is smooth function defined in the unit disk $D: x^2+y^2\le 1$, and consider the integral $$I=\int\_D f dxdy=\int\_0^1r \int\_0^{2\pi} f(re^{it})dt.$$


Then it is clear that for $r\in[0,1]$ there is $t\_r\in [0,2\pi]$ so that $$I=2\pi \int\_0^1 r f(re^{it\_r})dr.$$ My question is, can we choose $t\_r$ to depend smoothly on $r$.



A smooth curve and mean value theorem
You can even choose $t\_r$ to be independent of $r$. Indeed,
$$g(t):=\int\_0^1 rf(re^{it})\,dr,\qquad t\in[0,2\pi],$$
is a continuous function satisfying
$$I=\int\_0^{2\pi}g(t)\,dt.$$
Therefore, there exists $t\_0\in[0,2\pi]$ such that
$$I=2\pi\, g(t\_0)=2\pi \int\_0^1 r f(re^{it\_0})\,dr.$$


