Let's say that I am trying to calculate the average velocity of a fluid through a pipe and I only know the average velocity of a cross-section through the center, can I find the average velocity in the pipe as a whole if we know it is radially symmetric? How? I have made two attempts and failed both times. But I'll show my work below. Assume the pipe has radius R:
We are given the average velocity in the two dimensional slice, so we know $\bar{v}_{2D} = \frac{1}{R}\int_0^R v(r)\,dr $ **NOTE: we do not know v(r), just what this result is.
If we knew the function for the velocity profile (we don't) then the average velocity in the pipe would just be $ \frac{1}{\pi R^2}\int_0^{2\pi}\int_0^Rv(r)\,rdrd\theta$. At this point I am tempted to use the first relation to say that:
$\bar{v}_{2D} = \frac{1} {R}\int_0^R v(r)\,dr \longrightarrow \int_0^R v(r)\,dr=\bar{v}_{2D} R$ and substitute into the above equation but we can't because it is $rdr$ and not just $dr$.
My second line of reasoning is that this must be simpler than I am making. It has to be fundamentally tied with the relationship between area, $\pi$, and radius. I have a feeling that perhaps the ratio between the area of a right triangle and the area of the cone generated by rotating that triangle around its 90 degree junction in a full circle would be equal to the ratio of this average velocity to the average velocity over the pipe as a whole. Perhaps the ratio of every radially symmetric volume to a radial slice is the same? However this is just a guess and I have no Idea how to really verify it. Also, if this guess is wrong, then I have no idea how to proceed.
EDIT possible answer:
Using Peter's advice below I considered the following thought experiment to show this is not enough information to solve the problem: The ratio of the area of a rectangle to the volume of the cylinder it generates is different than the ratio of the area of right triangle to the volume of the cone it generates. However both are proportional to πr which is kind of cool!So the answer would probably be proportional to $\bar{v}_{2D} \pi r$ but we can't know how...
It is not possible in general. It will suffice to show that there exist two profiles $v(r)$ with the same $v_{2d}$ but different average velocity over the disk. For example
$$ v_1(r)=\frac{r}{R} \\ v_2(r)=1-\frac{r}{R} $$
Satisfy
$$ v_{2d}=\frac{1}{R}\int\limits_0^Rdr \ v_1(r)=\frac{1}{R}\int\limits_0^Rdr \ v_2(r)=\frac{1}{2} $$
However, the average velocity is different
$$ \bar{v}_1=\frac{2\pi}{\pi R^2}\int\limits_0^R dr \ rv_1(r)=\frac{2}{3} \\ \bar{v}_2=\frac{2\pi}{\pi R^2}\int\limits_0^R dr \ rv_2(r)=\frac{1}{3} \\ \bar{v}_1 \neq \bar{v}_2 $$
We conclude that in general the average velocity is not uniquely determined by $v_{2d}$.