# If I know the average value over a radial cross section, can I compute the average value over a disk (assuming radial symmetry)?

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.

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...

• If you do the calculation for 2 different velocity profiles and compare them, I would suspect (without actually doing it) that you would get 2 different answers. This would show the problem cannot be done with the information provided. Perhaps try a profile where the velocity is uniform across the cross section, and another where it is uniform across a small disk and zero elsewhere. Feb 12 at 5:34
• good points. However, if it is radially symmetric, can I not just rotate my cross-section and then have the same result? Which makes me wonder if the average wouldn't just remain the same. If I rotate that cross section and infinite number of times I have filled the disk...however the average shouldn't be the same because you have more and more points taking the values the further out radially you go. I imagine there is something really weird happening in the limits here? Feb 12 at 6:40
• radial symmetry is basically circular symmetry, right? For example a cone...Just confirming whether I got the question right...en.wikipedia.org/wiki/Circular_symmetry Feb 12 at 6:42
• That's correct! Or at least that is how I mean it :) Feb 12 at 6:47
• I think if you could provide a diagram it would be really easy for everyone to follow through. Also, so basically R is the radius of the cross-section through the center of the "pipe" right? And also could you please explain how you came up with that expression for $v_{2D}$ (in the first line of your second paragraph)? Feb 12 at 6:54

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 $$v_{2d}$$.

• Thank you, this is a more robust answer to my own thought experiment. Quick follow up question, I know this reasoning is wrong, but I am not sure why: If I take my radial slice and rotate it an infinite number of times I can completely "fill in" my disk. So then the average velocity of the disk is just $\frac{\infty \bar{v}_{2d}}{\infty} = \bar{v}_{2d}$ Is this reasoning wrong because the infinities cant cancel out because they approach infinity at different rates? Feb 12 at 16:11
• Actually when I calculated both your integrals I got 2/3 for both....Let me double check Feb 12 at 16:42
• You're welcome! Yes, I goofed with integrals (now corrected). I'm not entirely sure what you mean by 'rotating the radial slice', but note that if you did so (as I'm imagining), you wouldn't cover the disk uniformly: areas closer to $r=0$ would be 'denser' than those further away. And yes: infinities don't necessarily cancel out and it does depend on the rate of approach.
– Sal
Feb 12 at 16:53
• sweet thanks :) I feel like this helped so any nuanced rotational things click in my head Feb 12 at 17:19