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This might have been already asked in this site but I can't find it. So here's the integral:

$$\int_{r_\text{min}}^{r_\text{max}} \sqrt{\left(1-\frac{r_\text{min}}{r}\right)\left(\frac{r_\text{max}}{r}-1\right)}~dr=\pi \left(\frac{r_\text{min}+r_{\text{max}}}{2}-\sqrt{r_{\text{min}}r_{\text{max}}}\right)$$

Is there any geometric/"intuitive"/"insightful" proof of this given that the R.H.S. is precisely proportional to the difference of arithmetic mean and geometric means of the integral limits? For example, this appears in aspects of Kepler problem for finding the conserved quantity called "action" corresponding to radial motion (which justifies the choice for the above variables: $r_\text{min}$ and $r_\text{max}$ are the minimum and maximum values of radial coordinates in bounded motion with the potential $V=-\frac{\alpha}{r}$). This makes me wonder more if there is a geometric proof involving ellipses or something like that... But feel free to change the notation if the subscripts seem cumbersome.

P.S. I request everyone to suggest alternative titles so that this post can be found easily by future users. I couldn't think of anything better than the current one (v1).

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    $\begingroup$ So, in an ellipse, there is an alternative characterization of the semi-axes: as opposed to the shortest and longest distances from the center, we have the semi-major axis is the arithmetic mean of the longest and shortest distances from a fixed focus of the ellipse, and the semi-minor axis is the geometric mean of those same distances. I feel like this can be used somehow, but I'm not sure offhand. $\endgroup$
    – PrincessEev
    Commented Aug 4 at 7:53
  • $\begingroup$ Somewhat related: math.stackexchange.com/questions/4244874/… $\endgroup$
    – Hans Lundmark
    Commented Aug 4 at 8:54
  • $\begingroup$ Also related: math.stackexchange.com/questions/3549527/… $\endgroup$
    – Gonçalo
    Commented Aug 4 at 9:56
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    $\begingroup$ This particular answer converts the integral a subtraction of two parts, both of which are actually essentially the same, angular integrals getting $\pi$, but one multiplied by the arithmetic mean and the other by the geometric mean. However, that means it is not actually anything to do with the areas of ellipses. math.stackexchange.com/a/2815344 $\endgroup$
    – naturallyInconsistent
    Commented Aug 5 at 10:25

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