If we were to locate another intelligent lifeform, how could we then estimate the total number of intelligent lifeforms in the galaxy?

Given the vast size of the Milky Way, it is unlikely that we are the only intelligent lifeform to be found within it. Given that we only have one data point (the Earth), we are forced to use a long series of guesses to estimate the number of other intelligent lifeforms in the galaxy.

Let's say that we have made contact with an intelligent lifeform on another planet. How could we use that second data point to estimate the total number of lifeforms in the galaxy?

My approach is to assume that the Milky Way is a 2D disk, as it is $110 \text{ kly}$ in diameter but only $1 \text{ kly}$ thick. ($1 \text{ kly}$ = $1000 \text{ light-years}$). Also, I assume that the lifeform that we contact is the one that is closest to Earth.

Take the location of the nearest other civilization. Remove the z-coordinate so that it falls onto the 2D disk that I talked about earlier. Calculate the distance $D$ (measured in kly) to that planet from Earth. Since $1$ lifeform falls in the area given by $\pi D^2$, and the galaxy has a total area of $3025 \text{ kly}^2$, then the expected total number of lifeforms is given by

$$N = \frac{3025 \text{ kly}^2}{D^2}$$

First of all, is my solution correct? Second, is there a better way to do this?

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The reasoning sound reasonable but the modeling is a little bit off. It is known that the density of star is higher close to the galactic center. Weig hting the area by the density of stars should give a better estimate. Of course, the best way to get the answer is ask the alien! If human are their first extra-alien contact, it will significantly lower any estimate of existence of other aliens. – achille hui May 5 '13 at 14:00
@achillehui Yes, I'm aware that the stars are not distributed evenly, but I don't know how I could adjust my estimate to take that into account. – PhiNotPi May 5 '13 at 14:08

You are thinking along the right lines, but your logic can be made a lot tighter.
- The Milky Way is not a 2D disk
- The stars are not evenly distributed in the Milky Way.
- There is nothing indicated in the original question that indicates that the 2 life-bearing planets fall in $\pi D^2$ area. Even if we subscribe to the logic that we've looked in all directions evenly and ended up meeting these aliens at the last point, the area thus combed should be $\frac{D^2 (2\pi+3\sqrt3)}{6}$ and not $\pi D^2$ (assumption - the aliens must have been looking too)

I understand why you have made the assumptions counter to facts, but there is no need to do that.

Instead of distances, let's focus on number of stars. We know that the Milky Way has ~300 Billion stars. Let's say we have combed through A stars in search for life and the aliens have combed through B stars.

Then probable life-bearing planets

$$N = \frac{2 * 300 Billion}{AUB}$$

This can be further refined by using number of planets instead of number of stars.

Additionally, more exactitude can be imparted by adding the uncertainty that we have...we can represent N as a range, given that our estimates of number of stars in the Milky Way are a range (a pretty big range).

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@phinotpi If you are satisfied with the answer, don't forget to accept it by clicking on the check/tick so that it turns green :-) – OC2PS May 6 '13 at 15:01
Your equation assumes that the origin of the 2 lifeforms is independent. If we are both DNA based lifeforms then we are possibly derived from the same genesis event in this corner of the galaxy. We already know that there were 2 intelligent species on THIS planet - homo sapiens and homo neanderthalis. – Dale M May 7 '13 at 1:33