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According to wikipedia, here are the Cygnus X-1 vital stats:

Mass    14-16[7] M☉
Radius  20–22[8] R☉

A radius of 10 R☉ means a volume of 10^3 = 1000 Sols. Divided by 16 M☉ that means that Cyg X-1 is 60 time less dense than Sol. So how could it be a black hole?

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Cygnus X-1 is a binary system, and the radius you cite (taken presumably from Wikipedia article) is the radius of the binary system not the radius of the black hole. The Wikipedia article is highly misleading in this respect.

At least, I think it's the radius of the binary system. 20-22 R☉ is about 0.1 AU which I think is about the suggested spacing between the two stars.

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Thank you! I see it now mentions in the article that the radius of the event horizon is 26 km. – dotancohen Jul 18 '12 at 16:22
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As a side note (rather than addressing the real question as John did), the density of a black hole measured as $\frac{\text{mass}}{\text{volume inside the event horizon}}$ is not fixed and not required to be high. In fact it drops rapidly as the mass grows. For a Schwarzschild black hole

$$ \begin{array}0 \rho_{BH} & = \frac{M}{V_{EH}} \\ & = \frac{M}{\frac{4}{3}\pi R_{EH}^3} \\ & = \frac{3M}{4\pi \left( \frac{2GM}{c^2}\right)^3 } \\ & = \frac{3 c^6}{32 \pi G^3 M^2} \end{array} $$

Of course, for stellar mass black holes this is huge (on order of $10^{16}\text{ g/cm}^3$ for the sun), but it can be quite "reasonable" for the supermassive black holes at the center of large galaxies.

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Thank you, I did not realise how rapidly the density goes down with mass. A bit later I'll sit on that for a 1 million solar mass object, just to see what values of "reasonable" are! Thanks! – dotancohen Jul 19 '12 at 4:58
@dotancohen You can read it right off of the formula...you get a factor of $10^{12}$ vis-a-vis a 1 solar mass hole (mass squared, right?), so about $10^4$. – dmckee♦ Jul 19 '12 at 5:05
I seem to have a problem with the units. The initial formula gives (m/s)^6 / ( (m/s^2)^3 * g^2 ) which reduces to m^3 / g^2. What am I doing wrong? According to my finding, density cannot decrease exponentially with mass. – dotancohen Jul 19 '12 at 9:15
@dotancohen The units of $G$ are $\text{[length]}^3 \text{[mass]}^{-1} \text{[time]}^2$. That is, it is Newton's gravitational constant not the acceleration of gravity near the Earth's surface. Also, as a matter of vocabulary the final line shows that the density drops quadratically with mass not exponentially which would imply a form like $\rho \propto 1/\exp M$ – dmckee♦ Jul 19 '12 at 15:05
Thank you, I had a feeling that I was using the wrong units for G and you understood exactly why! As for quadratically / exponentially, wouldn't "dropping off quadratically" be 1/x^4 whereas "dropping off exponentially" would be 1/x^2? – dotancohen Jul 19 '12 at 17:25
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