Why aren't the hottest stars mostly invisible due to radiating mostly in ultra-violet? The hottest stars have surface temperatures in the range of 40,000K.  Wolfram Alpha says that such a star acting as a black body should radiate almost no energy in the visible spectrum.
Why then do such stars appear blue?  I know the composition of elements in a star can affect its color via absorption lines, but I don't see how that would affect actual radiated light.  And it doesn't seem that even these large stars are large or fast enough to red-shift their output by such a large amount.
Why aren't these stars invisible?  As we don't seem to observe it, is there a reason that it is not possible for a star that is hot enough to radiate mostly in ultra-violet?
 A: To calculate the spectral flux from a star, you multiply the Planck function by a factor proportional to its surface area.
The Planck functions for bodies at different temperatures never cross (see below, from http://bdaugherty.tripod.com/gcseAstronomy/science.html). Hence the Planck function, which is essentially the luminosity per unit area, of a hot blackbody is larger at all wavelengths than that of a cooler blackbody, even though the ratio becomes larger at smaller wavelengths.
The surface area of a hot main sequence star is also actually quite a bit larger than that of a cool main sequence star, and thus it means they radiate much more energy at all wavelengths, but especially in the UV.

A: Because having a peak in the UV does not means it stop radiating in the over wavelengthes. See the spectrum of black body emission here : https://en.wikipedia.org/wiki/Black_body
What you see is the visible part, i.e. what your eye can see. The blue side of the visible spectrum is closest to the peak, so more intensity is emitted here than on the red side, thus the perceived color.
Remarks: BTW, blue stars have a very imperfect black body spectrum because of the so many absorption spectral lines.
