Our goal is to "add up" all of the Iridium deposited on Earth during the major bolide impact 60MYA. We are given a function of the amount of Iridium in ng according to the distance from impact:
And we look to integrate this function over the surface of the Earth.
The main insight is to think of the function in terms of spherical coordinates (, , )—unfortunately mathematicians and physicists make things more difficult as they have opposite conventions for what symbol represents the vertical angle; here it is . Imagine the impact point is the north pole. is constant at 6371, as we are on the surface of the Earth. Distance to the north pole is independent of . So distance from that impact point will be determined only by the vertical angle , starting at 0. Instead of writing
we can write
Something I forgot to write when first documenting my thinking is how ends up as . It comes down to the designation of radians as units of angular measure: Circumference of an entire circle is and angular measure around that same span is assigned the value of . So fractions of either angular measure or arc length are computed as the same fraction of .
The job of adding up all of the Iridium across the surface of the Earth is then a surface integral:
where is the surface of the Earth.
It is a standard fact that . And integrating over the entire Earth means and . So the above integral becomes
Note that none of the integrand depends on . This makes the integration with respect to trivial, . And if we let , then
The integral that is left is not super obvious (I asked the computer to do it before deriving it myself), but it is certainly doable with only a bit of algebra and guessing.
Since and , this integral evaluates to , and
We can then use Sage to compute this number approximately.
Unfortunately, this isn't the number we want. We were hoping for something closer to ng. Certainly I could have made many errors. Most of the computations here, the number crunching and the finding of tricky integrals, I did in two separate ways and got the same answer, but I'm not the most historically accurate data entry clerk. I tried to spell everything out as clearly as possible so mistakes would be easier to find.
The earlier error, where we had a negative value for does not affect the problem much analytically. In general, we should expect a smaller answer than with positive, as the resulting unsigned exponent is larger than one. And in fact that is what we got. Instead of we now get .
Just to check that I haven't misplaced a parenthesis or something similar, let's do the calculation in parts.
First, the constant out in front:
Now the exponent
And now :
Putting it all together