Blake, 

   Your paper and presentation on Archimedes' "Quadrature of the Parabola" both contain
   a lot of good work.  It was very good that you worked out and included the modern proofs 
   for the preliminary propositions.  As we discussed, Proposition 3 is perhaps the most
   interesting one there since Archimedes seems to be getting very close to the 
   Cartesian equation y = x^2 for the parabola.  But after that point the paper is almost 
   entirely a more or less exact reproduction of the contents of that piece of ancient 
   mathematics, following the edition and translation published by T.L. Heath.  

   There would have been room for some more original discussion about the ways this work 
   was a precursor of later ideas in integral calculus.  I also think it would have been 
   instructive to actually work out the geometric series involved in the second proof 
   more explicitly.  I was trying to get you to do this with my question
   at your presentation.  But I don't think you really understood what I was getting
   at then.  In modern language, if A is the area of the triangle formed by the endpoints of 
   the parabolic segment and the vertex of the segment, what Archimedes is doing 
   in that proof amounts to summing:

   A + A/4 + A/4^2 + A/4^3 + ... 

   to get the area of the parabolic segment.

   Using the formula for the sum of a geometric series (with ratio r = 1/4 < 1, so 
   the series converges), 

   = A/(1 - 1/4) = 4A/3.

   To be more precise, Archimedes' proof amounts to giving a formula for the difference 
   between a partial sum of this series and the sum of the whole series and he shows
   (in modern language) that that difference goes to zero as you fill in the region between
   the triangle and the parabola with more and more smaller triangles.  

   I'll "cut you some slack," though, since you were working alone on one of the 
   more ambitious projects.   It was clear you really wanted to understand exactly 
   what Archimedes had done and the paper generally shows a good level of detailed 
   explanation about the steps in the various proofs.    

   A very small typographical point:  I don't know if this is a consistent issue, 
   but in looking at your paper in Google Docs, the square symbols marking the ends
   of proofs usually end up on a line after the last line of text in the proof.  It
   would be more usual to put them at the end of the last line of the proof itself
   so that they don't end up taking up so much space.

   Specific comments:

   Page 8: In the proof for Propositions 6 and 7, it would have been good to say
   more about the exact location of the centroid for a triangle and to relate that
   to the fact that appeared on the midterm exam for our course:  The centroid of
   a triangle is at the intersection of the medians of the triangle, and each median
   cuts the others into line segments with lengths in the ratio 1:2.  

   Pages 11 and 12:  I think the way Heath lays out his equivalent of Propositions
   14 and 15 on the page confused you a bit.  The statement you have on page 11 in
   italics is really something like a preliminary discussion.  The actual statements
   that are being proved here are the inequalities (1) and (2) on the next page.  

   Page 14:  Archimedes' proof of Proposition 16 is one of the "method of exhaustion"
   proofs that Brady Parsons discussed in his presentation.  See above for a more
   detailed presentation of the way we would express it using geometric series. 

   Page 23: Archimedes goes even farther to explain the "mechanical" method in
   the "Method of Mechanical Theorems" from George and Matt's presentation.  He
   gives yet another proof of the area formula for the parabolic segment there.
   The third proof is closely related to the first proof you discuss, but it's
   different because he talks about dividing the parabolic segment into lines
   (trapezoids with zero width) and balancing the whole collection of those 
   against a big triangle like the one in Proposition 16.  This is even closer 
   to modern integral calculus of course, where we take limits of Riemann sums 
   to get an integral.  It's possible to think about the limit process yielding 
   the integral as summing up infinitely many rectangles or trapezoids of width 
   zero.

   Final Project: 

   Annotated Bibliography -- 10/10
   
   Presentation -- 33/35  

   Paper -- 50/55

   Total -- 93/100   (letter: A-)