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In the proof of Theorem 5.5 on the below image, where does the n/6A term come from?

(click to see full size version)

enter image description here

The whole books might be download here: http://www.scribd.com/doc/95825150/Advanced-Calculus-of-Several-Variables

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    How did you beautify the picture, @Zev?2012-12-28
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    [I googled the first sentence](http://books.google.com/books?id=sZIFcJ8DJAIC&pg=PA252&lpg=PA252&dq=the+intuitive+content+of+theorem+5.4+multiple+integrals&source=bl&ots=OgIR0fxjVk&sig=TrRekYsihwN0r7VCpBGAy5abDVE&hl=en&sa=X&ei=5cbdUOOpJM6o0AHmnoCQDw&ved=0CDIQ6AEwAA#v=onepage&q=the%20intuitive%20content%20of%20theorem%205.4%20multiple%20integrals&f=false) to figure out what book this was, then ... ahem ... *scanned it myself* and uploaded the picture.2012-12-28
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    @ZevChonoles - Thank you so much!2012-12-28
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    Nice @Zev, although I am curious what "... ahem .." is. Victor, it seems like your question might be answered by reading Theorem 3.4 as pointed out in the proof.2012-12-28
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    @Tim The number 6 never appeared in Theorem 3.4 and its proof!2012-12-28
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    @Tim: It indicates I am clearing my throat to emphasize *just how much* I didn't in any way *get* this scan somewhere else, such as certain Russian book-sharing sites...2012-12-28
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    @Zev: I know that trick too. I was expecting that you had used some computer skills to improve the original scanned picture by Victor.2012-12-28
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    I'm guessing that somewhere a "$\forall \epsilon \gt 0$" is involved, and calling it $\eta / 6A$ is so it will come out nice later.2012-12-28

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Expanding on my comment, this is what I think it means. It would be useful to have Theorem 3.4 to be able to be completely sure, but I think the idea is that for any $\epsilon \gt 0$, $\exists \delta$ such that given any partition $P$ with $\text{mesh}\ P \lt \delta$, the difference between the Riemann sum associated with $P$ and the integral is less than $\epsilon$. Here, we take $\epsilon$ to be $\frac{\eta}{12A}$, and we know that $\exists \delta_2$ satisfying the above condition. Why $\frac{\eta}{12A}$? Without seeing the next page of the book, I can only guess that eventually that $12A$ will go away and we'll be left with $\eta$ alone. If you had started with $\eta$ instead of $\frac{\eta}{12A}$, we would end up with an extra $12A$ around.