Why are these two ways of measuring the length of the groove in a phonograph record different? I once heard about the following problem on a math exam for students in about grade 8 or 9.

A $33 \frac{1}{3}$ rpm record is 12 inches in diameter and a label diameter of 4 inches.  If the groove starts at the edge of the record and ends exactly at the label and plays for 18 minutes, how long is the groove?

The now-retired math teacher I know who came up with this problem figured that students would solve it by first calculating how many times the record would spin in 18 minutes at $33 \frac{1}{3}$ rpm, which is 600 turns, and multiplying that by what the teacher figured the average single turn groove length would be, which he took as being the circumference of a circle 8 inches in diameter (halfway between 4 and 12).
The answer, as computed this way, is exactly $4800\pi$ inches, or approximately 15079.64474 inches.
If, however, you solve this by integrating the length of the derivative from a point on the groove over the length of the grove using polar coordinates, then you get what looks like an entirely different answer of exactly 
$$\frac{1}{600\pi}\left(\begin{array}{c}
1800\,\pi\,\sqrt {3240000\,{\pi}^{2}+1}-600\pi\,\sqrt {360000\,{
\pi}^{2}+1} \\
+\ln  \left( \sqrt {3240000\,{\pi}^{2}+1}+1800\,\pi\right) -\ln  \left( \sqrt {360000\,{\pi}^{2}+1}+600\,\pi \right)
\end{array}\right)$$ inches, which is about $15079.64532$ inches.
Of course, integration is generally not expected of someone in grade 8 or grade 9 mathematics.  Some students may be able to do it, but it's definitely outside of the curriculum for that level.  Myself, I did not learn how to calculate length of an arc described by a function until I was quite well into my post secondary education, so I think it might not be reasonable to expect a student in junior high school to be able to solve it that way, and the teacher did not... he expected them to calculate it the first way (although by his account, none of the students who saw the question knew how to actually solve it using his method either).

If, as I suspect, the teacher's answer is actually wrong (even though it is an excellent approximation in this case), can someone explain why it is wrong, particularly given that the numbers are so close?  And is it the case that it would always be such a good estimate as in this question when calculating spiral length of a different number of turns and size?

 A: One way to look at it is the teacher is computing the total circumference of all the circles.  The difference in radius between two circles is $\frac {6-2}{600}=\frac 4{600}$ inch.  The outer groove is counted at radius $(6-\frac 2{600})\pi$, the second as $(6-\frac 6{600})\pi$ and so on until the last is at radius $(2+\frac 2{600})\pi$.  This gives the $600 \cdot 2\pi\cdot 4=4800\pi$ the teacher calculates. The outer groove, however, starts at a radius of $6$ and ends at a radius of $6-\frac 4{600}$.  A pretty good approximation to its length is as the hypotenuse of a right triangle with legs $(12-\frac 4{600})\pi$ and $\frac 4{600}$ where the first is the perimeter of a circle with the average radius and the second is the radial distance of one groove.  
I think this shows that the correct answer is slightly larger than the teacher's.  A simple approximation is to make an overall triangle with one leg $4800\pi$ and the other $4$.  This would give a length of $\sqrt{(4800\pi)^2+16}\approx15079.6452677$.  It is still a little low because if you unroll the spiral to a "right triangle" the "hypotenuse" is a little curved because the radial spacing is constant and becomes a larger fraction of the circumference.  I think this approach is within range for grades 8 or 9.
