How do I explain why $dA/dr = 2 \pi r$ geometrically? There's this question in my calculus book that goes something like this:
The derivative of the area of a circle with respect to its radius is equal to the circle's circumference ($dA/dr = 2 \pi r$). Give a geometric explanation of why this is the case.
To me this is really obvious, but I find it hard to put into words.  If you increased the radius of the circle by putting your finger inside it and pushing the edge outward, then you would have to push it around the whole circumference of the circle.  Heh.  I don't know.
 A: That's not too bad.  Try this one: 
If you want a larger circle with the same centre which has radius $r+\Delta r$ then the circumference of the bigger circle is $2\pi (r+\Delta r)$. The edge is a constant $\Delta r$ from the edge of the original circle all the way round.  
The area between the two circles is somewhere between that of a $\Delta r \times 2\pi r$ rectangle and a $\Delta r \times 2\pi (r+\Delta r)$  rectangle.  So the rate of change of this area as $\Delta r$ changes is between $2\pi r$ and $2\pi (r+\Delta r)$; take the limit as $\Delta r$ tends to $0$ to get the result.
A: It's because the area of a circle is really the sum of infinitely many circles' circumferences, which is the integral:
$$
\int^r_0 \! 2\pi{}t \, dt
$$
See http://betterexplained.com/articles/a-gentle-introduction-to-learning-calculus/ for a nice explanation
A: Draw a circle with area $A$ and radius $r$, then draw around it a larger circle. The distance from your first circle to the second is $dr$, and the difference of the areas is $dA$. As $dr \to 0$, $dA$ goes to the circumference of the interior circle, hence $dA \to 2\pi r\:dr$.
