This is actually my favorite example of why pi is wrong—it's the "exception" that proves the rule. To see why, set τ = C/r = 2 pi, and then consider the following chart of common quadratic forms:

  integral of u       1/2   u^2
  kinetic energy      1/2 m v^2
  distance fallen     1/2 g t^2
  spring energy       1/2 k x^2
  triangular area     1/2   b h
  circular area       1/2 τ r^2

N.B. I made a previous comment along these lines, but put it in the wrong place. If you're still a "pi is wrong" skeptic, considering the improvement in radian angle measure may yet convince you:

The explanations mapping complex exponentiation to rotations are basically right. In this context, it's worth noting that the conventional choice of circle constant is off by two. We should be using tau = τ = C/r as the circle constant, rather than pi = π = C/D. Read τ as "turn" and all those radian angle measures suddenly make sense. Ninety degrees? Instead of the confusing π/2 we have 90° = τ/4 = one quarter turn. And so on: 60° = τ/6 = one sixth of a turn, 180° = τ/2 = one half turn, etc.

In these terms, Euler's formula would be recast as

e^(i τ) = 1

That is, the exponential of the imaginary unit i times the circle constant τ is unity: one full rotation.

I've thought pi is horrible... and adopted using a loopy pi as 2pi. I loop the first vertical line over the horizontal, right, and then make it the horizontal, and so on. If that makes any sense.

I'm thinking of making a LaTeX package containing it... (and also some other personal conventions)

Also, to strengthen your point, 2pi makes the radian system a lot more natural.

N.B. I have a secret master plan to spread the use of τ, but this comment is too small to contain it. ;-)

So you're getting the area by integrating a linear thing.

Exactly. In symbols, this reads dA/dr = C ∝ r. Setting τ = C/r = 6.2831853…, we have dA = C dr = τ r dr => A = ½ τ r². Q.E.D.

dA = C dr = τ r dr => A = ½ τ r².

It was programming a lot that brought me to realize this. I got in the habit of always questioning the primitives I'm given, in the hope of simplifying the definitions built on them. It was a little boggling to notice that pi, something I'd grown up with, deserved the same kind of questioning.

And yes, every time you've done math with radians, that problem has haunted you. That's because we decided to use radians instead of "diameter-ans", to keep with the above definition.