How to calculate circumference of circle in $L^p$? I want to find the circumference of a circle as a function of its radius in $\mathbb{R}^2$ with $l_p$ metric. Is the following anywhere near correct for radius $r$?
$$ C_p(r) = 4\int_0^r \sqrt{1 + f_p'(t)^2}dt $$
where $f_p(t) = (r^p - t^p)^\frac{1}{p}$.
I get $$C_1(r) = 4\sqrt{2}r$$ and $$C_2(r) = 2\pi r$$
$C_2$ is obviously correct but shouldn't $C_1(r) = 8r$?
Thanks.
Edit:
So, the correct form is:
$$ C_p(r) = 4r\int_0^1 \left(1 +\left|\frac{d(1-t^p)^{1/p}}{dt}\right|^p\right)^{\frac{1}{p}}dt$$
 A: If $p=1$, then the "circle" is a square with corners $(r,0)$, $(0,r)$, $(-r,0)$, and $(0,-r)$.  To see this, note for example that the line through $(r,0)$ and $(0,r)$ has equation $x+y=r$.  For non-negative $x$ and $y$, this says precisely that the Manhattan distance from the origin to $(x,y)$ is $r$. 
Each side of our square has length $\sqrt{2}r$, so its Euclidean "circumference" is indeed $4\sqrt{2} r$. 
Comment: We calculated the "real" (meaning, Euclidean) circumference of the circle. The Manhattan metric circumference is different. Note that the Manhattan distance between $(r,0)$ and $(0,r)$ is $2r$, so the circumference under the Manhattan metric is $8r$. Your formula
$$ C_p(r) = 4\int_0^r \sqrt{1 + f_p'(t)^2}dt $$
has the built-in assumption that we are calculating the Euclidean perimeter.  (Just recall the derivation of the arclength formula, or note the giveaway square and square root in the formula.)  If Euclidean circumference is not what we want to compute, then in parametric form we need instead to integrate $\left(\left|\frac{dx}{dt}\right|^p + \left|\frac{dy}{dt}\right|^p\right)^{1/p}$.
When $p=1$, the integrand is simply $2$, and we do obtain $8r$.  
A: If you calculate it numerically, it seems that $C_p(r)$ = $C_q(r)$ when $p^{-1}+q^{-1}=1$.
That is a nice connection between dual spaces, which includes $C_1(r)=8r=C_{\infty}(r)$.
Also, when $p$ is large, $C_p(1) = 8-8p^{-1}\ln 2 + O(p^{-2})$
I don't know of a closed form for the answer.
