Find the centroid $C=(\bar{x},\bar{y})$ of the parabolic arc $y=16-x^2$ over $[-4,4]$.
From symmetry, $$\bar{x}=0$$ To find $\bar{y}$, substitute $\tilde{y}=y$, $dL=\sqrt{1+4x^2} dx$ in $$\frac{\int_{-4}^{4} {\tilde{y} dL}}{\int_{-4}^{4} {dL}}=\frac{\int_{-4}^{4} {(16-x^2) \sqrt{1+4x^2} dx}}{\int_{-4}^{4} {\sqrt{1+4x^2} dx}}$$
This is in rectangular coordinates. Is it correct? Also, how would I do this in polar coordinates?
Your setup in rectangular coordinates is correct.
To get the same integral in polar coordinates, let
$$\begin{cases}x(\theta) = r(\theta) \cos\theta \\ y(\theta) = r(\theta) \sin\theta\end{cases}$$
where $r(\theta)$ is the polar representation of the curve, which we can solve for upon replacing all the variables:
$$y = 16-x^2 \\ \implies r\sin\theta = 16 - r^2\cos^2\theta \\ \implies r = \frac{-\sin\theta \pm \sqrt{1+63\cos^2\theta}}{2\cos^2\theta}$$
Choose the solution with the positive square root, which traces out the same parabola when $\theta\in[0,\pi]$,
$$r(\theta) = \frac{-\sin\theta + \sqrt{1+63\cos^2\theta}}{2\cos^2\theta}$$
By the chain rule, $\dfrac{dy}{dx}=\dfrac{dy}{d\theta}\cdot\dfrac{d\theta}{dx}$, so we can rewrite the line element as
$$\begin{align*} dL &= \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \\ &= \sqrt{1 + \left(\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\right)^2} \, \frac{dx}{d\theta} \, d\theta \\ &= \sqrt{\left(\frac{dr}{d\theta}\cos\theta-r(\theta)\sin\theta\right)^2 + \left(\frac{dr}{d\theta}\sin\theta+r(\theta)\cos\theta\right)^2} \, d\theta \end{align*}$$
Plugging in $r(\theta)$ makes quite the mess, so I won't bother simplifying this any further.
The integrals for arc length and each moment are given by
$$\begin{align*} L &= \int_0^\pi dL \\ L_r &= \int_0^\pi \left(16-\left(r(\theta)\cos\theta\right)^2\right) \, dL \\ L_\theta &= \int_0^\pi \theta \, dL \end{align*}$$
so that the arc length is $L\approx33.637$, and the polar centroid of the curve is the point
$$\left(\bar r,\bar\theta\right)=\left(\frac{L_r}L,\frac{L_\theta}L\right)\approx\left(8.273,1.571\right)=\left(\ldots,\dfrac\pi2\right)$$
The $r$-coordinate is the same as the $y$-coordinate of the rectangular centroid so one can determine its value exactly if desired.