What's wrong with my calculation for the area of a circle? This is my calculation for a simple circle of radius 1. Take a pizza slice of the circle. From geometry, the cone has an area of $r^2\pi * 2\pi / \theta$ 
Now to get the area of the circle, $$2\pi^2 \int_0^{2\pi} 1/\theta \, d\theta$$ which yields a nonsensical answer.
 A: You got "cone" and "sector" mixed up. The area of a sector of radius r and central angle $\theta$ (in radians) is not $\frac{r^2\pi^2}\theta$ but $\frac{r^2\theta}2$.
The circle you want to find the area of has the polar equation $r=1$ with $\theta$ from 0 to $2\pi$. The correct integral therefore does not have any $\theta$ in the integrand:
$$A=\frac12\int_0^{2\pi}r^2\ d\theta=\frac12\int_0^{2\pi}1\ d\theta=\frac12[\theta]_0^{2\pi}=\pi$$
A: The area of a single slice is $\pi r^2 \frac{\theta}{2\pi}=\frac{r^2}{2}$, not the reciprocal as you have. 
These are the slices we need to sum up over $\theta$, by computing the integral
$$
1/2r^2\int_{0}^{2\pi}d\theta=\pi r^2
$$
A: The slices (sectors) are (approximated as) triangles of base $r\Delta \theta$ and height $r$.   The area is thus calculated as:
$$\begin{align}A~=~&\int_0^{2\pi} \dfrac{r^2\operatorname d \theta}2 \\[1ex] =~& \left.\dfrac{r^2\theta}{2}\right\rvert_{\theta=0}^{\theta=2\pi}\\[1ex] =~& \pi r^2\end{align}$$
No cones were harmed in the integration of this area; only pizza was sliced into infinitely thin sectors.
A: Using your arc length formula $ l = r\theta $, we can approximate the 'slice of pizza' as a triangle with base $\Delta\theta$ and height of $r$  (we are allowed to do such a thing as we are taking very small slices). 
And so the area of each triangle, we will express as: $$\delta A = \frac{r^2\Delta\theta}{2}$$
Thus $$A = \int_{0}^{2\pi}\frac{r^2}{2} \ d\theta = \pi r^2 $$
I would observe that your method a bit cyclic, as your calculation of the sector involves assuming the area of the circle to begin with! 
