Where is the seem in my proof? I tried to derive the area of a circle using proper calculus, but when I did so, I messed something up and got $A = 2 \pi r^2$, which is obviously wrong, but I can't figure out why this derivation fails.
Suppose there exists a circle with radius $r$.  The area of this circle can be represented as the sum of its sectors, where, when the circle is divided into $n$ sectors, each with area $A_m$, $A_{total} = \sum_{i=1}^n A_i$.
The area of a sector may be approximated by a rectangle whose sides are the arc length of the sector and the radius, meaning $A_{\theta} \approx \theta r^2$.  When $\theta = 0$, this approximate becomes accurate as a line segment may be thought of as a rectangle, in this case with dimensions $0 \times r$.  Therefore:
$$A_{total} = \lim_{||\theta|| \to 0} \sum_{i = 1}^{n} \theta r^2$$
$$= \int_0^{2 \pi}r^2d \theta $$
$$= \theta r^2 \bigg|_0^{2 \pi}=2 \pi r^2$$
What happened?  Why doesn't this derivation work?
Edit: I understand the approximating the area of a sector as a triangle with $base = \theta r$ and height $r$ is better, but why does the rectangle based derivation I did fail?
 A: Your are trying to estimate the area of the circle by adding up the areas of the rectangles.  

But the area of the circle is not the sum of the rectangles.
It just isn't.  It's not that it is better to do it with triangles.  It's that it is possible to do it with triangles.  It is not possible to do it with rectangles.  
The rectangles don't add up to the circle.
A: Simple: the initial formula is wrong. The area of a small sector can be approximated by the area of a triangle, with vertex  at the centre of the circle. The formula for the area of a triangle is half the area  of your rectangle, so there remains no factor $2$.
A: Your basic misconception is to try to approximate a line segment (which is one-dimensional) with a rectangle (which is two-dimensional):

When $\theta=0$, this approximate becomes accurate as a line segment may be thought of as a rectangle.

A line segment cannot be thought of as a rectangle here. Instead, you would have to look at the limits of the areas of the sector and the rectangle as the angle approaches zero and show that these limits are equal; as others have pointed out, they are not. (The ratio of the areas approaches $2$, not $1$.)
If you need further convincing, then using your logic, we could say that the line segment could be represented by a cuboid of arbitrarily high dimension, or by an ellipse, or ..., each of which would lead to a different result.
