The books use the area of sectors to derive the formula for the area under the polar curves. I couldn't understand how when the number of sectors reach infinity, the area of the region created by each sector would be the same as the area of the region under the polar curve. Wouldn't it be an overestimate? I know its the same concept as when we derived the area under a curve for rectangle coordinates but I cant understand it for some reason.
If the books used the area of a triangle instead of the area of sectors, then I would understand it. I even derived the same formula using the area of triangles instead of the area of sectors. For some reason I cant understand how the area of a sector can be used to drive the formula. In my mind it would always be an overestimate because of the curvature of the sector. If the answer is that the sector acts as a triangle when the number of sectors reaches infinite then why not use the area of triangle to begin with?
Either scheme works. But when it comes to estimating the area with a finite number of triangles or sectors, one method will outperform the other with respect to precision.
Let $\Delta A(\theta)=A(\theta+\Delta\theta)-A(\theta)$ denote the area of whichever shape you choose to approximate the region. With respect to the figures below, $A(\theta)$ is the area "under" the curve as $\theta$ sweeps out counterclockwise from the positive $x$ axis up to points on the curve labeled $r(\theta)$; $A(\theta+\Delta\theta)$ is the same area, plus whatever extra area is covered up to the point that is $r(\theta+\Delta\theta)$ units away from the origin.
Using triangles:
The area of one such triangle (recall the formula for the SAS case) is
$$\Delta A(\theta) = \frac12 r(\theta) r(\theta+\Delta\theta) \sin\Delta\theta$$
Divide both sides $\Delta\theta$ and let it approach $0$:
$$\frac{dA}{d\theta} = \lim_{\Delta\theta\to0} \frac{\Delta A(\theta)}{\Delta\theta} = \frac12 r(\theta) \lim_{\Delta\theta\to0} \left[r(\theta+\Delta\theta) \frac{\sin\Delta\theta}{\Delta\theta}\right] = \frac12 r(\theta)^2$$
Using sectors:
Consider a sector subtended by a central angle $\Delta\theta$. Its area is proportional to the total area of the circle with radius $r(\theta)$, such that
$$\frac{\Delta A(\theta)}{\Delta\theta} = \frac{\pi r(\theta)^2}{2\pi} = \frac12 r(\theta)^2$$
Letting $\Delta\theta\to0$ gives the same result. Either way, you end up with the same integral formula for the total area bounded by $r(\theta)$.