Question on integral for area in polar coordinates

Question

My book defines this:

$a(S) = \frac 1 2 \sum_k^n (s_k)^2 (\theta_k - \theta_{k-1}) = \frac 1 2 \int_a^b s^2(\theta) d\theta$

where $a(S)$ is area of a sector, $s_k$ is a partition index $s(\theta) = s_k$.

The following picture describes this. My question is why is $s^2$ vs $s$? Visually it seems like the area of a sector is $\frac 1 2 b*h$ of a triangle, where $b,h$ are interchanged with $s,\theta$.

Answer 1

The area of a circle of radius $r$ is $\pi \, r^2$. The area of a sector of a circle with angle $\theta$ is $\left(\frac{\theta}{2\pi}\right) \pi \, r^2$, since the ratio $\theta/2\pi$ is the fraction of the whole circle that the sector takes up.

In each sector, using $s_k$ as the radius, and $\theta_{k+1}-\theta_k$ in place of what I called just $\theta$ above, you get the formula:

$$ \mathrm{area} = \frac{\theta_{k+1}-\theta_k}{2\pi} \cdot \pi s_k^2 = \frac{\theta_{k+1}-\theta_k}{2} \, s_k^2 $$

The $\pi$'s cancel out, which is why they don't appear in the formula.

Answer 2

$$ \begin{gathered} \Delta s = \frac{1}{2}rr'\sin \Delta \theta \hfill \\ \Delta \theta \to 0;r' \to r;\sin \Delta \theta \to d\theta \hfill \\ ds = \frac{1}{2}r^2 d\theta \hfill \\ S=\int_{\alpha}^{\beta}\frac12 r^2(\theta)\,d\theta\hfill\\ \end{gathered} $$