# Why is the integral of the arc length in polar form not similar to the length of the arc of a circular sector?

So I learned that the area enclosed by a polar function is computed by $$A = \int \frac{r(\theta)^2}{2}d\theta.$$ Which, I learned, comes somewhat from the formula for the area of a circular sector $$A_{sector}= \frac{r^2\theta}{2}.$$

So I expected the integral for the arc length to be $$S=\int r(\theta)d\theta$$ which is similar to the length of the arc of a circular sector $$S_{sector} = r\theta.$$

But then I learned it is actually $$S = \int \sqrt {r(\theta)^2+\left(\frac {dr(\theta)}{d\theta}\right)^2}d\theta.$$

I was confused why this was the case, and I did some searching and I found that the change in $r$ should be taken into account and that $S=\int r(\theta)d\theta$ only works if r is the radius of curvature. So my question is why can can the area computation use the similarity with the circular sector and the arc length computation can't?

The area under a curve $y=f(x)$ $\> (a\leq x\leq b)$ is given by the integral $$\int_a^b f(x)\ dx\ ,$$ which "comes somewhat" from the formula for the area of a rectangle $$A_{\rm rectangle}= {\rm height}\cdot{\rm width}\ .$$ So one could expect that the integral for the arc length would be $$L=\int_a^b dx\ ,$$ which is similar to the arc length of the top edge of the rectangle: $$L_{\rm top\ edge}={\rm width}\ .$$ But we all know that the correct formula for the arc length is $$\int_a^b\sqrt{1+f'^2(x)}\ dx\ ;$$ the reason being that the projection of a line element $\Delta s$ onto the $x$-axis is shorter than $\Delta s$ by a factor of $\cos\phi$, and this factor does not go away by making $\Delta s$ shorter.