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?


You have the same phenomenon in rectangular coordinates:

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.

Same thing in polar coordinates.


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