Theory

Theory 1

Polar arclength formula

The arclength of the polar graph of $r(\theta )$, for $\theta \in [\alpha ,\beta ]$:

$$L={\int }_{\alpha }^{\beta }\sqrt{r^{\prime }{(\theta )}^2+r{(\theta )}^2}d\theta$$

To derive this formula, convert to Cartesian with parameter $\theta$:

$$r=r(\theta )\gg \gg (x,y)=(r\cos \theta ,r\sin \theta )$$

From here you can apply the familiar arclength formula with $\theta$ in the place of $t$.

Extra - Derivation of polar arclength formula

Let $r=r(\theta )$ and convert to parametric Cartesian, so:

$$\begin{array}{cc}x(\theta )& =r(\theta )\cos \theta \\ y(\theta )& =r(\theta )\sin \theta \end{array}$$

Then:

$$\begin{array}{c}ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}d\theta \\ \\ x^{\prime }={(r\cos \theta )}^{\prime }\gg \gg r^{\prime }\cos \theta -r\sin \theta \\ y^{\prime }={(r\sin \theta )}^{\prime }\gg \gg r^{\prime }\sin \theta +r\cos \theta \end{array}$$

Therefore:

$$\begin{array}{cc}{(x^{\prime })}^2+{(y^{\prime })}^2\gg \gg & \phantom{,+}{r}^{\prime 2}{\cos }^2\theta -2r{r}^{\prime }\cos \theta \sin \theta +r^2{\sin }^2\theta \\ & +r^{\prime 2}{\sin }^2\theta +2r{r}^{\prime }\sin \theta \cos \theta +r^2{\cos }^2\theta \\ & \\ & =r^{\prime 2}+r^2\end{array}$$

Therefore:

$$ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}d\theta \gg \gg \sqrt{r^{\prime 2}+r^2}d\theta$$

Theory 2

Sectorial area from polar curve

$$A={\int }_{\alpha }^{\beta }\frac{1}{2}r{(\theta )}^{2}d\theta$$

The “area under the curve” concept for graphs of functions in Cartesian coordinates translates to a “sectorial area” concept for polar graphs. To compute this area using an integral, we divide the region into Riemann sums of small sector slices.

To obtain a formula for the whole area, we need a formula for the area of each sector slice.

Area of sector slice

Let us verify that the area of a sector slice is ${\displaystyle \frac{1}{2}{r}^2\theta }$.

Take the angle $\theta$ in radians and divide by $2\pi$ to get the fraction of the whole disk.

Then multiply this fraction by $\pi r^2$ (whole disk area) to get the area of the sector slice.

$$\left(\frac{\theta }{2\pi }\right)\left(\pi r^2\right)\gg \gg \frac{1}{2}{r}^2\theta$$

Now use $d\theta$ and $r(\theta )$ for an infinitesimal sector slice, and integrate these to get the total area formula:

$$A={\int }_{\alpha }^{\beta }\frac{1}{2}r{(\theta )}^{2}d\theta$$

---

One easily verifies this formula for a circle.

Let $r(\theta )=R$ be a constant. Then:

$$\text{Area of circle}={\int }_0^{2\pi }\frac{1}{2}{R}^{2}d\theta \gg \gg \frac{1}{2}{R}^2\theta {|}_0^{2\pi }\gg \gg R^2\pi$$

---

The sectorial area between curves:

Sectorial area between polar curves

$$A={\int }_{\alpha }^{\beta }\frac{1}{2}(r_1(\theta {)}^2-r_0(\theta {)}^2)d\theta$$

Subtract after squaring, not before!

This aspect is not similar to the Cartesian version: ${\displaystyle \int f-gdx}$