A Date in the Polar Coordinate System (Part 3)

The arc length and the area of the figure has to equal. As derived in part 2 , the area of the shape is equal to $\dfrac{3\pi k^2}{2}$ .

To find the arc length, we the formula from $\theta = 0$ to $2\pi$ .

$\begin{aligned} \text{Arc length} & = \displaystyle \int _{\alpha} ^{\beta} \sqrt{r^2 +\left( \frac{\mathrm{d}r}{\mathrm{d}\theta} \right) ^2 } ~~\mathrm{d} \theta\\ & = \displaystyle \int _{0} ^{2 \pi} \sqrt{r^2 +\left( \frac{\mathrm{d}r}{\mathrm{d}\theta} \right) ^2 } ~~\mathrm{d} \theta \\\end{aligned}$

We know that $r = k + k\sin \theta$ , and $\dfrac{\mathrm{d}\theta}{\mathrm{d}\theta} = k \cos \theta$ . We substitute these values in the last step.

$\begin{aligned} \text{Arc length} & = \displaystyle \int _{0} ^{2\pi} \sqrt{(k + k \sin \theta)^2 +(k \cos \theta ) ^2 } ~~\mathrm{d} \theta\\ & = |k| \displaystyle \int _{0} ^{2\pi} \sqrt{(1 + \sin \theta)^2 +(\cos \theta ) ^2 } ~~\mathrm{d} \theta\\ & = |k| \displaystyle \int _{0} ^{2\pi} \sqrt{1 + \sin^2 \theta + 2\sin \theta+\cos^2 \theta} ~~\mathrm{d} \theta\\ & = |k |\displaystyle \int _{0} ^{2\pi} \sqrt{1 + \sin^2 \theta + 2\sin \theta+\cos^2 \theta} ~~\mathrm{d} \theta\\ \end{aligned}$

We use the identity $\sin^2 \theta + \cos^2 \theta = 1$ and take out $\sqrt{2}$ common.

$\begin{aligned}\text{Arc length} & = |k| \displaystyle \int _{0} ^{2\pi} \sqrt{2 + 2\sin \theta} ~~\mathrm{d} \theta\\ & = |k| \sqrt{2} \displaystyle \int _{0} ^{2\pi} \sqrt{1 + \sin \theta} ~~\mathrm{d} \theta\\ \end{aligned}$

To integrate $1 + \sin \theta$ , we first multiply-divide by $1 - \sin \theta$ .

$\begin{aligned}\text{Arc length} & = |k| \sqrt{2} \displaystyle \int _{0} ^{2\pi} \sqrt{\frac{(1 + \sin \theta)(1 - \sin \theta)}{1 - \sin \theta}} ~~\mathrm{d} \theta\\ & = k \sqrt{2} \displaystyle \int _{0} ^{2\pi} \sqrt{\frac{\cos^{2} \theta}{1 - \sin \theta}} ~~\mathrm{d} \theta\\ & = |k| \sqrt{2} \displaystyle \int _{0} ^{2\pi}\frac{|\cos \theta|}{ \sqrt{1 - \sin \theta}} ~~\mathrm{d} \theta\\ & = |k| \sqrt{2} \left( \displaystyle \int _{0} ^{\frac{\pi}{2}}\frac{\cos \theta}{ \sqrt{1 - \sin \theta}} ~~\mathrm{d} \theta + \displaystyle \int _{\frac{\pi}{2}} ^{\frac{3\pi}{2}}\frac{- \cos \theta}{ \sqrt{1 - \sin \theta}} ~~\mathrm{d} \theta + \displaystyle \int _{\frac{3\pi}{2}} ^{2\pi}\frac{\cos \theta}{ \sqrt{1 - \sin \theta}} ~~\mathrm{d} \theta \right) \\ \end{aligned}$

To solve this, we substitute $1 - \sin \theta = t$ , and therefore $\cos \theta ~~ \mathrm{d} \theta = - \mathrm{d} t$ . The limits of integration also change accordingly.

$\begin{aligned} \text{Arc length} & = |k| \sqrt{2} \left( \displaystyle \int _{1} ^{0}\frac{-1}{ \sqrt{t}} ~~\mathrm{d} t + \displaystyle \int _{0} ^{2}\frac{1}{ \sqrt{t}} ~~\mathrm{d} t + \displaystyle \int _{2} ^{1}\frac{-1}{ \sqrt{t}} ~~\mathrm{d} t \right) \\ \end{aligned}$

Integration of $\dfrac{1}{\sqrt{t}}$ is equal to $2 \sqrt{t}$ . After integrating and applying second fundamental theorem of calculus , we get

$\begin{aligned} \text{Arc length} & = |k| \sqrt{2} \left((0 - (- 2)) + (2\sqrt{2} - 0) + ((-2) - (-2 \sqrt{2}) \right) \\ & = |k| \sqrt{2} \left(2 + 2\sqrt{2} - 2 + 2 \sqrt{2} \right)\\ & = |k|\sqrt{2} \times 4 \sqrt{2} \\ & = 8 |k| \\ \end{aligned}$

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We now equate this with the area

$\frac{3\pi k^2}{2} = 8 |k|$

$k^2 =\frac{16 |k| }{3 \pi}$

$|k| = \frac{16 }{3 \pi}$

$k = \pm \frac{16 }{3 \pi}$

However, since it is given that $k$ is negative, we will take the negative value. Therefore

$k = \boxed{\dfrac{- 16}{3 \pi}} ~~~ _\square$