A Date in the Polar Coordinate System (Part 2)

To find the area of the shape, we will use area of polar equation from $\theta = 0$ to $2\pi$

$\begin{aligned} \text{Area} & = \frac{1}{2} \displaystyle \int _{0} ^{2\pi} (k + k \sin \theta)^2 \mathrm{d} \theta\\ & = \frac{k^{2}}{2} \displaystyle \int _{0} ^{2\pi} (1 + \sin \theta)^2 ~~ \mathrm{d} \theta\\ & = \frac{k^{2}}{2} \displaystyle \int _{0} ^{2\pi} 1 + \sin^2 \theta + 2\sin \theta ~~ \mathrm{d} \theta\\ & = \frac{k^{2}}{2} \displaystyle \int _{0} ^{2\pi} 1 + \frac{1 - \cos 2\theta }{2} + 2\sin \theta ~~ \mathrm{d} \theta\\ & = \frac{k^{2}}{2} \left[ \frac{3}{2}\theta + \frac{\sin 2\theta }{4} - 2\cos \theta ~~ \right] _{0} ^{2\pi} \\ & = \frac{k^{2}}{2} \left( 3 \pi + 0 - 0\right) \\ & = \frac{3 \pi k^{2}}{2} \end{aligned}$

Since we know the area is $18 \pi$ , $\frac{3 \pi k^{2}}{2} = 18 \pi$ $k^{2} = 12$

Since it is given that $k$ is negative, we will take the negative square root.

$k = \boxed{ - 2 \sqrt{3} } ~~~ _\square$