Question

Find the area of the region that lies inside the first curve and outside the second curve. r = 6 − 6 sin θ, r = 6

Answer 1

Each curve completes one loop over the interval $0\le t\le2\pi$. Find the intersections of the curves within this interval.

$6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi$

The region of interest has an area given by the double integral

$\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta$

equivalent to the single integral

$\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta$

which evaluates to $9\pi+72$.