Question

Find the area of the region that lies inside the first curve and outside the second curve. r = 9 cos θ, r = 4 + cos θ

Answer 1

$9\cos\theta=4+\cos\theta\implies \cos\theta=\dfrac12\implies\theta=\pm\dfrac\pi3$

The area is given by

$\displaystyle\int_{-\pi/3}^{\pi/3}\int_{4+\cos\theta}^{9\cos\theta}r\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac12\int_{-\pi/3}^{\pi/3}\int\bigg((9\cos\theta)^2-(4+\cos\theta)^2\bigg)\,\mathrm d\theta$
$=\displaystyle\int_0^{\pi/3}(80\cos^2\theta-8\cos\theta-16)\,\mathrm d\theta$
$=8\pi+6\sqrt3$