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=\frac12\implies\theta=\pm\frac\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`