[tex]9\cos\theta=4+\cos\theta\implies \cos\theta=\dfrac12\implies\theta=\pm\dfrac\pi3[/tex]
The area is given by
[tex]\displaystyle\int_{-\pi/3}^{\pi/3}\int_{4+\cos\theta}^{9\cos\theta}r\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\frac12\int_{-\pi/3}^{\pi/3}\int\bigg((9\cos\theta)^2-(4+\cos\theta)^2\bigg)\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_0^{\pi/3}(80\cos^2\theta-8\cos\theta-16)\,\mathrm d\theta[/tex]
[tex]=8\pi+6\sqrt3[/tex]
