I have been crazy finding the area of the region inside $r=\cos{\theta}$ but outside of $r=4\cos{3\theta}$. I can't decide the integral bounds
[Math] Area of the region inside $r=\cos{\theta}$ but outside of $r=4\cos{3\theta}$.
areaintegrationmultivariable-calculuspolar coordinates
Related Solutions
HINT...You need to establish where the curves intersect in order to determine the limits, so solve $3\cos \theta=1+\cos \theta$.
So your limits are $\pm\frac {\pi}{3}$
It can be verified that in $[0,2\pi]$ the inequality $$\rho_1(\theta):=4-3\sin(\theta)\leq \rho_2(\theta):=5\sin(\theta)$$ is verified in $[\theta_1,\theta_2]=[\pi/6,5\pi/6]$.
Thus by using the given here, $$\mbox{Area}=\frac{1}{2}\int_{\theta_1}^{\theta_2}\rho_2^2(\theta) d \theta-\frac{1}{2}\int_{\theta_1}^{\theta_2}\rho_1^2(\theta) d \theta\\ =\frac{1}{2}\int_{\pi/6}^{5\pi/6}\left((25-9)\sin^2(\theta)-16+24\sin(\theta)\right)d \theta. $$ The final result is $14\sqrt{3}-\frac{8\pi}{3}$.
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Best Answer
The above answer is incorrect (it equals approximately $0.53$ which is larger than half the area of the circle, which is impossible).
The area inside a loop of $r=4\cos3\theta$ equals $$ A_1=\int_{-\pi/6}^{\pi/6}\int_0^{4\cos3\theta}r\; drd\theta =\frac{4\pi}{3} $$ The area inside a loop of $r=4\cos3\theta$ and inside the circle $r=\cos\theta$ is given by $$ A_2=A_1-2\int_0^{\cos^{-1}\frac{\sqrt{13}}{4}}\int_{\cos\theta}^{4\cos3\theta}r\; drd\theta = \frac{\sqrt{39}}{16}+\frac{15\cos^{-1}(\frac{\sqrt{13}}{4})}{2} $$ And finally, the wanted area is the area of the circle $r=\cos\theta$ minus $A_2$: $$ A=\frac{\pi}{4}-A_2 =\frac{8\pi}{3}-\frac{\sqrt{39}}{16}+\frac{15\cos^{-1}(\frac{\sqrt{13}}{4})}{2} \approx 0.346 $$
The answer seems reasonable, as it is approximately have the area of the circle.