Area of the region inside $r=\cos{\theta}$ but outside of $r=4\cos{3\theta}$.

Question

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

Answer 1

We can use a triple angle formula $\cos(3\theta)= 4\cos^3(\theta)-3\cos(\theta)$ to find the coordinates at which the the two curves intersect in the first and fourth quadrant: $$ \cos(\theta) = 4\cos(3\theta)= 4(4\cos^3(\theta) - 3\cos(\theta))=16\cos^3(\theta) -12\cos(\theta) $$ so $$ 16\cos^3(\theta) - 13\cos(\theta)=0 \implies \cos(\theta)(16\cos^2(\theta)-13) = 0 $$ which is true at either $\theta= \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$ or when $$ \cos^2(\theta) = \frac{13}{16} \implies \boxed{\theta = \cos^{-1}(\sqrt{13}/4)}. $$ As the range of inverse cosine is restricted to $0<\cos^{-1}(\theta)<\frac{\pi}{2}$ This will give you two answers. The only between $0$ and $\frac{\pi}{2}$ is obviously the one you're looking for, along with it's negative evil-twin in the fourth quadrant.

From there you really just need to do: $$ 4\int_{-\cos^{-1}(\sqrt{13}/4)}^{\cos^{-1}(\sqrt{13}/4)}\int_{0}^{\cos(\theta)} r\cos(3\theta) \ d\theta \ dr $$ which will be the area of $r\cos(\theta)$ not highlighted. Subtract that from the area of $r\cos(\theta)$ (very easy to calculate :D ) and you've got it.

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EDIT:

Actually I'm not sure you need to find the point of intersection at all. Since the furthest away things ever get in the r-direction is $\cos(\theta)$ I'm pretty sure you can just do the area of the $\cos(\theta)$ (1), minus this integral: $$ 4\int_{-\pi/2}^{\pi/2} \int_{0}^{\cos(\theta} r\cos(3\theta) \ dr \ d\theta $$

Answer 2

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.