Finding area bound by polar graph

Question

Find the area on the inside of the curve $ r^2 = 4\cos(14\theta)$, but on the outside of the unit circle $r = 1.$

I tried finding the area of the part of the leaves which is outside the unit circle, by solving the following integral:

$$ \frac12 14\int_\frac{-π}{42}^\frac{π}{42} 4\cos(14\theta)\,\text{d}\theta $$

but I obtained the wrong answer.

Answer 1

Let $\alpha=\arccos(1/4)/14$ so that $4\cos(14\theta)\geq 1$ for $-\alpha\leq\theta\leq\alpha$. To the integral $$\frac 12\int_{-\alpha}^\alpha r^2\,\mathrm d\theta$$ you have to subtract the area of circular sector $ACB$ which equals $\alpha$. Thus the required area is given by \begin{align} 14\left(\frac 12\int_{-\alpha}^\alpha r^2\,\mathrm d\theta-\alpha\right) &=14\left(\frac{\sqrt{15}}{14}-\alpha\right)\\ &=\sqrt{15}-14\alpha\\ &=\sqrt{15}-\arccos(1/4) \end{align}