Find the area inside the lemniscate $r^2 = 8 cos 2\theta$ and outside the circle $r = 2$.
Emily Wong 
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I think the formula is
$$A = \frac 1 2 \int_{\alpha}^{\beta} (\text{outer})^2 - (\text{inner})^2 d\theta$$
where $\alpha, \beta$ are where they intersect in $[0, 2\pi]$.
This is what I got based on that
By symmetry, we have
$$\frac A 4 = \frac 1 2 \int_{0}^{\pi/6} (8 cos 2\theta) - (2)^2 d\theta$$
Is that right?
$\endgroup$ 41 Answer
$\begingroup$Which one is in your book? $$\frac A4=\int_0^{\frac{\pi}6}\int_2^{\sqrt{8\cos(2\theta)}}rdrd\theta=\int_0^{\frac{\pi}6}\frac12(8\cos(2\theta)-2^2)d\theta=2\sin(\frac{\pi}3)-\frac{\pi}3=0.68$$
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