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How do you find the area of the region bounded by the polar curves $r=4\sin(2\theta)$ for $0\le\theta \le 2\pi$?

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In polar coordinates we have:

$$\mathcal A={1\over2}\int_0^{2\pi}r^2d\theta={1\over2}\int_0^{2\pi}16\sin^2(2\theta) d\theta=8\int_0^{2\pi}{1-\cos(4\theta)\over2}=4\left[\theta-{\sin(4\theta)\over4}\right]_0^{2\pi}=8\pi$$

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  • $\begingroup$ Thank you! I am taking Cal_2 and this solution is perfect! $\endgroup$ – Liana Emelyanova Dec 11 '16 at 4:35
  • $\begingroup$ @LianaEmelyanova Not at all!! Good luck;-) $\endgroup$ – MattG88 Dec 11 '16 at 4:36
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Let's find the area in the first quadrant:

We have $0 \leq \theta \leq \frac{\pi}{2}$, and $0 \leq r \leq 4\sin(2\theta)$.

So our area is going to simply be $$\int_0^{\frac{\pi}{2}} \int_0^{4\sin(2\theta)} r dr d\theta$$

which evaluates to $2\pi$. Since the region is identical is in each quadrant, we multiply by $4$ to get a final answer of $8\pi$.

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  • $\begingroup$ we did not study double integrals,yet. It is cal_3. But, thank you very much! I ll keep it for my reference! $\endgroup$ – Liana Emelyanova Dec 11 '16 at 4:33

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