How do you find the area of the region bounded by the polar curves

Question

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

Answer 1

In polar coordinates we have:

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

Answer 2

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 $8\pi$.