Find the area bounded by $r=6\sin(2\theta)$ Winplot plot:

I tried this:
$$A = 4 \cdot \frac 1 2 \int_0^{\pi/2} (6\sin(2\theta))^2 d\theta$$
Is that right?
How about
$$A = 8 \cdot \frac 1 2 \int_0^{\pi/4} (6\sin(2\theta))^2 d\theta$$
 A: To calculate the area in polar coordinates we split the figure into tiny triangles of angle $\text{d}\theta$, such that their base is $r\text{d}\theta$ and their height is $r(\theta)$. Using the formula of the area of a triangle $A=\frac{1}{2}b\cdot h$ we have tha the area of the infinitesimal triangles is
$$\text{d}A(\theta)=\frac{1}{2}r(\theta)\cdot r(\theta)\text{d}\theta$$
And now, integrating over $\theta$ we get the area of the full figure
$$A=\int\text{dA}=\frac{1}{2}\int_{\theta_0}^{\theta_f}\left[r(\theta)\right]^2\text{d}\theta$$
In your case, sing the symmetry of the problem (seen from the figure you posted), the result is, for instance, twice the integral in the first and second quadrants, 4 times the integral in the first (or any) quadrant, or 8 times the integral from $0$ to $\pi/4$, namely
$$ A=2\cdot\frac{1}{2}\int_0^{\pi}\text{d}\theta\,36\sin^2(2\theta)=36\cdot\frac{\pi}{2}=18\pi$$
$$ A=4\cdot\frac{1}{2}\int_0^{\pi/2}\text{d}\theta\,36\sin^2(2\theta)=2\cdot 36\cdot\frac{\pi}{4}=18\pi$$
$$ A=8\cdot\frac{1}{2}\int_0^{\pi/4}\text{d}\theta\,36\sin^2(2\theta)=4\cdot 36\cdot\frac{\pi}{8}=18\pi$$
So, indeed, the expressions you are using are correct.
