This question is very difficult for me. To answer this question, I started with graphing the polar equations:
I also shaded the area I am going to find.
Based on the comments below, I will first find the intersection the circle makes with the upper petal. I found it by doing the following:
$2 sin\Theta = 2 sin(2\Theta )$
$2sin\Theta =4sin\Theta cos\Theta $
$\frac{1}{2} = cos\Theta $
$cos^{-1}(\frac{1}{2}) = \Theta $
$\frac{\pi }{3} = \Theta$
Next, I chose r = 2 (hence, $\frac{\pi }{4}$) as my starting point to the point where it intersects. Then, I multiplied it to 4 since there are four halves (not necessarily geometrical halves; I only concerned myself with the assumption that I must start tracing from $\frac{\pi }{4}$ to the point where it intersects, and there are two possible points.
I can now be able to obtain the area of the two cut petals:
$A = 4(\frac{1}{2})\int_{\frac{\pi}{4}}^{\frac{\pi }{3}}(2sin\Theta )^{2}d\Theta $
After that, I computed for the area of the two full petals at the bottom.
$4(\frac{1}{2})\int_{\frac{-\pi }{4}}^{0}(2sin\Theta )^{2}d\Theta $
Please take note that for the bounds I used above, I started from the negative counterpart of $\frac{7\pi }{4}$ to the pole (or origin; r = 0). I found the value of the theta at the pole by equating $2sin2\Theta $ to 0. The answer from that calculation is 0.
Adding both areas will give me 5.06 units squared.
Is this now correct?
Best Answer
Note that the curves intersect at $$\theta =0,\pi/3$$
Using the formula for the area bounded by polar curves we have the area of the two top regions found by $$2\int _0^{\pi /3} (1/2) (r_1^2-r_2^2)d\theta $$ where $$r_1=2\sin \theta, r_2=2\sin 2\theta$$
The area of the two bottom loops are found by $$2\int _0^{\pi/2}(1/2) r^2 d\theta $$ where $$r=2\sin 2\theta$$