# Find the area outside $r = 2\sin\theta$ and inside $r = 2 \sin(2\theta )$

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?

• Is there an inside missing in the title after and? Commented Jan 23, 2020 at 16:17
• Yes. I edited the question accordingly. Commented Jan 23, 2020 at 16:18
• There are 2 types of shapes involved. The petal and the cut petal. To deal with the cut petal, find the intersection region between the petal and the circle [requires piecewise] Commented Jan 23, 2020 at 16:18
• Is it correct to assume that for the circle, the radius is equal to 1 and I should use the well-known formula pi*r^2? In this case, it will be pi units squared. Commented Jan 23, 2020 at 16:19
• Why do you care about the whole circle? The area that you want to calculate is the area of two full petals plus the area of two cut petals. So concentrate on calculating the area of a full petal (hint: what is the range of $\theta$ that will give you just one petal?) and then calculate the area of a cut petal (hint: at what values of $\theta$ does the circle intersect the petal in the first quadrant? Then you can calculate the area of a cut petal as the difference of two areas - do you see how?). Divide and conquer! Commented Jan 23, 2020 at 16:36

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 $$r=2\sin 2\theta$$