# What is the “Full Cycle” of a Polar Curve?

In order to find the arc length or area etc of a polar curve, you must integrate from $$\theta_1$$ to $$\theta_2$$. However, I'm having trouble finding the values of $$\theta_1$$ and $$\theta_2$$.

I know that they must constitute of one and only one full cycle of the curve. $$r = \sin\theta$$ completes its cycle when $$\theta = 2\pi$$ while $$r = \cos\theta$$ is done by $$\pi$$.

However, I'm confused as to where these numbers come from.

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I was told that I could find the values of $$\theta$$ by drawing the polar curves. In drawing $$r = \cos\theta$$, I found that once $$\theta = \pi$$, I had gotten my full circle. However, when drawing $$r = \sin\theta$$, I have the full circle by the time I get to $$\theta = \pi$$. Going to $$2\pi$$ only traces out the curve once more time. So why then is $$2\pi$$ considered the "full cycle" of sin?

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Also, how do the values of these full cycles change and what determines this change?

For example, I was trying to find the area of the region bounded by $$r= 6\cos8\theta$$ (this is the only information given. I originally tried $$\int_0^\pi {(6\cos(6\theta))^2}d\theta$$ but the right answer is found by taking the integral to $$2\pi$$ instead. But doesn't the cycle of cosine end at $$\pi$$?? When I found the area in $$r=3\cos(5\theta)$$, going to $$\pi$$ worked!

• Your intuition is correct. Who told you that you needed a full $2\pi$ to graph $r=\sin{\theta}$? – John Douma Apr 1 at 20:28
• For finding the area of $r = 8sin(2\theta)$, I had to go to $2\pi$ for it to be correct. However, looking back now, I see that it must have something to do with the 2\theta part. – CodingMee Apr 2 at 0:48
• If you multiply the angle by $2$ then you end up with different numbers of petals of the flower because the radius can go from $0$ to $r$ in an angle of $\frac{\pi}{4}$ instead of $\frac{\pi}{2}$. – John Douma Apr 2 at 1:32

## 1 Answer

The correct way to do it would be to choose only the domain where $$r\ge0$$. That means for $$r=\sin\theta$$ you go from $$0$$ to $$\pi$$, while for $$r=\cos\theta$$ you choose $$-\pi/2\le\theta\le\pi/2$$.

For the other curves, you need to add together all pieces of the integral where $$r>0$$. For $$r=6\cos(8\theta)$$ you integrate $$\theta$$ between $$-\frac{\pi}{16}$$ and $$\frac{\pi}{16}$$, then from $$\frac{3\pi}{16}$$ to $$\frac{5\pi}{16}$$, and $$\frac{7\pi}{16}$$ to $$\frac{9\pi}{16}$$, and so on, until the upper limit is still a number less than $$2\pi$$. This way you have only one curve.

The reason your integration works from $$0$$ to $$\pi$$ and sometimes from $$0$$ to $$2\pi$$ is given by how many times you overlap the contour.

• I did not do the integration by hand. I used a calculator specifically for the probability of my flawed human self making some mistakes. Either way, thank you. I'll see if this works. – CodingMee Apr 1 at 20:33
• I've made one small mistake, but I'm going to fix my answer. You need to add all regions where $r>0$, and I've picked only one. – Andrei Apr 1 at 20:36