# [SOLVED]Calculate area with double integral.

##### Member
Hello all,

I haven't been on here for a while. I'm glad to see that everything is picking up nicely.
Anyway, I have a question that I see the answer to, but I am not understanding the concept.

Find the area of the region bounded by all leaves of the rose $$r=2\cos(3\theta)$$

The thing I am having a hard time grasping is the region of integration wrt $$\theta$$. It appears that it is going from $$0$$ to $$\pi$$, but it seems to me that it should be $$0$$ to $$2\pi$$. However, that isn't correct.

Can anyone explain to me why the entire bottom half of the rose isn't included in the integration?

Thanks much,
Mac

Last edited:

#### MarkFL

Staff member
It only takes $\displaystyle \pi$ radians to make a complete circuit.

I would use symmetry and multiply the integral by 6 and integrate from 0 to $\displaystyle \frac{\pi}{6}$ (over 1/2 a petal).

##### Member
It only takes $\displaystyle \pi$ radians to make a complete circuit.

I would use symmetry and multiply the integral by 6 and integrate from 0 to $\displaystyle \frac{\pi}{6}$ (over 1/2 a petal).
Note that I made a mistake in the original problem. It should be $$r=2\cos(3\theta)$$, not $$r=\cos(3\theta)$$. I have made the correction.

I still don't think I understand. If the region of integration starts at the origin, and heads out r, then proceeds to sweep around the full angle, then it should be 2pi.

There is a concept here I am failing to grasp.

#### MarkFL

Staff member
Your revision doesn't affect anything conceptually. It will quadruple the area though.

Trace your way around the rose. You will find it only takes $\displaystyle \frac{\pi}{6}$ radians to travel from the tip of a petal to the origin, or $\displaystyle \frac{\pi}{3}$ radians to travel all the way around a petal. There are 3 petals, so it only takes $\displaystyle \pi$ radians to travel all the way around the rose 1 time.

##### Member
Your revision doesn't affect anything conceptually. It will quadruple the area though.

Trace your way around the rose. You will find it only takes $\displaystyle \frac{\pi}{6}$ radians to travel from the tip of a petal to the origin, or $\displaystyle \frac{\pi}{3}$ radians to travel all the way around a petal. There are 3 petals, so it only takes $\displaystyle \pi$ radians to travel all the way around the rose 1 time.
Oh, well then. That's simple. For some reason I thought the equation would automatically account for the "vacant" areas, but I guess not.

Thanks, I appreciate the assistance.

Mac

#### Opalg

##### MHB Oldtimer
Staff member
Find the area of the region bounded by all leaves of the rose $$r=2\cos(3\theta)$$

The thing I am having a hard time grasping is the region of integration wrt $$\theta$$. It appears that it is going from $$0$$ to $$\pi$$, but it seems to me that it should be $$0$$ to $$2\pi$$. However, that isn't correct.

Can anyone explain to me why the entire bottom half of the rose isn't included in the integration?
The point is that for values of $\theta$ where $\cos(3\theta)$ is negative, $r$ is also negative. In that case, the corresponding point on the curve will be on the opposite side of the origin to the direction in which $\theta$ is pointing. For example, as $\theta$ goes from $\pi/6$ to $\pi/2$, the value of $r=2\cos(3\theta)$ goes from 0 to $-2$ and then back to 0, and the curve will trace out one of the petals in "the bottom half of the rose".