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[SOLVED] Calculate area with double integral.

MacLaddy

Member
Jan 29, 2012
52
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

Administrator
Staff member
Feb 24, 2012
13,775
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).
 

MacLaddy

Member
Jan 29, 2012
52
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

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

MacLaddy

Member
Jan 29, 2012
52
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
Feb 7, 2012
2,708
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".