# Area bounded by polar curves

1. Jul 15, 2005

### itzela

I am having trouble finding the area between 2 polar curves..... I have the procedure down, but the bounds are throwing me off. Any help with understanding how to bound would be great appreciated!

I have attatched one problem that I am having hard time with and the work I have done. I know that I am doing something wrong because I am getting a negative number for an area (which shouldn't be).

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2. Jul 15, 2005

### qbert

At Pi/2 your lower limit changes to 0. (Otherwise
you're integrating from the bottom half of the circle
up to the 1 + cos(theta) curve.)

Last edited: Jul 15, 2005
3. Jul 15, 2005

That was tricky, but I figured out what the problem is. The two graphs don't intersect where they appear to intersect. They do intersect at $\frac {\pi} {3}[/tex], but they don't really intersect at the origin. There [itex]\theta = \pi [/tex] for [itex] r=1+cos(\theta)[/tex], but [itex]\theta= \frac {\pi} {2}[/tex] for [itex] r=3cos(\theta)[/tex]. What you need to do is two seperate integrals with different limits. Find the whole area for [itex] r=1+cos(\theta)[/tex] and then subtract out the area for [itex] r=3cos(\theta)[/tex]. 4. Jul 15, 2005 ### Knavish Find and double the area from [itex]\pi/2$ to $\pi$ for the $1+cos(\theta)$ graph. Then, find the area from $\pi/3$ to $\pi/2$ for $[(1+cos(\theta))^2 - (3cos(\theta))^2]$; double this as well. The total area is the sum of the two.

Last edited: Jul 15, 2005
5. Jul 15, 2005

### itzela

Thanks a bunch guys =)

What I did was:

1. found the area from $\pi$ to $\pi/3$ for the graph of $1+cos(\theta)$ -- doubled it

2. found the area from $\pi/3$ to $\pi/2$ for the graph $3cos(\theta)$ -- doubled it

3. Substracted the value i got in #2 from #1... and that was my answer.

... Does it sound on the right path?

6. Jul 15, 2005

### Knavish

Yup, that works.