# Find the area inside r=3+2sin(theta) and r=2

1. May 7, 2012

### penetrated

1. The problem statement, all variables and given/known data

find the area inside r=3+2sin(theta) and r=2

2. Relevant equations

c above

3. The attempt at a solution
I knew how to do inside the polar curve and outside the circle, well, i guess everyone know how to solve what i knew. but not both inside to me.
also, I am kind of stuck at the range of integral...
∏=7∏/6 and 11∏/6

2. May 7, 2012

### clamtrox

Start out by finding when $2 > 3+2 \sin(\theta)$ and use that to write the integration limits. You should be finding something like

$$A = \int_{C_1} d\theta \int_2^{3+2 \sin(\theta)} dr + \int_{C_2} d\theta \int^2_{3+2 \sin(\theta)} dr$$

3. May 7, 2012

### cjc0117

Okay, so you want the area inside $r=3+2sinθ$ and outside $r=2$. You found the intersections by plugging $r=2$ into $r=3+2sinθ$:

$2=3+2sinθ\rightarrow sinθ=-\frac{1}{2}\rightarrowθ=\frac{7π}{6},\frac{11π}{6}$

But on the interval $(\frac{7π}{6},\frac{11π}{6})$, $r=3+2sinθ$ is less than $r=2$. For instance, $3+2sin(\frac{3π}{2})=1$, which is less than 2.

So you can't use $\frac{7π}{6}$ as your smaller angle and $\frac{11π}{6}$ as your larger angle. Can you think of another way to represent one of these angles so that $\frac{7π}{6}$ is the larger angle?

4. May 7, 2012

### sharks

Personally, i found the OP's username inappropriate and offensive. Reported.