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

[penetrated]

Homework Statement

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

Homework Equations

c above

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

 

[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

 

[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?

 

[DryRun]

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