# Finding the area enclosed by r=3sin theta

1. ### grog

23
1. The problem statement, all variables and given/known data

Find the area enclosed inside r=3 sin (theta)

2. Relevant equations

integral?

3. The attempt at a solution

basically, I took $$\int3sin\Theta$$ from 0 to 2pi, then pulled the 3 out to get

$$3\int sin\Theta$$ from 0 to 2pi and then

$$3[-cos(\Theta)]$$ evaluated from 0 to 2pi.

that seems too easy. what am I missing?

Last edited: Dec 15, 2008
2. ### Dick

25,887
Who said everything has to be super hard? But if you work that out, you'll get 0 for the area. Is that right? It might help to draw a picture. And, hey, area in polar coordinates isn't the integral of r dtheta, is it? Would you look up the right formula for area?

Last edited: Dec 15, 2008
3. ### grog

23
ah. that's what it was. I forgot about the formula for area. : (

Thanks!

4. ### HallsofIvy

40,689
Staff Emeritus
Well, actually, Dick, area in polar coordinates is r dtheta! You didn't say quite what you meant to, did you?

5. ### Dick

25,887
Are you SURE?

6. ### Dick

25,887
The integral of r*dtheta*dr is the area. Not the integral of r*dtheta. I missed it at my first reading as well.