Finding the area enclosed by r=3sin theta

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Homework Statement



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

Homework Equations



integral?

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

Thanks!
 
Dick said:
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?
Well, actually, Dick, area in polar coordinates is r dtheta! You didn't say quite what you meant to, did you?

Would you look up the right formula for area?
 
HallsofIvy said:
Well, actually, Dick, area in polar coordinates is r dtheta! You didn't say quite what you meant to, did you?

Are you SURE?
 
The integral of r*dtheta*dr is the area. Not the integral of r*dtheta. I missed it at my first reading as well.
 
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