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Finding the area enclosed by r=3sin theta

by grog
Tags: area, polar coordinates
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grog
#1
Dec15-08, 01:43 PM
P: 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 [tex]\int3sin\Theta[/tex] from 0 to 2pi, then pulled the 3 out to get

[tex]3\int sin\Theta[/tex] from 0 to 2pi and then

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

that seems too easy. what am I missing?
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Dick
#2
Dec15-08, 01:58 PM
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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?
grog
#3
Dec15-08, 04:00 PM
P: 23
ah. that's what it was. I forgot about the formula for area. : (

Thanks!

HallsofIvy
#4
Dec15-08, 04:55 PM
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Finding the area enclosed by r=3sin theta

Quote Quote by Dick View Post
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?
Dick
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Dec15-08, 05:00 PM
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Quote Quote by HallsofIvy View Post
Well, actually, Dick, area in polar coordinates is r dtheta! You didn't say quite what you meant to, did you?
Are you SURE?
Dick
#6
Dec15-08, 10:15 PM
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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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