Double integral finding the area

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Homework Statement


Use a double integral to find the area of the region bounded by the curve r= 1+sin(theta)?


Homework Equations





The Attempt at a Solution

I can't figure out what theta is intregrated from. I've tried from -(pi)/2 -> +(pi)/2 and that doesnt work. I've also tried from 0-> 2(pi) and that doesnt work. I have no clue what im supposed to integrate theta to. I would really appreciate some help with this.
 

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  • #2
SammyS
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Homework Statement


Use a double integral to find the area of the region bounded by the curve r= 1+sin(theta)?

Homework Equations



The Attempt at a Solution

I can't figure out what theta is integrated from. I've tried from -(pi)/2 -> +(pi)/2 and that doesn't work. I've also tried from 0-> 2(pi) and that doesnt work. I have no clue what I'm supposed to integrate theta to. I would really appreciate some help with this.
Have you done a polar plot?

0 → 2π should give the right answer.

What does your integral look like ?
 
  • #3
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i put a picture up
 

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  • #4
SammyS
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i put a picture up
Your integral looks good:
[itex]\displaystyle \int_0^{2\pi}\int_0^{1+\sin(\theta)}r\,dr\,d\theta[/itex]​

What do you get for tour final answer?
 
  • #5
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I get (3(pi) + 2)/2
 
  • #6
SammyS
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I get (3(pi) + 2)/2
Where does the " + 2 " come from ... as in
(3(π) + 2)/2​
?

What do you get upon integration w.r.t. r ?
 
  • #7
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Oh I see what ive been doing now. When I evaluated theta at 0 I was getting -2. I forgot to multiply this by 1/2 which would have given me 3(pi)(-2+2)/2 or 3(pi)/2 for my final answer. Thank you so much for all of your help I was stuck on that problem for awhile. :smile:
 

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