Find the area of the bounded region

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Homework Help Overview

The discussion revolves around finding the area of a region bounded by a polar function, specifically where \(0 \leq \theta \leq \pi\) and \(r = \frac{1}{\sqrt{1+\theta}}\). Participants are exploring the setup of double integrals in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of using double integrals and express concerns about the graph of the function, noting its looping behavior. There are inquiries about the correctness of the integral setup.

Discussion Status

Some participants have provided feedback on the initial setup of the integrals, indicating that the method appears to be on the right track. However, there is still uncertainty regarding the graphical representation and the implications of the function's behavior.

Contextual Notes

Participants mention the requirement to use double integrals and express confusion about the function's graph, which may affect their understanding of the bounded area.

tix24
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Hi guys I am very new here this is my second post. (sorry in advance i don't know how to use the functions of the site fully yet)

i think this is the correct method to follow, some feedback or hints would be great thanks in advance!

1. Homework Statement

Find the area bounded by where 0≤theta≤pie

r=1/√(1+theta)

Homework Equations

The Attempt at a Solution


(∫ dtheta ) (∫rdr)

bounds of integration ∫dtheta from o to pie

bounds of integration ∫rdr from 0 to r=1/√(1+theta)
 
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Good start, now you just need to perform the integrals.
 
Orodruin said:
Good start, now you just need to perform the integrals.
im more worried about the method, we are suppose to use double integrals, but i got confused over the graph of this function. I checked it in wolfram and it was looping in it self, that is why i don't know if the integrals are set up correctly or not.
any tips regarding the set up of the integral it self?
 
You already did. Naturally, if you give the radius as a function of an angle, you will get some sort of loop or spiral unless you bound the argument. In your case it is bounded to be between 0 and π.
 
Orodruin said:
You already did. Naturally, if you give the radius as a function of an angle, you will get some sort of loop or spiral unless you bound the argument. In your case it is bounded to be between 0 and π.
thank you very much, it was very helpful of you
 

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