Double integral over a region needing polar coordinates.

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Homework Help Overview

The problem involves evaluating a double integral of the function arctan(y/x) by converting to polar coordinates over a specified region defined by inequalities related to x and y. The subject area pertains to multivariable calculus and the use of polar coordinates in integration.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to convert the integral to polar coordinates but expresses uncertainty about the bounds of integration due to the region's definition. Participants discuss the implications of the inequalities on the shape of the region and clarify the effect of the condition 0≤y≤x.

Discussion Status

The discussion is ongoing, with participants providing clarifications about the geometric interpretation of the region in polar coordinates. Some guidance has been offered regarding the bounds of integration, particularly concerning the angle θ.

Contextual Notes

The original poster notes confusion regarding the boundaries of the region defined by the inequalities, particularly how the condition 0≤y≤x interacts with the circular bounds given by 1≤x^2+y^2≤4.

Tropicalism
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1. Evaluate the double integral ∫∫arctan(y/x) dA by converting to polar coordinates over the Region R= { (x,y) | 1≤x^2+y^2≤4 , 0≤y≤x }


My attempt at solving

Converting to polar using x=rcosθ and y=rsinθ I get
∫∫arctan(tan(θ))r drdθ


I understand that I have to integrate first with respect to r and then with respect to θ but I'm not sure what the region is supposed to look like over which I am integrating so I don't know what my bounds should be. I understand that 1≤x^2+y^2≤4 is basically 2 circles, one with radius 1, and one with radius 2 and my region is between them. I don't know how 0≤y≤x affects it though, I thought it would just make it the top right portion of the circle but that's incorrect.
Any help is greatly appreciated, thanks.
 
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Hi Tropicalism! Welcome to PF! :smile:
Tropicalism said:
I don't know how 0≤y≤x affects it though, I thought it would just make it the top right portion of the circle but that's incorrect.

"0≤y and 0≤x" would be top right portion (the first quadrant :wink:)

"0≤y≤x" is the half of it under the diagonal x=y :smile:
 
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tiny-tim said:
Hi Tropicalism! Welcome to PF! :smile:


"0≤y and 0≤x" would be top right portion (the first quadrant :wink:)

"0≤y≤x" is the half of it under the diagonal x=y :smile:
Oh, ok thank you. So using polar that portion would end at θ=∏/4 correct?
 
yup! :biggrin:
 
tiny-tim said:
yup! :biggrin:

Thanks! Thats all I needed.
 

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