Double integrals using polar co-ordinates

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Homework Help Overview

The discussion revolves around evaluating a double integral using polar coordinates, specifically for the function ∫∫e^(-x^2-y^2) dA over the region R defined by the circle x^2+y^2=1. Participants are exploring the conversion of Cartesian coordinates to polar coordinates and the implications for the limits of integration.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the transformation of the region defined by the circle into polar coordinates, questioning the appropriate limits for r. There is a focus on understanding why r is constrained to nonnegative values and how this affects the integration limits.

Discussion Status

Some participants have provided insights regarding the choice of r as a nonnegative number, clarifying that the limits 0≤r≤1 cover the unit disk without redundancy. There is an ongoing exploration of similar scenarios, such as changing the radius of the circle, which prompts further questions about the limits of r.

Contextual Notes

Participants are considering the implications of the chosen limits for r in the context of polar coordinates and the nature of the region being integrated over. The discussion reflects on how different representations of the same geometric region can lead to varying interpretations of the limits.

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Homework Statement


∫∫e-(x^2+y^2) dA
R

Where R is the region enclosed by the circle x2+y2=1

First thing I did was graph the region where the function was enclosed. I saw that they didnt give a limitation to where the circle lied. So I automatically knew that d(theta) would lie on the interval 0≤theta≤2π.

Here is the part where things get a bit tricky for me, I need to find what dr is

So my thought process automatically told me to change x2+y2=1

Into polar coordinates.

Using the following relations x=rcos(theta); y=rsin(theta) and plugging in I get the relation

r^2=1

So I would think dr would lie on the interval from -1≤r≤1

However what the book says is that r lies on 0≤r≤1.

This I do not understand.

Thanks

Higgenz

Homework Equations





The Attempt at a Solution

 
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Mdhiggenz said:

Homework Statement


∫∫e-(x^2+y^2) dA
R

Where R is the region enclosed by the circle x2+y2=1

First thing I did was graph the region where the function was enclosed. I saw that they didnt give a limitation to where the circle lied. So I automatically knew that d(theta) would lie on the interval 0≤theta≤2π.

Here is the part where things get a bit tricky for me, I need to find what dr is

So my thought process automatically told me to change x2+y2=1

Into polar coordinates.

Using the following relations x=rcos(theta); y=rsin(theta) and plugging in I get the relation

r^2=1

So I would think dr would lie on the interval from -1≤r≤1

However what the book says is that r lies on 0≤r≤1.

This I do not understand.

Thanks

Higgenz

Homework Equations





The Attempt at a Solution


r generally chosen to be a nonnegative number. Then 0<=r<=1 and 0<=theta<2*pi covers the unit disk once. If you let -1<=r<=1 then you have covered the unit disk twice. Changing theta->theta+pi and r->(-r) wouldn't change x and y.
 
Wow thanks dick. However, what if for example we had a the same circle

X^2+y^2=4

Would then go from 0≤r≤2 ?
 
Mdhiggenz said:
Wow thanks dick. However, what if for example we had a the same circle

X^2+y^2=4

Would then go from 0≤r≤2 ?

Sure. Same thing.
 
Awesome thank you
 

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