Xy coordinates to polar coordinates for double integral. hepl please

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Homework Help Overview

The problem involves converting a specified region in Cartesian coordinates to polar coordinates for the purpose of evaluating a double integral. The region is defined by inequalities involving \(x\) and \(y\), and the integrand is a function of the arctangent of the ratio of \(y\) to \(x\).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conversion of the region to polar coordinates, with one participant providing their limits of integration. There is also a mention of a potential mistake in evaluating the integral, particularly regarding the function being integrated.

Discussion Status

The discussion is ongoing, with participants exploring the correct setup for the polar coordinates and questioning the evaluation of the integral. Some guidance has been offered regarding the function to integrate, but no consensus has been reached on the final evaluation.

Contextual Notes

There is a noted discrepancy in the equations for converting Cartesian coordinates to polar coordinates, which may affect the understanding of the problem. Participants are also reflecting on the limits of integration and the nature of the function being integrated.

Andrew123
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Homework Statement


ok change the region R = { (x,y) | 1 <= X^2 + y^2 <= 4 , 0 <= y <= x } to polar region and perform the double integral over region R of z=arctan(y/x)dA


Homework Equations


r^2 = x^2 + y^2, x = r*sin(@), y = r * cos (@)


The Attempt at a Solution



i got R = { (rcos(@), rsin(@) | 1 <= r <= 2 , 0 <= @ <= pi/4 }

and 3/8 * pi ^2 answer in back of book is 3/64 * pi ^2


thankyou for your time!
 
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Andrew123 said:

Homework Statement


ok change the region R = { (x,y) | 1 <= X^2 + y^2 <= 4 , 0 <= y <= x } to polar region and perform the double integral over region R of z=arctan(y/x)dA


Homework Equations


r^2 = x^2 + y^2, x = r*sin(@), y = r * cos (@)


The Attempt at a Solution



i got R = { (rcos(@), rsin(@) | 1 <= r <= 2 , 0 <= @ <= pi/4 }

and 3/8 * pi ^2 answer in back of book is 3/64 * pi ^2


thankyou for your time!

You've correctly converted to polar coordinates and found the limits of integration, but you somehow made a mistake evaluating the integral...Did you by chance forget that you are integrating the function [itex]\tan^{-1}\left(\frac{y}{x}\right)=\theta[/itex] over this region, andf just find the area of the region instead?:wink:
 
thankyou veeery much!
 
Andrew123;2056564[h2 said:
Homework Equations[/h2]
r^2 = x^2 + y^2, x = r*sin(@), y = r * cos (@)

Not sure this made a difference in your answer, but the equations for x and y above are wrong. They should be
x = r*cos(theta)
y = r*sin(theta)
 

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