Converting to polar double integrals

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Homework Help Overview

The problem involves converting a double integral from Cartesian to polar coordinates. The original integral is defined over a specific region and includes a rational function of the variables x and y.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the validity of shifting the coordinate system and its implications on the integral's values. There is exploration of the conversion process to polar coordinates and the resulting limits of integration.

Discussion Status

Some participants have provided insights on the implications of changing the coordinate system and the correct limits for the polar integral. Others are questioning the effects of this shift on the integrand and the overall setup of the integral.

Contextual Notes

There are concerns about the limits of integration and the behavior of the integrand when the origin is moved. Additionally, there is mention of potential undefined expressions arising during the integration process.

cos(e)
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Homework Statement


we are given the intgral from 0 to 2( the integral from 0 to sqrt(1-(x-1)^2) of ((x+y)/(x^2+y^2))dydx, so convert to polar integral and solve


Homework Equations





The Attempt at a Solution



i got integral from 0 to pi/2(integral 0 to 2cos(theta) of (x+y)/r dr d(theta), and not sure if I am right and don't know how to procede. i let x=1+cos(theta) and y=sin(theta) and r^2=x^2+y^2. sorry bout the ugly format but i don't know how to write it out in maths form.

Thanks in advanced
 
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Strictly speaking, you have shifted the coordinate system so the origin is at the old (1, 0). That perfectly valid but might violate instruction to just change to polar coordinates.

If [itex]x= 1+ cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex] then [itex]\sqrt{1- (x-1)^2}= \sqrt{1- r^2cos^2(\theta)}= sin(\theta)[/itex] so your equation becomes [itex]y= r sin(\theta)= sin(\theta)[/itex] and [itex]r= 1[/itex], a circle about the origin with radius 1 (the graph in the original coordinates is, of course, a circle about (1, 0) with radius 1).

But the fact that y is equal to the positive square root means you are only getting the upper half of the circle. The integral is from [itex]\theta= 0[/itex] to [itex]\pi[/itex], not [itex]2\pi[/itex]. And, in polar coordinates, the "differential of area", dxdy, becomes [itex]rdrd\theta[/itex].

Your integrand is [itex](x+y)/(x^2+ y^2)=[/itex]
[tex]\int_{r=0}^1\int_{\theta= 0}^\pi (r cos(\theta)+ r sin(\theta))/r^2= (1/r)(cos(\theta)+ sin(\theta)) r dr\theta[/itex][/tex]
 
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ok i see, i thought about moving the orgin but i thought that would change my values of (x+y)/(x^2+ y^2) when integrating and screw up my integral? So in general, can i move the origin and the function in the double integral (in this case (x+y)/(x^2+ y^2) ) will always fix itself?
 
okay i did your double integral of sin(theta) + cos(theta) drd(theta) and i got 2 yet the answers in the book have pi/2+1.

i tried letting x=1+rcos(theta) and y=rsin(theta), r^2=x^2+y^2, and i get the double integral of 1/r+cos(theta)+sin(theta), but when i antiderive with respect to r i get log|r|+... and when i sub in the terminals for r(0 and 1) i get log|0| which is undefined.

any help please?
 
Good question Cos(e).

Worked Solution attached as Word Document. Fingers Crossed the file opens.
 

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