Converting to polar double integrals

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SUMMARY

The discussion focuses on converting a double integral from Cartesian to polar coordinates. The integral in question is defined as the integral from 0 to 2 of the integral from 0 to sqrt(1-(x-1)^2) of ((x+y)/(x^2+y^2)) dy dx. The user attempts to convert this to polar coordinates using the substitutions x = 1 + cos(θ) and y = sin(θ), leading to an integral from 0 to π of (1/r)(cos(θ) + sin(θ)) r dr dθ. The correct limits for θ are established as 0 to π, and the differential area element is confirmed as r dr dθ. The user encounters confusion regarding the origin shift and the evaluation of the integral, ultimately leading to a discrepancy with the expected answer of π/2 + 1.

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  • Understanding of double integrals in calculus
  • Familiarity with polar coordinates and their conversion from Cartesian coordinates
  • Knowledge of integration techniques, including handling limits and area differentials
  • Proficiency in trigonometric identities and their application in integrals
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Students and educators in calculus, particularly those focusing on multivariable calculus and integral transformations. This discussion is beneficial for anyone seeking to deepen their understanding of polar integrals and coordinate transformations.

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 x= 1+ cos(\theta) and y= sin(\theta) then \sqrt{1- (x-1)^2}= \sqrt{1- r^2cos^2(\theta)}= sin(\theta) so your equation becomes y= r sin(\theta)= sin(\theta) and r= 1, 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 \theta= 0 to \pi, not 2\pi. And, in polar coordinates, the "differential of area", dxdy, becomes rdrd\theta.

Your integrand is (x+y)/(x^2+ y^2)=
\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]
 
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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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