Evaluate the iterated intergal by converting to polar coordinate?

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Discussion Overview

The discussion revolves around evaluating an iterated integral by converting it to polar coordinates. Participants are focused on the mathematical transformation of the integral and the appropriate limits for the variables involved.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the integral \(\int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2} \, dx \, dy\) and suggests converting to polar coordinates.
  • Another participant assumes familiarity with the conversion of \(dx \, dy\) into \(r\) and \(\theta\) and emphasizes the need to convert the limits of integration as well.
  • Limits of integration are proposed as \(0 \leq y \leq 1\), \(0 \leq x \leq 2\), \(0 \leq r \leq 2\), and \(0 \leq \theta \leq \frac{\pi}{2}\), but one participant notes that the upper limit of \(r\) will depend on \(\theta\).
  • There is a repeated emphasis on converting the inequality \(y^2 \leq 2x - x^2\) into polar coordinates.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the upper limit of \(r\) and the correct conversion of the inequality into polar coordinates, indicating ongoing disagreement and uncertainty in the discussion.

Contextual Notes

The discussion includes assumptions about the participants' familiarity with polar coordinates and does not resolve the mathematical steps necessary for the conversion.

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ZuzooVn said:
Help me Evaluate the iterated intergal by converting to polar coordinate:


http://www.ziddu.com/gallery/4894419/Untitled.jpg.html

hmm … that's \int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2} dxdy

ok … I assume you know how to convert dxdy into r and θ

and for the limits, convert y2 ≤ 2x - x2 into r and θ also :wink:
 
tiny-tim said:
hmm … that's \int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2} dxdy

ok … I assume you know how to convert dxdy into r and θ

and for the limits, convert y2 ≤ 2x - x2 into r and θ also :wink:

0≤y≤1
0≤x≤2
0≤r≤2
0≤θ≤ pi/2
:smile:
 
ZuzooVn said:
0≤y≤1
0≤x≤2
0≤r≤2
0≤θ≤ pi/2
:smile:

(have a pi: π :wink:)

No, the upper limit of r will depend on θ.

I repeat … convert y2 ≤ 2x - x2 into r and θ
 

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