Domain for Polar Coordinate Part 2

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Homework Help Overview

The discussion revolves around the evaluation of a function in polar coordinates, specifically focusing on the domain defined by the inequality x² + y² ≤ R. The original poster attempts to express the function f(x,y) = e^(x²+y²) in polar coordinates and is exploring the implications of the domain on the limits of integration.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the domain in polar coordinates, particularly questioning the non-negativity of the radial coordinate r and its implications for setting limits of integration. The original poster expresses confusion regarding the limits and attempts to apply a substitution method.

Discussion Status

Some participants have provided clarifications regarding the constraints of polar coordinates, specifically that r must be non-negative. The original poster acknowledges this clarification and appears to adjust their understanding accordingly. There is an ongoing exploration of the correct formulation for the integral.

Contextual Notes

There is a noted emphasis on the correct interpretation of the limits for r in the context of polar coordinates, with participants reflecting on the implications of this constraint for their calculations.

DrunkApple
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Homework Statement


f(x,y) = [itex]e^{x^2+y^2}[/itex]
[itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ R

Homework Equations





The Attempt at a Solution


I believe this is a circle.

[itex]f^{2pi}_{0}[/itex][itex]^{sqrt(R)}_{-sqrt(R)}[/itex][itex]e^{x^2+y^2}[/itex]*r*dr*dθ

= [itex]f^{2pi}_{0}[/itex][itex]f^{sqrt(R)}_{-sqrt(R)}[/itex][itex]e^{r^2}[/itex]*r*dr*dθ

after u substitution...

= [itex]\frac{1}{2}[/itex][itex]f^{2pi}_{0}[/itex][itex]f^{R}_{R}[/itex][itex]e^{u}[/itex]*du

Does this makes sense?

Of course not... r domain are same...
 
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Polar coordinates are defined on the domain r>=0 and 0<=theta<2*pi. r is nonnegative. Your lower limit for r should be 0.
 
ohhh so r can never be negative so they are always equal to or bigger than 0?
ok I got it.
Thank u

So the correct one is

= [itex]\frac{1}{2}[/itex][itex]f^{2pi}_{0}[/itex][itex]f^{R}_{0}[/itex][itex]e^{u}[/itex]*du
 
DrunkApple said:
ohhh so r can never be negative so they are always equal to or bigger than 0?
ok I got it.
Thank u

So the correct one is

= [itex]\frac{1}{2}[/itex][itex]f^{2pi}_{0}[/itex][itex]f^{R}_{0}[/itex][itex]e^{u}[/itex]*du

Use \int for a [itex]\int[/itex] in LaTeX.
 

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