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Domain for Polar Coordinate Part 2

  • Thread starter DrunkApple
  • Start date
  • #1
111
0

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...
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 
  • #3
111
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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
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,810
934
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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