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Integration in Polar - polar coordinate

  • Thread starter DrunkApple
  • Start date
  • #1
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Homework Statement


f(x,y) = xy
x ≥ 0, y ≥ 0, [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 4


Homework Equations





The Attempt at a Solution


[itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] rcosθ * rsinθ * r drdθ

I just wanted to check... is this right?? cuz I really don't think it is
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


f(x,y) = xy
x ≥ 0, y ≥ 0, [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 4


Homework Equations





The Attempt at a Solution


[itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] rcosθ * rsinθ * r drdθ

I just wanted to check... is this right?? cuz I really don't think it is
Why are you putting the upper limit for r to be 2*sec(theta)? The outer boundary is just a circle, isn't it?
 
  • #3
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Why are you putting the upper limit for r to be 2*sec(theta)? The outer boundary is just a circle, isn't it?
right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?
OH WAIT
is it just bounded by 0 to 2?

OH OH WAIT I JUST HAD AN EPIC EPIPHANY!!! THIS IS U SUBSTITUTION!!!!
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
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right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?
OH WAIT
is it just bounded by 0 to 2?

OH OH WAIT I JUST HAD AN EPIC EPIPHANY!!! THIS IS U SUBSTITUTION!!!!
Uh, the boundary isn't x=2. It's x^2+y^2=4. u-substitution?? I think you are overcomplicating this.
 
  • #5
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[itex]f^{pi/2}_{0}[/itex][itex]f^{2}_{0}[/itex] rcosθ * rsinθ * r drdθ
So it's not this?
 
  • #6
Dick
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[itex]f^{pi/2}_{0}[/itex][itex]f^{2}_{0}[/itex] rcosθ * rsinθ * r drdθ
So it's not this?
That's it.
 
  • #7
111
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yes so I'll go head and calculate all this cuz there seems to be a problem and I don't know if it's calculation error or book error

[itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] [itex]r^{3}[/itex]cosθ * rsinθ * drdθ

=[itex]f^{pi/2}_{0}[/itex]([itex]r^{4}[/itex]cosθsinθ)/4 * drdθ

=4[itex]f^{pi/2}_{0}[/itex]cosθsinθ dθ

After u substitution...

=4[itex]f^{1}_{0}[/itex]u du

=4?
 
  • #8
Dick
Science Advisor
Homework Helper
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The integral from 0 to 1 of u*du isn't 1.
 
  • #9
111
0
oh oh wow... How could I miss that...
Thank you very much ;D
 

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