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Domain for Polar Coordinate

  • Thread starter DrunkApple
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Homework Statement


f(x,y) = y([itex]x^{2}[/itex] + [itex]y^{2}[/itex])^-1
y ≥ [itex]\frac{1}{2}[/itex], [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 1


Homework Equations





The Attempt at a Solution


Would you check my domain please?

[itex]f^{pi}_{0}[/itex][itex]f^{sqrt(3)/2}_{-sqrt(3)/2}[/itex] sinθ drdθ
 

Answers and Replies

  • #2
Dick
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Nope, all wrong. Did you draw a sketch of the region? The angle doesn't go from 0 to pi. And your r limits look more like x limits.
 
  • #3
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Yes I did and it looks like the attachment. and the Coordinate was (-[itex]\sqrt{3}[/itex]/2, [itex]\frac{1}{2}[/itex]) to ([itex]\sqrt{3}[/itex]/2, [itex]\frac{1}{2}[/itex]) is it not?
 

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  • #4
Dick
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Yes I did and it looks like the attachment. and the Coordinate was (-[itex]\sqrt{3}[/itex]/2, [itex]\frac{1}{2}[/itex]) to ([itex]\sqrt{3}[/itex]/2, [itex]\frac{1}{2}[/itex]) is it not?
Nice sketch! And like I said the sqrt(3)/2 stuff is the x limits. It's not the starting angle and ending angle. Those would be the angles your endpoints make with the origin, right?
 
  • #5
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for dθ, do I have to use anti tangent(I think I forgot the proper name) of the coordinate to get the angle?

for dr... I am lost on dr cuz... wait... is dr the distance from the origin to the each of the coordinate?

edit - so dr domain is -1 to 1 right?
 
  • #6
Dick
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for dθ, do I have to use anti tangent(I think I forgot the proper name) of the coordinate to get the angle?

for dr... I am lost on dr cuz... wait... is dr the distance from the origin to the each of the coordinate?

edit - so dr domain is -1 to 1 right?
Sure, you can use arctan to find the angles. Now to find the r limits think about a line through the origin at some angle theta. What's the radius value where it hits the circle x^2+y^2=1? That's easy. Now what's the radius value where it hits the line y=2? Time to use y=r*sin(theta).
 
  • #7
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Sure, you can use arctan to find the angles. Now to find the r limits think about a line through the origin at some angle theta. What's the radius value where it hits the circle x^2+y^2=1? That's easy. Now what's the radius value where it hits the line y=2? Time to use y=r*sin(theta).
ok so for dθ, its pi/6 to 5pi/6

for dr, the radius of x^2 + y^2 = 1 is 1 obviously. But where did y = 2 come from? Did you mean y = 1/2? Isn't dr domain based on the distance from the origin? like in the picture?
 

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  • #8
Dick
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ok so for dθ, its pi/6 to 5pi/6

for dr, the radius of x^2 + y^2 = 1 is 1 obviously. But where did y = 2 come from? Did you mean y = 1/2? Isn't dr domain based on the distance from the origin? like in the picture?
Sure. I meant y=(1/2). Sorry. And the r limit (that's not the same as 'dr') is the distance from the origin to boundary curve along the angle theta. So upper limit for r is 1, right.
 
  • #9
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Sure. I meant y=(1/2). Sorry. And the r limit (that's not the same as 'dr') is the distance from the origin to boundary curve along the angle theta. So upper limit for r is 1, right.
and the lower limit is -1 I hope?? :uhh:
 
  • #10
Dick
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and the lower limit is -1 I hope?? :uhh:
Noo. y=(1/2)=r*sin(theta). Solve for r. The lower limit depends on theta, doesn't it? Can't you see that from your sketch? As theta changes the distance from the origin to the point on the line changes.
 
  • #11
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Noo. y=(1/2)=r*sin(theta). Solve for r. The lower limit depends on theta, doesn't it? Can't you see that from your sketch? As theta changes the distance from the origin to the point on the line changes.
No this one I really don't get. So lower limit always depends on theta?
 
  • #12
Dick
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No this one I really don't get. So lower limit always depends on theta?
Not always! As theta changes the distance to the circle x^2+y^2 stays 1. The distance to the line y=(1/2) does. If r is the distance along the angle theta from the origin to y=(1/2), what's r at theta=pi/2? What's r at theta=pi/6?
 
  • #13
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Not always! As theta changes the distance to the circle x^2+y^2 stays 1. The distance to the line y=(1/2) does. If r is the distance along the angle theta from the origin to y=(1/2), what's r at theta=pi/2? What's r at theta=pi/6?
well r at theta = pi/6 is 1
r at theta = pi/2 is 0
if not f*** me
 
  • #14
SammyS
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r at theta = pi/2 is 0
if not f*** me
r=0 is the origin. That's clearly not in the domain.

(No intention to fool you.)
 
  • #15
Dick
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well r at theta = pi/6 is 1
r at theta = pi/2 is 0
if not f*** me
You are really missing some essential part of the picture here. And I can't figure out what it is. If you draw a line from the origin at an angle of pi/2, where does it hit the line y=(1/2)??
 
  • #16
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You are really missing some essential part of the picture here. And I can't figure out what it is. If you draw a line from the origin at an angle of pi/2, where does it hit the line y=(1/2)??
are you asking for this green part?
 

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  • #17
Dick
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are you asking for this green part?
Exactly right. It's not 0, is it? And that's only the lower bound for r at theta=pi/2. Like I said before the lower bound is a function of theta. Can you write an expression for it?
 
  • #18
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No it is not. I believe the distance from the origin to y = 1/2 is 1/2... that's the green part
 
  • #19
Dick
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No it is not. I believe the distance from the origin to y = 1/2 is 1/2... that's the green part
Ok, so the lower bound of r at pi/2 is 1/2 and the lower bound for r at pi/6 is 1. Now what is an expression for it at an arbitrary value of theta?? Once again, y=(1/2)=r*sin(theta).
 
  • #20
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OHOHOHOHOHOHOHO i got it now I fully get it now Now I know
THank you very much
I'll buy you a beer when we meet :D
 

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