# Domain for Polar Coordinate

1. Nov 13, 2011

### DrunkApple

1. The problem statement, all variables and given/known data
f(x,y) = y($x^{2}$ + $y^{2}$)^-1
y ≥ $\frac{1}{2}$, $x^{2}$ + $y^{2}$ ≤ 1

2. Relevant equations

3. The attempt at a solution
Would you check my domain please?

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

2. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

Yes I did and it looks like the attachment. and the Coordinate was (-$\sqrt{3}$/2, $\frac{1}{2}$) to ($\sqrt{3}$/2, $\frac{1}{2}$) is it not?

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4. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

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. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

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?

#### Attached Files:

• ###### calc 3.png
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8. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

and the lower limit is -1 I hope?? :uhh:

10. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

No this one I really don't get. So lower limit always depends on theta?

12. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

well r at theta = pi/6 is 1
r at theta = pi/2 is 0
if not f*** me

14. Nov 13, 2011

### SammyS

Staff Emeritus
r=0 is the origin. That's clearly not in the domain.

(No intention to fool you.)

15. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

are you asking for this green part?

#### Attached Files:

• ###### calc 3.png
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17. Nov 13, 2011

### Dick

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. Nov 13, 2011

### DrunkApple

No it is not. I believe the distance from the origin to y = 1/2 is 1/2... that's the green part

19. Nov 13, 2011

### Dick

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. Nov 14, 2011

### DrunkApple

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