# Domain for Polar Coordinate

## Homework Statement

f(x,y) = y($x^{2}$ + $y^{2}$)^-1
y ≥ $\frac{1}{2}$, $x^{2}$ + $y^{2}$ ≤ 1

## The Attempt at a Solution

Would you check my domain please?

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

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
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.

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?

#### Attachments

• 7.6 KB Views: 303
Dick
Homework Helper
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?
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?

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?

Dick
Homework Helper
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).

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?

#### Attachments

• 8.4 KB Views: 270
Dick
Homework Helper
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.

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:

Dick
Homework Helper
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.

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?

Dick
Homework Helper
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?

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
...
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.)

Dick
Homework Helper
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)??

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?

#### Attachments

• 8.5 KB Views: 327
Dick
Homework Helper
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?

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

Dick