Computing Line Integrals Related to Vector Field F in R2

[cereeal]

We are given a vector field:
F=$$\frac{-y}{x^2+y^2}$$ , $$\frac{x}{x^2+y^2}$$
Then asked if F is conservative on R² \ (0,0). I just solved the partial derivatives of each part of the vector field and they did indeed equal each other, but I don't under stand what the "\(0,0)" part means.

We are then asked to compute the line integral where C is an arbitrary closed smooth contour that encloses the origin. I used Green's Theorom to solve this and got 2pi.
We are then asked to compute the line integral where C is an arbitrary closed smooth contour that does not enclose the origin. I have no idea where to start on this one.

We are then given the function arctan(y/x) and asked if it is a potential of F. I did this easy enough, but then we are asked if that function is a potential of F on each quadrant (not including the coordinate axes). I didn't quite understand how it wouldn't be, perhaps because the function doesn't exist when x=0?

Lastly, and sorry about all the questions, let C be a smooth curve in R² that does not passes through the origin. Find all possible values of the line integral.

Thanks for any help!

 

[Dick]

F is undefined at (0,0). R/(0,0) is the plane without the origin. And you can't use Greens theorem directly for a curve that encloses the origin. What did you do? You CAN use it if the curve doesn't enclose the origin. What are you double integrating in that case? And yes, the problem with the potential arctan(y/x) is that it isn't defined at x=0. So you can only use it as a potential for regions that don't cross the y axis. Finally for the last part, what do you get for the line integral if C is a curve that goes around the origin twice counterclockwise? What about clockwise?

 

[cereeal]

For the Green's Theorem I made an arbitrary counterclockwise circle (C') within the enclosed region (D). This gave me:
$$\int _{C} P dx + Q dy + \int_{-C'} P dx + Q dy = \int \int_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA = 0$$

Therefor the two integrals on the left side of the equation are equal. You can then fill in the parametrization values and end up with 2pi.

 

[Dick]

For the Green's Theorem I made an arbitrary counterclockwise circle (C') within the enclosed region (D). This gave me:
$$\int _{C} P dx + Q dy + \int_{-C'} P dx + Q dy = \int \int_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA = 0$$

Therefor the two integrals on the right side of the equation are equal. You can then fill in the parametrization values and end up 2pi.

That's fine. If C doesn't enclose the origin, then it's even easier. You don't need the inner circle.

 

[cereeal]

Thanks, couldn't get those integrals working for some reason.
And that's true, I don't know why I didn't think of using the same application on the closed shape not at the origin, I was stuck on using the inner circle.

So for my last question would I still just be applying Green's theorem to it?

 

[Dick]

Thanks, couldn't get those integrals working for some reason.
And that's true, I don't know why I didn't think of using the same application on the closed shape not at the origin, I was stuck on using the inner circle.

So for my last question would I still just be applying Green's theorem to it?

Sure, if it doesn't enclose the origin just use Green's theorem directly. For curves that wrap, say twice counterclockwise around the origin, just work on the special case of a circle and do the path integral directly. To wrap twice around the origin just change your parameter limits to 0 to 4pi instead of 0 to 2pi.

 

[modtor]

Just one question: When C doesn't enclose the origin, are you supposed to find a constant value like the 2pi you get when C was enclosing the origin?

 

[Dick]

Just one question: When C doesn't enclose the origin, are you supposed to find a constant value like the 2pi you get when C was enclosing the origin?

Yes. You get a constant. What is it??

 

[modtor]

It is alright the answer was simple... my mistake

 

[modtor]

Explanation: Suppose that (x0,y0) and (x1,y1) are 2 points in C. We connect the 2 points by 2 different paths C1 and C2
Let C=C2-C1
Then the very left side of your integral is equal to 0
$$\int _{C} P dx + Q dy = 0$$

 

[modtor]

Another question:
Let Ω= R^2\{(0,y);y≥0}. Find a closed form formula for the potential of F on Ω
I don't know how to start the calculation

 

[itsagoal89]

yaaaa peter SHI!

 

[modtor]

MTH 254... Oh Yeah
For the formula for the potential of F, I am stuck. I get something like:
[tex] f_{x} = \frac{-y}{x^2 + y^2}[\tex]
[tex] f_{y} = \frac{x}{x^2 + y^2}[\tex]
We have to take care of the restrictions {(0,y);y≥0}
Do you have an idea how to go farther?

 

[Dick]

MTH 254... Oh Yeah
For the formula for the potential of F, I am stuck. I get something like:
[tex] f_{x} = \frac{-y}{x^2 + y^2}[\tex]
[tex] f_{y} = \frac{x}{x^2 + y^2}[\tex]
We have to take care of the restrictions {(0,y);y≥0}
Do you have an idea how to go farther?

So you characters are all in the same class? Then put your heads together. If you've figured out arctan(y/x) is a potential function, then you should realize that's an angle. Think about it for a bit.

 

[itsagoal89]

modtor, for g) when you integrate it without any parameters you get just t as the indefinite answer,, does that mean all possible values are between 0 and 2pi or if not, than what?

 

[modtor]

Ok for g), there is 2 possibilities: as long as the smooth curve does not pass through the origin, either C encloses the origin or C doesn't enclose the origin. So you have to use your answer from question c) and d) to answer to this one...
Does it look good?

 

[cereeal]

What parametrization did you use for C? I tried thinking about what Dick said a few posts ago and it seems the line integral increases the more turns you go around a path so wouldn't that mean all values are possible?

And for the last question, because arctan(y/x) isn't defined on the y-axis and we know it is a potential of F, wouldn't that be the formula for the potential. I'm not sure I exactly am getting your hint Dick, but again thank you for the help.

 

[modtor]

For which question do you want to know my parametrization?

 

[itsagoal89]

ok yea, for c its 2pi, and d its zero, so g) i put 0=<t=<2pi.



or wait, your saying, that since it doesn't pass through, it either encloses, or it doesnt,. so there are 2 possible answers, 0 and 2pi, 2pi if it is enclosed, 0 if it is not enclosed?

 

[modtor]

Exactly

 

[modtor]

Only 2 answers 0 or 2pi

 

[cereeal]

Nevermind, I agree with that modtor. I'm not getting any progress on (h) though.

 

[Dick]

What parametrization did you use for C? I tried thinking about what Dick said a few posts ago and it seems the line integral increases the more turns you go around a path so wouldn't that mean all values are possible?

And for the last question, because arctan(y/x) isn't defined on the y-axis and we know it is a potential of F, wouldn't that be the formula for the potential. I'm not sure I exactly am getting your hint Dick, but again thank you for the help.

arctan(y/x) is the angle measured from the x-axis in first quadrant. It's the negative of the angle from the x-axis in the fourth quadrant. If you patch those together then it's a continuous function for x>0. It's NOT continuous between the fourth quadrant and the third. Fundamentally, the potential function is just an angle. You just have to redefine the angle by shifting it by a constant to make it continuous except at points along the positive y axis.

 

[itsagoal89]

yeah, I can't figure out part (h) either, any ideas?

Let Ω = R2 \ {(0,y); y ≥ 0}. Find a closed form formula for the potential of F on Ω.

I don't even know where to start with this one

 

[cereeal]

If anyone else could help it would be greatly appreciated.
Would the previous arctan(y/x) function be the answer? Either way, I also do know where to start.

 

[Dick]

If anyone else could help it would be greatly appreciated.
Would the previous arctan(y/x) function be the answer? Either way, I also do know where to start.

arctan(y/x) is discontinuous along the negative y-axis. You want it to be continuous there. Define the potential by quadrant. You can always add or subtract pi to make the values match up there.