Finding a line integral

[Robam]

Hello, I'm having a bit of trouble with this problem:

​< link to objectionable page removed >

\(\displaystyle \mbox{Calculate }\, \displaystyle{\oint\limits_C}\, \hat{F}\, \dot\, d\hat{R}\)

\(\displaystyle \mbox{where }\, \hat{F}(x,\, y)\, =\, (x\, -\, y)\hat{i}\, +\, (x\, +\, y^3)\hat{j}\)

So far, I've gotten that ∮_C F∙dR = ∮_C (x-y)dx + (x+y³)dy = ∬_D 2 dA (by Green's Theorem)
How do I find the bounds of integration to evaluate the double integral?

 

[Romsek]

Hello, I'm having a bit of trouble with this problem: <link removed>

\(\displaystyle \mbox{Calculate }\, \displaystyle{\oint\limits_C}\, \hat{F}\, \dot\, d\hat{R}\)

\(\displaystyle \mbox{where }\, \hat{F}(x,\, y)\, =\, (x\, -\, y)\hat{i}\, +\, (x\, +\, y^3)\hat{j}\)

So far, I've gotten that ∮_C F∙dR = ∮_C (x-y)dx + (x+y³)dy = ∬_D 2 dA (by Green's Theorem)
How do I find the bounds of integration to evaluate the double integral?

If your F(x,y) were the gradient of a scalar function then this integral would have a solution independent of the chosen contour.
A quick toying around with it doesn't obviously show it as the gradient of some scalar function.

That means the value of the integral is going to depend on the contour, which they haven't given you.

if the (x-y) term were (x+y) then F would be the gradient of (1/2 x^2 + 1/4 y^4 + xy) and you could apply what you know about the integral over closed contours of the gradient of a scalar function. Maybe this is a typo?

 

[Robam]

If your F(x,y) were the gradient of a scalar function then this integral would have a solution independent of the chosen contour.
A quick toying around with it doesn't obviously show it as the gradient of some scalar function.

That means the value of the integral is going to depend on the contour, which they haven't given you.

if the (x-y) term were (x+y) then F would be the gradient of (1/2 x^2 + 1/4 y^4 + xy) and you could apply what you know about the integral over closed contours of the gradient of a scalar function. Maybe this is a typo?

I'm pretty sure it's not a typo, as I emailed my professor and he said the question looks correct. If it isn't correct though, are you referring to using the fundamental theorem of line integrals to solve? Wouldn't I need the end points of the curve to be able to use that?

 

[Romsek]

I'm pretty sure it's not a typo, as I emailed my professor and he said the question looks correct. If it isn't correct though, are you referring to using the fundamental theorem of line integrals to solve? Wouldn't I need the end points of the curve to be able to use that?

I am. But you have a circle on your integral sign and that usually means a closed contour, i.e. the endpoints are the same point.

The integral along a closed contour of a field that's the gradient of a scalar function is 0. The field is called conservative etc.

Given that your F doesn't seem to be the gradient of a scalar function, and given that there is no specification for the contour, I don't see how you can proceed.

Maybe someone else has a better idea.

 

[HallsofIvy]

It is impossible to do the problem as stated without know precisely what the contour is. As Robam said, if this \(\displaystyle
F\cdot dR\) were an "exact differential" then the integral around any closed path would be trivially 0. Since it is not, the integral depends upon exactly what the path is- and that is not given.

(By Green's theorem, the integral is twice the area inside the closed boundary- but we don't know that area. If it were me and my teacher insisted upon an answer, that is what I would say- "twice the area.")