Green's theorem - confused about orientation

[Feodalherren]

Homework Statement

∫Fdr
Over C where C is the cirlce (x-3)^2+ (y+4)^2=4
F=<y-cosy, xsiny>

Homework Equations

The Attempt at a Solution

So I applied Green's theorem and converted to polar and ended up with -4π, it should be positive.
The orientation confused me since day one with this theorem. When I parametrize C I get <2+2cost,-4+2sint> which IS counter clockwise, which DOES curl in the positive Z direction, so when why does my book say that -C gives the positive orientation?

 

[LCKurtz]

You probably mean <3+2cost,-4+2sint> for the parameterization. How do you know the answer should be positive? That doesn't follow just from going counterclockwise. It depends on the field.

 

[Feodalherren]

Yes that's what I meant :). The solutions manual says it should be positive.
It says that the curve is oriented negatively to begin with so they say -C must be used for Green's Theorem. I'm confused since it looks to me, from the parametrization, that the orientation is counter-clockwise to begin with and hence in the positive Z so C shouldn't be multiplied by -1.

 

[LCKurtz]

Yes that's what I meant :). The solutions manual says it should be positive.

The solutions manual says what should be positive? The answer??

It says that the curve is oriented negatively to begin with so they say -C must be used for Green's Theorem. I'm confused since it looks to me, from the parametrization, that the orientation is counter-clockwise to begin with and hence in the positive Z so C shouldn't be multiplied by -1.

Your initial statement of the problem doesn't give any orientation unless you assume the standard counterclockwise orientation. If you integrate in that direction the answer is $-4\pi$ as you have correctly calculated with Green's theorem. Unless there is something you haven't told us, the solution manual likely just has a typo in the answer.