What is the value of the line integral $\oint\limits_{C}(sin(y)dx+xcos(y)dy)$ , where C consists of the circle $x^2+y^2=2 $ from $(-1,1)$ to $(1,1)$ and the line segment from $(1,1)$ to $(-1,1)$?
What I have done so far:
Using Green's Theorem I have $$\oint\limits_{C}(sin(y)dx+xcos(y)dy)= \iint\limits_{\Omega}2cos(y)dxdy$$
However I don't know what the limits should be for the integrals.
On
The vector field $(\sin y, x \cos y)$ is conservative. You can check this by noting that $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. So, the value of the integral will be independent from the path and, in particular, if you end up returning to the starting point, the integral is zero.
You have incorrectly done Greens.
$P= \sin(y)$ $ Q = x \cos(y)$
$$ \frac{ \partial Q}{\partial x} = \cos(y)$$
$$ \frac{ \partial P}{\partial y} = \cos(y)$$
Now greens say:
$$ \oint Pdx + Qdy = \int \frac{ \partial Q}{\partial x} - \frac{\partial P}{\partial y} dA= \int 0 dA$$
Comment:
The line integral being zero for any loop is equivalent to field being conservative, for any loop we could have done the green's conversion is zero, hence the criteria for if vector field is conservative or not is that the:
$$ \frac{ \partial Q}{\partial x} - \frac{ \partial P}{\partial y} = 0$$
As mentioned by @PierreCarre