$$ \newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\dx}{\text{ dx}}
\newcommand{\rang}{\text{rang}}
\newcommand{\s}{\ \ \ \ \ \ }
\newcommand{\arrows}{\s \Leftrightarrow \s}
\newcommand{\Arrows}{\s \Longleftrightarrow \s}
\newcommand{\arrow}{\s \Rightarrow \s}
\newcommand{\c}{\bcancel}
\newcommand{\v}[2]{
\begin{pmatrix}
#1 \\
#2 \\
\end{pmatrix}
}
\newcommand{\vt}[3]{
\begin{pmatrix}
#1 \\
#2 \\
#3 \\
\end{pmatrix}
}
\newcommand{\stack}[2]{
\substack{
#1 \\
#2
}
}
\newcommand{\atom}[3]{
\substack{
#1 \\
#2
}
\ce{#3}
}
$$
Green’s Theorem
$$ \oint_{\mathcal{C}} \vec{F}(x,y) \ \mathrm{d\vec{r}} = \int_{a}^{b}\vec{F}(x(t),y(t)) \cdot \vec{r}'(t) \mathrm{dt} = \oint_\mathcal{C} f_{1}(x,y) \mathrm{dx} + f_{2}(x,y) \mathrm{dy} = \iint_{R} \frac{\partial f_{2}}{\partial x} - \frac{\partial f_{1}}{\partial y}\ \mathrm{dA} $$
Curve segments are always integrated counter clockwise. Meaning the bounded region is always in the left.
You can go clockwise by simply making the expression negative.
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