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Let \(F (x,y) \) be a planar vector field, and \(p\) a point in the domain of \(F\). Let \(C\subset \mathbb {R}^2 \) be a very small
circle around \(p\), positively oriented (counterclockwise). Then the integral
\[ \int _C F \cdot d \gamma \]
measures
how much counterclockwise rotation around \(p\) is generated by \(F\). It turns out we have
another quantity that measures precisely that:
\[ \text {curl}(F)(p) . \]
Overall, if we have a region \(D\) with
boundary \(\partial D\) oriented positively, the two integrals
\begin{gather*} \int _{\partial D} F \cdot d \gamma \\ \iint _{D} \text {curl}(F) \, dA \end{gather*}
measure the overall amount of
counterclockwise rotation generated by \(F\) in the region \(D\). Green’s Theorem states that
they agree.
Green’s Theorem Let \(F(x,y)\) be a planar vector field, \(D \subset \mathbb {R}^2\) a region inside the domain of \(F\), and \(\partial D\)
its boundary oriented positively. Then
\[ \int _{\partial D} F \cdot d \gamma = \iint _{D} \text {curl}(F) \, dA \]
Let \(F = \langle P, Q \rangle \) and \(D = [a_1, b_1] \times [a_2, b_2]\) the rectangle with vertices \((a_1,b_1)\), \((a_1, b_2)\), \((a_2, b_2)\), \((a_2, b_1)\). Let \(C_1\), \(C_2\), \(C_3\), \(C_4\) be the bottom, right, top, and left
sides of the rectangle, respectively, oriented counterclockwise. Then using the
Fundamental Theorem of Calculus,
\begin{align*} \iint _{D} \text {curl}(F) \, dA & = \int _{a_1}^{b_1} \int _{a_2} ^{b_2} \left [ \frac {\partial Q }{\partial x} - \frac {\partial P}{\partial y} \right ] dydx \\ & = \int _{a_2}^{b_2} \int _{a_1} ^{b_1} \frac {\partial Q }{\partial x} dxdy - \int _{a_1}^{b_1} \int _{a_2} ^{b_2} \frac {\partial P}{\partial y} dydx \\ & = \int _{a_2} ^{b_2} [ Q ( b_1 , y ) - Q(a_1 , y) ] dy - \int _{a_1} ^{b_1} [ P ( x , b_2 ) - P(x, a_2 ) ] dx \\ & = \int _{C_2} F \cdot d \gamma _2 + \int _{C_4} F \cdot d \gamma _4 + \int _{C_3} F \cdot d \gamma _3 + \int _{C_1} F \cdot d \gamma _1 \\ & = \int _{\partial D } F \cdot d \gamma \end{align*}
Let \(C\subset \mathbb {R}^2\) be the curve that travels along straight lines from \((-3,0)\) to \((2,0)\), then from \((2,0) \) to \((0,4)\), and
then from \((0,4)\) back to \((-3,0)\). Compute
\[ \int _C \langle 2y + x^2 \cos x - 5 , y^2e^y - 3x \rangle \cdot d \gamma \]
Applying Green’s Theorem, the integral above
coincides with the integral
\[ \iint _D \text {curl}(F) \, dA \]
where \(F = \langle 2y + x^2 \cos x - 5 , y^2e^y - 3x \rangle \) and \(D\) is the interior of the triangle. So we compute
\begin{align*} \text {Area}(D) & = \iint _D 1 \, dA \\ & = \iint _D \text {curl}(\langle 0,x \rangle ) \, dA \\ & = \int _{\partial D } \langle 0,x \rangle \cdot d \gamma \end{align*}
\begin{align*} \text {Area}(D) & = \iint _D 1 \, dA \\ & = - \iint _D \text {curl}(\langle y , 0 \rangle ) \, dA \\ & = - \int _{\partial D } \langle y , 0 \rangle \cdot d \gamma \end{align*}
When an oriented curve \(C\) is the boundary of a region, and \(C\) is oriented in such a
way that it has the region on its left, for a planar vector field \(F\) we denote
\[ \oint _C F \cdot d \gamma = \int _C F \cdot d \gamma \]
MTH 255H