es

Green's theorems

Semi-infinite strip

Polar coordinates

Curvilinear coordinates

General case

 

Green's Theorem in Rectangular Domain

We start with the circulation form of Green’s theorem for plane vector field F = (P(x, y), Q(x, y)). From our previous discussion, we know that
\[ \oint_{C} \nabla f \bullet {\text d}{\bf r} = 0 \]
for any smooth function f and simple closed path C. So Green's theorem extend thiis formula for arbitrary vector fields, not necessary of gradient form.
Theorem (Green): Let ∂R be a positively oriented, piecewise smooth, simple closed curve in a plane, and let R be the region bounded by ∂R. If P and Q are functions of (x, y) defined on an open region containing R and have continuous partial derivatives there, then \begin{equation} \label{EqGreen.1} \oint_{\partial R} \left( P\,{\text d}x + Q\,{\text d}y \right) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) {\text d}x\,{\text d}y , \end{equation} where the path of integration along ∂R is counterclockwise.
We consider a rectangle R in ℝ² oriented along Cartersian axes.
original = Polygon[{{0, 0}, {1.5, 0}, {1.5, 1}, {0, 1}}]; rec = Graphics[{EdgeForm[{Thick, Black}], Opacity[.3], Black, original, LightRed}, Axes -> False]; arx = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{-0.7, -0.3}, {2, -0.3}}]}]; ary = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{-0.4, -0.6}, {-0.4, 1.3}}]}]; txt = Graphics[{Black, Text[Style["A", FontSize -> 15, Bold], {0, -0.15}], Text[Style["x-axis", FontSize -> 18, Bold], {2.0, -0.15}], Text[Style["B", FontSize -> 15, Bold], {1.5, -0.15}], Text[Style["y-axis", FontSize -> 18, Bold], {-0.1, 1.4}], Text[Style["C", FontSize -> 15, Bold], {1.5, 1.15}], Text[Style["D", FontSize -> 15, Bold], {0, 1.15}]}]; Show[rec, arx, ary, txt] Show[rec, arx, ary, txt]
Figure 1: Rectangular domain.

We prove this theorem for F = (P, Q) and R = [𝑎, b] × [c, d]. We parameterize each side of R as follows:

\begin{align*} \overline{AB} \ &: \quad \mathbf{r}_1 (t) = (t,c), \quad a \le t \le b , \\ \overline{BC} \ &: \quad \mathbf{r}_2 (t) = (b, t), \quad c \le t \le d , \\ \overline{CD} \ &: \quad \mathbf{r}_3 (t) = (t,d), \quad a \le t \le b , \\ \overline{DA} \ &: \quad \mathbf{r}_4 (t) = (a, t), \quad c \le t \le d . \end{align*}
Then
\begin{align*} \int_{\partial R} \mathbf{F} \bullet {\text d}{\bf r} &= \int_{\overline{AB}} \mathbf{F} \bullet {\text d}\mathbf{r} + \int_{\overline{BC}} \mathbf{F} \bullet {\text d}\mathbf{r} + \int_{\overline{CD}} \mathbf{F} \bullet {\text d}\mathbf{r} + \int_{\overline{DA}} \mathbf{F} \bullet {\text d}\mathbf{r} \\ &= \int_{\overline{AB}} \mathbf{F} \bullet {\text d}\mathbf{r} + \int_{\overline{BC}} \mathbf{F} \bullet {\text d}\mathbf{r} - \int_{\overline{DC}} \mathbf{F} \bullet {\text d}\mathbf{r} - \int_{\overline{AD}} \mathbf{F} \bullet {\text d}\mathbf{r} \\ &= \int_a^b \mathbf{F}\left( \mathbf{r}_1 (t) \right) \bullet \mathbf{r}'_1 (t) \,{\text d}t + \int_c^d \mathbf{F}\left( \mathbf{r}_2 (t) \right) \bullet \mathbf{r}'_2 (t) \,{\text d}t - \int_1^b \mathbf{F}\left( \mathbf{r}_3 (t) \right) \bullet \mathbf{r}'_3 (t) \,{\text d}t - \int_c^d \mathbf{F}\left( \mathbf{r}_4 (t) \right) \bullet \mathbf{r}'_4 (t) \,{\text d}t \\ &= \int_a^b P(t,c)\, {\text d}t + \int_c^d Q(b,t) \, {\text d}t - \int_a^b P(t,d)\, {\text d}t - \int_c^d Q(a,t) \, {\text d}t \\ &= \int_a^b \left[ P(t,c) - P(t,d) \right] {\text d}t + \int_c^d \left[ Q(b,t) - Q(a,t) \right] {\text d}t \\ &= - \int_a^b \left[ P(t,d) - P(t,c) \right] {\text d}t + \int_c^d \left[ Q(b,t) - Q(a,t) \right] {\text d}t . \end{align*}
By the Fundamental Theorem of Calculus,
\[ P(t,d) - P(t,c) = \int_c^d \frac{\partial}{\partial y}\,P(t, y)\,{\text d}y \]
and
\[ Q(b,t) - Q(a,t) = \int_a^b \frac{\partial}{\partial x}\,Q(x, t)\,{\text d}x . \]
Therefore,
\[ - \int_a^b \left[ P(t,d) - P(t,c) \right] {\text d}t + \int_c^d \left[ Q(b,t) - Q(a,t) \right] {\text d}t = - \int_a^b {\text d}t \int_c^d {\text d}y \,\frac{\partial}{\partial y}\,P(t, y) + \int_c^d {\text d}t \int_a^b {\text d}x \, \frac{\partial}{\partial x}\,Q(x, t) . \]
However,
\begin{align*} - \int_a^b {\text d}t \int_c^d {\text d}y \,\frac{\partial}{\partial y}\,P(t, y) + \int_c^d {\text d}t \int_a^b {\text d}x \, \frac{\partial}{\partial x}\,Q(x, t) &= - \int_a^b {\text d}x \int_c^d {\text d}y \, \frac{\partial}{\partial y}\,P(x, y) + \int_c^d {\text d}y \int_a^b {\text d}x\, \frac{\partial}{\partial x}\,Q(x, y) \\ &= \int_a^b {\text d}x \int_c^d {\text d}y\left[ Q_x - P_y \right] \\ &= \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) {\text d} A . \end{align*}
   

Example 1: Let us evaluate the line integral \[ \oint_C x\,y^3 {\text d}x + \left( y^2 -4 \right) {\text d} y , \] where 𝐶 is a rectangle with vertices (1, 2), (4, 2), (4, 5), and (1, 5) oriented counterclockwise.

If we were to evaluate this line integral without using Green’s theorem, we would need to parameterize each side of the rectangle, break the line integral into four separate line integrals, and use the methods from the section titled Line Integrals to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green’s theorem makes the calculation much simpler.

Let F = (x y³, y² − 4).    ■

End of Example 1

Now we extend our previous results for triangular domains.

triangle = Polygon[{{0, 0}, {1.5, 0}, {2, 1}}]; rec = Graphics[{EdgeForm[{Thick, Black}], Opacity[.3], Black, triangle, LightRed}, Axes -> False]; arx = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{-0.7, -0.3}, {2.2, -0.3}}]}]; ary = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{-0.4, -0.6}, {-0.4, 1.2}}]}]; txt = Graphics[{Black, Text[Style["A", FontSize -> 15, Bold], {0, -0.15}], Text[Style["x-axis", FontSize -> 18, Bold], {2.0, -0.15}], Text[Style["B", FontSize -> 15, Bold], {1.5, -0.15}], Text[Style["y-axis", FontSize -> 18, Bold], {-0.1, 1.3}], Text[Style["C", FontSize -> 15, Bold], {2.1, 1.15}], Text[Style["0", FontSize -> 15, Bold], {-0.5, -0.45}]}]; Show[txt, arx, ary, rec]
Figure 2: Triangular domain.

We demotstrate the validity of Green's formula \eqref{EqGreen.1} for triangular domain in the following example.    

Example 2:    ■

End of Example 2

Finally, we consider arbitrarily oriented rectangular domain.

trans = Polygon[{{0, 0}, {1.5*Sqrt[3]/2, 1.5/2}, {0.799, 1.6160254037844}, {-1/2, Sqrt[3]/2}}] recT = Graphics[{EdgeForm[{Thick, Black}], Opacity[.3], Black, trans, Blue}, Axes -> False] txt = Graphics[{Black, Text[Style["A", FontSize -> 15, Bold], {0, -0.15}], Text[Style["x-axis", FontSize -> 18, Bold], {2.0, -0.15}], Text[Style["B", FontSize -> 15, Bold], {1.4, 0.75}], Text[Style["y-axis", FontSize -> 18, Bold], {-0.1, 1.4}], Text[Style["C", FontSize -> 15, Bold], {0.9, 1.65}], Text[Style["D", FontSize -> 15, Bold], {-0.58, 0.8}]}]; Show[recT, arx, ary, txt]
Figure 3: Arbitrarily oriented rectangular domain.

   
Example 3:    ■
End of Example 3
Theorem 2 (Flux form of Green's theorem): Let R be a simply connected region with a boundary curve 𝐶 = ∂R that is a piecewise smooth, simple closed curve that is oriented counterclockwise (Figure 16.4.7). Let F = (P, Q) be a vector field with component functions that have continuous partial derivatives on an open region containing R. Then \begin{equation} \label{EqGreen.2} \oint_{\partial R} \left( P \bullet \hat{\bf n} \right) {\text d}s = \iint_R \left( \frac{\partial Q}{\partial y} + \frac{\partial P}{\partial x} \right) {\text d}A, \end{equation} where \( \displaystyle \quad \hat{\bf n} \ \) is the unit vector normal to the boundary ∂R directed outside R.
RJB Figure 16.4.7 in
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/16%3A_Vector_Calculus/16.04%3A_Greens_Theorem

   

Example 4:    ■

End of Example 4

 

  1. Find the area of the region enclosed by the curve with parameterization r(t) = (sint cost, sint).
  2. Use Green’s theorem to calculate line integral \( \displaystyle \quad \oint_C \sin \left( x^2 \right) {\text d}x + \left( 3 x - y^3 \right) {\text d} y , \quad \) where C is a right triangle with vertices (−2, 2), (4, 2), and (4, 7) oriented counterclockwise.

 

  1. Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
  2. Beezer, R., A First Course in Linear Algebra, 2015.
  3. Beezer, R., A Second Course in Linear Algebra, 2013.