{"id":1392,"date":"2025-10-17T14:46:05","date_gmt":"2025-10-17T14:46:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1392"},"modified":"2025-11-06T13:59:39","modified_gmt":"2025-11-06T13:59:39","slug":"greens-theorem-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/greens-theorem-learn-it-2\/","title":{"raw":"Green\u2019s Theorem: Learn It 2","rendered":"Green\u2019s Theorem: Learn It 2"},"content":{"raw":"

Circulation Form of Green\u2019s Theorem<\/h2>\r\nThe first form of Green's theorem that we examine is the circulation form. This form relates the vector line integral over a simple, closed plane curve [latex]C[\/latex] to a double integral over the region enclosed by [latex]C[\/latex]. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.\r\n\r\n
\r\n

Green's theorem, circulation form<\/h3>\r\nLet [latex]D[\/latex] be an open, simply connected region with a boundary curve [latex]C[\/latex] that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure 2). Let [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] be a vector field with component functions that have continuous partial derivatives on [latex]D[\/latex]. Then,\r\n

[latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\oint_CPdx+Qdy=\\displaystyle\\iint_D(Q_x-P_y)dA[\/latex].<\/p>\r\n\r\n<\/section>[caption id=\"attachment_1400\" align=\"aligncenter\" width=\"493\"]\"A Figure 2. The circulation form of Green\u2019s theorem relates a line integral over curve [latex]C[\/latex] to a double integral over region [latex]D[\/latex].[\/caption]

Green's theorem can only be used for a two-dimensional vector field [latex]\\mathbf{F}[\/latex]. If [latex]\\mathbf{F}[\/latex] is a three-dimensional field, then Green's theorem does not apply.[latex]\\\\[\/latex]Since [latex]\\displaystyle\\int_CPdx+Qdy=\\displaystyle\\int_C\\mathbf{F}\\cdot\\mathbf{T}ds[\/latex], this version of Green's theorem is sometimes called the tangential form<\/em> of Green's theorem.<\/section>\r\n

The proof of Green\u2019s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that [latex]D[\/latex] is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which [latex]{\\bf{F}}\\langle{P},Q\\rangle[\/latex] is conservative. In this case,<\/p>\r\n

[latex]\\large{\\displaystyle\\oint_CPdx+Qdy=0}[\/latex]<\/p>\r\n

because the circulation is zero in conservative vector fields. By the cross-partial property of conservative fields theorem, [latex]{\\bf{F}}[\/latex] satisfies the cross-partial condition, so [latex]P_y=Q_x[\/latex]. Therefore,<\/p>\r\n

[latex]\\large{\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\iint_D0dA=0=\\displaystyle\\oint_CPdx+Qdy}[\/latex],<\/p>\r\n

which confirms Green\u2019s theorem in the case of conservative vector fields.<\/p>\r\n\r\n

\r\n

Proof<\/strong><\/p>\r\n\r\n\r\n

\r\n\r\n
\r\n

Let\u2019s now prove that the circulation form of Green\u2019s theorem is true when the region [latex]D[\/latex] is a rectangle. Let [latex]D[\/latex] be the rectangle [latex][a,b]\\times[c,d][\/latex] oriented counterclockwise. Then, the boundary [latex]C[\/latex] of [latex]D[\/latex] consists of four piecewise smooth pieces [latex]C_1[\/latex], [latex]C_2[\/latex], [latex]C_3[\/latex], and [latex]C_4[\/latex] (Figure 3). We parameterize each side of [latex]D[\/latex] as follows:<\/p>\r\n

[latex]\\begin{aligned}\r\nC_1:{\\bf{r}}_1(t)&=\\langle{t},c\\rangle,a\\leq{t}\\leq{b} \\\\\r\nC_2:{\\bf{r}}_2(t)&=\\langle{b},t\\rangle,c\\leq{t}\\leq{d} \\\\\r\n-C_3:{\\bf{r}}_3(t)&=\\langle{t},d\\rangle,a\\leq{t}\\leq{b} \\\\\r\n-C_4:{\\bf{r}}_4(t)&=\\langle{a},t\\rangle,c\\leq{t}\\leq{d}\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n<\/section><\/div>\r\n

[caption id=\"attachment_1402\" align=\"aligncenter\" width=\"337\"]\"A Figure 3. Rectangle [latex]D[\/latex] is oriented counterclockwise.[\/caption]<\/div>\r\n
\r\n\r\nThen,<\/span>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}&=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\\r\n&=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_1(t))\\bullet{\\bf{r}}_1(t)dt+\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_2(t))\\bullet{\\bf{r}}_2(t)dt-\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_3(t))\\bullet{\\bf{r}}_3(t)dt-\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_4(t))\\bullet{\\bf{r}}_4(t)dt \\\\\r\n&=\\displaystyle\\int_a^bP(t,c)dt+\\displaystyle\\int_c^dQ(b,t)dt-\\displaystyle\\int_a^bP(t,d)dt-\\displaystyle\\int_c^dQ(a,t)dt \\\\\r\n&=\\displaystyle\\int_a^b(P(t,c)-P(t,d))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\\r\n&=-\\displaystyle\\int_a^b(P(t,d)-P(t,c))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt\r\n\\end{aligned}[\/latex].<\/p>\r\n

By the Fundamental Theorem of Calculus,<\/p>\r\n

[latex]\\large{P(t,d)-P(t,c)=\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dy\\text{ and }Q(b,t)-Q(a,t)=\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dx}[\/latex].<\/p>\r\n

Therefore,<\/p>\r\n

[latex]\\begin{aligned}\r\n&-\\displaystyle\\int_a^b(P(t,d)-P(t,c)dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\\r\n&=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt\r\n\\end{aligned}[\/latex].<\/p>\r\n

But,<\/p>\r\n

[latex]\\begin{aligned}\r\n-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt&=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(x,y)dydx+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdy \\\\\r\n&=\\displaystyle\\int_a^b\\displaystyle\\int_c^d(Q_x-P_y)dydx \\\\\r\n&=\\displaystyle\\int \\ \\displaystyle\\int_D(Q_x-P_y)dA\r\n\\end{aligned}[\/latex].<\/p>\r\n

Therefore, [latex]\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}=\\int\\int_D(Q_{x}-P_{y})dA[\/latex] and we have proved Green\u2019s theorem in the case of a rectangle.<\/p>\r\n

To prove Green\u2019s theorem over a general region [latex]D[\/latex], we can decompose [latex]D[\/latex] into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/div>\r\n<\/section>

\r\n

Calculate the line integral<\/p>\r\n

[latex]\\displaystyle\\oint_Cx^2ydx+(y-3)dy[\/latex]<\/p>\r\n

where [latex]C[\/latex] is a rectangle with vertices [latex](1, 1)[\/latex], [latex](4, 1)[\/latex], [latex](4, 5)[\/latex], and [latex](1, 5)[\/latex] oriented counterclockwise.<\/p>\r\n[reveal-answer q=\"384752098\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384752098\"]\r\n

Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle=\\langle{x}^2y,y-3\\rangle[\/latex]. Then, [latex]Q_x=0[\/latex] and [latex]P_y=x^{2}[\/latex]. Therefore, [latex]W_x-P_y=-x^{2}[\/latex].<\/p>\r\n

Let [latex]D[\/latex] be the rectangular region enclosed by [latex]C[\/latex] (Figure 4). By Green\u2019s theorem,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\oint_Cx^2ydx+(y-3)dy&=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\\r\n&=\\displaystyle\\int \\ \\displaystyle\\int_D-x^2dA=\\displaystyle\\int_1^5 \\ \\displaystyle\\int_1^4-x^2dxdy \\\\\r\n&=\\displaystyle\\int_1^5-21dy = -84\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1404\" align=\"aligncenter\" width=\"719\"]\"A Figure 4. The line integral over the boundary of the rectangle can be transformed into a double integral over the rectangle.[\/caption]\r\n

Analysis<\/span><\/strong><\/p>\r\n

If we were to evaluate this line integral without using Green\u2019s 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\u00a0line integrals\u00a0to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green\u2019s theorem makes the calculation much simpler.<\/p>\r\n\r\n

\r\n

[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/section>

\r\n

Calculate the work done on a particle by force field<\/p>\r\n

[latex]{\\bf{F}}(x,y)=\\langle{y}+\\sin{x},e^y-x\\rangle[\/latex]<\/p>\r\n

as the particle traverses circle [latex]x^2+y^2=4[\/latex] exactly once in the counterclockwise direction, starting and ending at point [latex](2, 0)[\/latex].<\/p>\r\n[reveal-answer q=\"187491994\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"187491994\"]\r\n

Let [latex]C[\/latex] denote the circle and let [latex]D[\/latex] be the disk enclosed by [latex]C[\/latex]. The work done on the particle is<\/p>\r\n

[latex]W=\\displaystyle\\oint_C({y}+\\sin{x})dx+(e^y-x)dy[\/latex].<\/p>\r\n

As with\u00a0Example \"Applying Green\u2019s Theorem over a Rectangle\", this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green\u2019s theorem (Figure 5).<\/p>\r\n

Let [latex]{\\bf{F}}(x,y)=\\langle{P}(x,y),Q(x,y)\\rangle=\\langle{y}+\\sin{x},e^y-x\\rangle[\/latex]. Then, [latex]Q_x=-1\\text{ and }P_y=1[\/latex]. Therefore, [latex]Q_x-P_y=-2[\/latex].<\/p>\r\n

By Green\u2019s theorem,<\/p>\r\n

[latex]\\begin{aligned}\r\nW&=\\displaystyle\\oint_C({y}+\\sin{x})dx+(e^y-x)dy \\\\\r\n&=\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\iint_D-2dA \\\\\r\n&=-2(\\text{area}(D))=-2\\pi(2^2)=-8\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"attachment_1405\" align=\"aligncenter\" width=\"717\"]\"A Figure 5. The line integral over the boundary circle can be transformed into a double integral over the disk enclosed by the circle.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>In the preceding two examples, the double integral in Green\u2019s theorem was easier to calculate than the line integral, so we used the theorem to calculate the line integral. In the next example, the double integral is more difficult to calculate than the line integral, so we use Green\u2019s theorem to translate a double integral into a line integral.\r\n\r\n

Calculate the area enclosed by ellipse [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1[\/latex] (Figure 6).[caption id=\"attachment_1406\" align=\"aligncenter\" width=\"349\"]\"A Figure 6. Ellipse [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1[\/latex] is denoted by [latex]C[\/latex].[\/caption][reveal-answer q=\"837599384\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"837599384\"]\r\n

Let [latex]C[\/latex] denote the ellipse and let [latex]D[\/latex] be the region enclosed by [latex]C[\/latex]. Recall that ellipse [latex]C[\/latex] can be parameterized by<\/p>\r\n

[latex]x=a\\cos{t},y=b\\sin{t},0\\leq{t}\\leq2\\pi[\/latex].<\/p>\r\n

Calculating the area of [latex]D[\/latex] is equivalent to computing double integral [latex]\\displaystyle\\iint_DdA[\/latex]. To calculate this integral without Green\u2019s theorem, we would need to divide [latex]D[\/latex] into two regions: the region above the\u00a0x<\/em>-axis and the region below. The area of the ellipse is<\/p>\r\n

[latex]\\displaystyle\\int_{-a}^a\\displaystyle\\int_0^{\\sqrt{b^2-(bx\/a)^2}}dydx+\\displaystyle\\int_{-a}^a\\displaystyle\\int_{\\sqrt{b^2-(bx\/a)^2}}^0dydx[\/latex].<\/p>\r\n

These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green\u2019s theorem to transform [latex]\\displaystyle\\iint_DdA[\/latex] into a line integral around the boundary [latex]C[\/latex].<\/p>\r\n

Consider vector field<\/p>\r\n

[latex]{\\bf{F}}(x,y)=\\langle{P},Q\\rangle=\\left\\langle-\\frac{y}2,\\frac{x}2\\right\\rangle[\/latex].<\/p>\r\n

Then, [latex]Q_x=\\frac12[\/latex] and [latex]P_y=-\\frac12[\/latex], and therefore [latex]Q_x-P_y=1[\/latex]. Notice that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0was chosen to have the property that [latex]Q_x-P_y=1[\/latex]. Since this is the case, Green\u2019s theorem transforms the line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0over [latex]C[\/latex] into the double integral of 1 over [latex]D[\/latex].<\/p>\r\n

By Green\u2019s theorem,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\iint_DdA&=\\displaystyle\\iint_D(Q_x-P_y)dA \\\\\r\n&=\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}=\\frac12\\displaystyle\\int_C-ydx+xdy \\\\\r\n&=\\frac12\\displaystyle\\int_0^{2\\pi}-b\\sin{t}(-a\\sin{t})+a(\\cos{t})b\\cos{t}dt \\\\\r\n&=\\frac12\\displaystyle\\int_0^{2\\pi}ab\\cos^2t+ab\\sin^2tdt=\\frac12\\displaystyle\\int_0^{2\\pi}abdt=\\pi{a}b\r\n\\end{aligned}[\/latex].<\/p>\r\n

Therefore, the area of the ellipse is\u00a0[latex]\\pi{a}b[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

In the above example, we used vector field [latex]\\mathbf{F}(x,y)=\\langle P,Q\\rangle=\\left\\langle-\\frac{y}{2},\\frac{x}{2}\\right\\rangle[\/latex] to find the area of any ellipse. The logic of the previous example can be extended to derive a formula for the area of any region [latex]D[\/latex]. Let [latex]D[\/latex] be any region with a boundary that is a simple closed curve [latex]C[\/latex] oriented counterclockwise. If [latex]\\mathbf{F}(x,y)=\\langle P,Q\\rangle=\\left\\langle-\\frac{y}{2},\\frac{x}{2}\\right\\rangle[\/latex], then by the same logic as in the previous example,<\/p>\r\n

[latex]\\text{area of }D=\\displaystyle\\iint_DdA=\\frac12\\displaystyle\\oint_C-ydx+xdy[\/latex].<\/span><\/p>\r\n\r\n

\u00a0If [latex]\\mathbf{F}=\\langle P,Q\\rangle[\/latex] is any vector field with [latex]Q_x-P_y=1[\/latex], then the logic of the previous paragraph works. So, the equation above is not the only equation that uses a vector field's mixed partials to get the area of a region.<\/section>","rendered":"

Circulation Form of Green\u2019s Theorem<\/h2>\n

The first form of Green’s theorem that we examine is the circulation form. This form relates the vector line integral over a simple, closed plane curve [latex]C[\/latex] to a double integral over the region enclosed by [latex]C[\/latex]. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.<\/p>\n

\n

Green’s theorem, circulation form<\/h3>\n

Let [latex]D[\/latex] be an open, simply connected region with a boundary curve [latex]C[\/latex] that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure 2). Let [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] be a vector field with component functions that have continuous partial derivatives on [latex]D[\/latex]. Then,<\/p>\n

[latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\oint_CPdx+Qdy=\\displaystyle\\iint_D(Q_x-P_y)dA[\/latex].<\/p>\n<\/section>\n

\"A
Figure 2. The circulation form of Green\u2019s theorem relates a line integral over curve [latex]C[\/latex] to a double integral over region [latex]D[\/latex].<\/figcaption><\/figure>\n
Green’s theorem can only be used for a two-dimensional vector field [latex]\\mathbf{F}[\/latex]. If [latex]\\mathbf{F}[\/latex] is a three-dimensional field, then Green’s theorem does not apply.[latex]\\\\[\/latex]Since [latex]\\displaystyle\\int_CPdx+Qdy=\\displaystyle\\int_C\\mathbf{F}\\cdot\\mathbf{T}ds[\/latex], this version of Green’s theorem is sometimes called the tangential form<\/em> of Green’s theorem.<\/section>\n

The proof of Green\u2019s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that [latex]D[\/latex] is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which [latex]{\\bf{F}}\\langle{P},Q\\rangle[\/latex] is conservative. In this case,<\/p>\n

[latex]\\large{\\displaystyle\\oint_CPdx+Qdy=0}[\/latex]<\/p>\n

because the circulation is zero in conservative vector fields. By the cross-partial property of conservative fields theorem, [latex]{\\bf{F}}[\/latex] satisfies the cross-partial condition, so [latex]P_y=Q_x[\/latex]. Therefore,<\/p>\n

[latex]\\large{\\displaystyle\\iint_D(Q_x-P_y)dA=\\displaystyle\\iint_D0dA=0=\\displaystyle\\oint_CPdx+Qdy}[\/latex],<\/p>\n

which confirms Green\u2019s theorem in the case of conservative vector fields.<\/p>\n

\n

Proof<\/strong><\/p>\n

\n
\n
\n

Let\u2019s now prove that the circulation form of Green\u2019s theorem is true when the region [latex]D[\/latex] is a rectangle. Let [latex]D[\/latex] be the rectangle [latex][a,b]\\times[c,d][\/latex] oriented counterclockwise. Then, the boundary [latex]C[\/latex] of [latex]D[\/latex] consists of four piecewise smooth pieces [latex]C_1[\/latex], [latex]C_2[\/latex], [latex]C_3[\/latex], and [latex]C_4[\/latex] (Figure 3). We parameterize each side of [latex]D[\/latex] as follows:<\/p>\n

[latex]\\begin{aligned} C_1:{\\bf{r}}_1(t)&=\\langle{t},c\\rangle,a\\leq{t}\\leq{b} \\\\ C_2:{\\bf{r}}_2(t)&=\\langle{b},t\\rangle,c\\leq{t}\\leq{d} \\\\ -C_3:{\\bf{r}}_3(t)&=\\langle{t},d\\rangle,a\\leq{t}\\leq{b} \\\\ -C_4:{\\bf{r}}_4(t)&=\\langle{a},t\\rangle,c\\leq{t}\\leq{d} \\end{aligned}[\/latex].<\/p>\n<\/section>\n<\/div>\n

\n
\"A
Figure 3. Rectangle [latex]D[\/latex] is oriented counterclockwise.<\/figcaption><\/figure>\n<\/div>\n
\n

Then,<\/span><\/p>\n

[latex]\\begin{aligned} \\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}&=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\ &=\\displaystyle\\int_{C_1}{\\bf{F}}\\bullet{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_3}{\\bf{F}}\\bullet{d}{\\bf{r}}-\\displaystyle\\int_{-C_4}{\\bf{F}}\\bullet{d}{\\bf{r}} \\\\ &=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_1(t))\\bullet{\\bf{r}}_1(t)dt+\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_2(t))\\bullet{\\bf{r}}_2(t)dt-\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}_3(t))\\bullet{\\bf{r}}_3(t)dt-\\displaystyle\\int_c^d{\\bf{F}}({\\bf{r}}_4(t))\\bullet{\\bf{r}}_4(t)dt \\\\ &=\\displaystyle\\int_a^bP(t,c)dt+\\displaystyle\\int_c^dQ(b,t)dt-\\displaystyle\\int_a^bP(t,d)dt-\\displaystyle\\int_c^dQ(a,t)dt \\\\ &=\\displaystyle\\int_a^b(P(t,c)-P(t,d))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\ &=-\\displaystyle\\int_a^b(P(t,d)-P(t,c))dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\end{aligned}[\/latex].<\/p>\n

By the Fundamental Theorem of Calculus,<\/p>\n

[latex]\\large{P(t,d)-P(t,c)=\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dy\\text{ and }Q(b,t)-Q(a,t)=\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dx}[\/latex].<\/p>\n

Therefore,<\/p>\n

[latex]\\begin{aligned} &-\\displaystyle\\int_a^b(P(t,d)-P(t,c)dt+\\displaystyle\\int_c^d(Q(b,t)-Q(a,t))dt \\\\ &=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt \\end{aligned}[\/latex].<\/p>\n

But,<\/p>\n

[latex]\\begin{aligned} -\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(t,y)dydt+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdt&=-\\displaystyle\\int_a^b\\displaystyle\\int_c^d\\frac{\\partial}{\\partial{y}}P(x,y)dydx+\\displaystyle\\int_c^d\\displaystyle\\int_a^b\\frac{\\partial}{\\partial{x}}Q(x,t)dxdy \\\\ &=\\displaystyle\\int_a^b\\displaystyle\\int_c^d(Q_x-P_y)dydx \\\\ &=\\displaystyle\\int \\ \\displaystyle\\int_D(Q_x-P_y)dA \\end{aligned}[\/latex].<\/p>\n

Therefore, [latex]\\displaystyle\\int_C{\\bf{F}}\\bullet{d}{\\bf{r}}=\\int\\int_D(Q_{x}-P_{y})dA[\/latex] and we have proved Green\u2019s theorem in the case of a rectangle.<\/p>\n

To prove Green\u2019s theorem over a general region [latex]D[\/latex], we can decompose [latex]D[\/latex] into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.<\/p>\n

[latex]_\\blacksquare[\/latex]<\/p>\n<\/div>\n<\/section>\n

\n

Calculate the line integral<\/p>\n

[latex]\\displaystyle\\oint_Cx^2ydx+(y-3)dy[\/latex]<\/p>\n

where [latex]C[\/latex] is a rectangle with vertices [latex](1, 1)[\/latex], [latex](4, 1)[\/latex], [latex](4, 5)[\/latex], and [latex](1, 5)[\/latex] oriented counterclockwise.<\/p>\n