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17.4 Green's Theorem

Let \(\vec{F} = \langle f, g \rangle\)

The vector quantity \(\left( \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right) \vec{k} = \langle 0, 0, g_x - f_y \rangle\) is called the curl of \(\vec{F}\), written as \(\text{curl } \vec{F}\).

\(|\text{curl } \vec{F}| = |g_x - f_y|\) is a measure of rotation in \(\vec{F}\).

For example, \(\vec{F} = \langle x, y \rangle\)

Vector field plot of F = <x, y> showing arrows pointing radially outward from the origin with no rotation. — Figure: Vector field plot of F = showing arrows pointing radially outward from the origin with no rotation.

showing arrows pointing radially outward from the origin with no rotation." class="mx-auto">

\(\text{curl } \vec{F} = \langle 0, 0, 0 - 0 \rangle = \vec{0}\)

\(|\text{curl } \vec{F}| = 0\)

indicating no rotation
confirmed by visual inspection

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\(\vec{F} = \langle -y, x \rangle\)

Figure: Vector field plot of F = <-y, x> showing arrows circulating counter-clockwise around the origin.

showing arrows circulating counter-clockwise around the origin." class="mx-auto">

we see rotation

\(\text{curl } \vec{F} = \langle 0, 0, g_x - f_y \rangle = \langle 0, 0, 1 - (-1) \rangle = \langle 0, 0, 2 \rangle\)

\(|\text{curl } \vec{F}| = 2 \neq 0\), indicating a rotation

From last time,

if \(\vec{F} = \langle f, g \rangle\), then \(\vec{F}\) is conservative

if \(g_x = f_y \rightarrow g_x - f_y = 0\)

this means if \(\vec{F}\) is conservative, then \(|\text{curl } \vec{F}| = 0\)

or a conservative vector field is irrotational

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Green's Theorem

If \(\vec{F} = \langle f, g \rangle\) and \(C\) is a simple closed path traversed once in the counterclockwise direction, then

\[ \oint_C \vec{F} \cdot d\vec{r} = \oint_C f dx + g dy = \iint_R \left( \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right) dA \]

Circle means closed path

Figure: A region R in the xy-plane bounded by a closed curve C with counterclockwise orientation arrows.

Why is this true?

\(\oint_C \vec{F} \cdot d\vec{r}\) is a line integral accumulating \(\vec{F}\) along the path.

→ circulation on the boundary

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right side: \(\iint_R (g_x - f_y) dA\)

\(| \text{curl } \vec{F} |\)

So this is accumulating rotation on region \(R\).

inside each grid there is rotation

Figure: A region in the xy-plane divided into a grid, with small red circular arrows indicating rotation in each cell.

rotations cancel on boundary of two boxes

Figure: Diagram showing four adjacent square cells with internal rotations where internal boundary arrows cancel out.

after canceling inside, ends up like

Figure: A single square boundary with net counterclockwise flow after internal cancellations.

Figure: Final region in xy-plane showing only the boundary circulation remaining.

accumulates \(\vec{F}\) along boundary → left side of theorem

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Example: Line Integral and Green's Theorem

Given the vector field:

\[ \vec{F} = \langle y+2, x^2+1 \rangle \]

And the curve \( C \): from \( (-1, 1) \) to \( (1, 1) \) along \( y = x^2 \) then back to \( (-1, 1) \) along \( y = 2 - x^2 \).

Figure: Coordinate graph showing a closed region bounded by y=x^2 (C1) and y=2-x^2 (C2) from x=-1 to x=1.

Let's do this as a line integral, then try the Green's Theorem way.

Method 1: Line Integral

\( C_1: \vec{r}(t) = \langle t, t^2 \rangle \) for \( -1 \le t \le 1 \)
\( C_2: \vec{r}(t) = \langle -t, 2-t^2 \rangle \) for \( -1 \le t \le 1 \)

\[ \int_C \vec{F} \cdot \vec{r}' dt = \underbrace{\int_{-1}^{1} \langle t^2+2, t^2+1 \rangle \cdot \langle 1, 2t \rangle dt}_{C_1} + \underbrace{\int_{-1}^{1} \langle 4-t^2, t^2+1 \rangle \cdot \langle -1, -2t \rangle dt}_{C_2} \]

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\[ = \int_{-1}^{1} (2t^3 + t^2 + 2t + 2) dt + \int_{-1}^{1} (-2t^3 + t^2 - 2t - 4) dt = \dots = \frac{14}{3} - \frac{22}{3} = \boxed{-\frac{8}{3}} \]

Method 2: Green's Theorem

Green's Theorem says we can trade a line integral for a double integral over the region \( R \):

\[ \iint_R (g_x - f_y) dA \]

\( \vec{F} = \langle y+2, x^2+1 \rangle \)

\( g_x - f_y = 2x - 1 \)

Region \( R \):

\( -1 \le x \le 1 \)
\( x^2 \le y \le 2 - x^2 \)

\[ \iint_R (g_x - f_y) dA = \int_{-1}^{1} \int_{x^2}^{2-x^2} (2x - 1) dy dx \]\[ = \int_{-1}^{1} (2x - 1)(2 - 2x^2) dx = \int_{-1}^{1} (-4x^3 + 2x^2 + 4x - 2) dx \]\[ = \dots = \boxed{-\frac{8}{3}} \]

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Green's Theorem for Area

Green's Theorem can also find area of \( R \) as a line integral:

\[ \oint f dx + g dy = \iint_R (g_x - f_y) dA \]

If \( g_x - f_y = 1 \), then the right side is \( \iint_R dA = \text{area of } R \).

Now come up with a vector field \( \vec{F} = \langle f, g \rangle \) such that \( g_x - f_y = 1 \).

One possibility: \( \vec{F} = \langle -\frac{1}{2}y, \frac{1}{2}x \rangle \)

Note: Here \( f = -\frac{1}{2}y \) and \( g = \frac{1}{2}x \).

Use these on the left side:

\[ \oint f dx + g dy = \oint -\frac{1}{2}y dx + \frac{1}{2}x dy \]

\[ = \frac{1}{2} \oint x dy - y dx \]

This gives area of \( R \)

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Green's Theorem: Alternative Form

Green's Theorem \( \oint f dx + g dy = \iint_R (g_x - f_y) dA \) is based on \( \int \vec{F} \cdot \vec{T} ds \) on the left side, where \( \vec{T} \) is the unit tangent.

If we use the unit normal \( \vec{N} \) instead of unit tangent \( \vec{T} \), we end up with:

\[ \oint f dy - g dx = \iint_R (f_x + g_y) dA \]

This is called the Divergence of \( \vec{F} = \text{div } \vec{F} \).

(In "standard" Green's Theorem, the right side is \( \text{curl } \vec{F} \)).

The divergence measures the change of a small volume as it travels in the vector field, and in this alternate Green's Theorem form we get the flux.

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Illustration of Divergence

\(\vec{F} = \langle x, 0 \rangle\)

\(\text{div } \vec{F} = f_x + g_y = 1 + 0 = 1\)

\(\rightarrow\) this means the rate of increase in volume of a small box as it moves in \(\vec{F}\) is positive

Vector field plot of \vec{F} = \langle x, 0 \rangle with horizontal arrows growing longer as they move right. — Figure: Vector field plot of \(\vec{F} = \langle x, 0 \rangle\) with horizontal arrows growing longer as they move right.

Small magnitude on left than right so it ends being stretched, so volume increases (positive divergence)

Figure: A small blue box is shown being stretched into a longer rectangle by a larger vector on the right side.