Lesson 29: Greens theorem (17.4)

Accessible transcription generated on 4/1/2026

Original Notes

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Lesson 29: Greens theorem (17.4)

Evaluate $\int_C \vec{F} \cdot d\vec{r}$, where $C: \vec{r}(t), a \le t \le b$

Is $\vec{F}$ conservative? i.e., $\vec{F} = \nabla \phi$

Yes: Apply FTCLI

\[\int_C \vec{F} \cdot d\vec{r} = \int_C \nabla \phi \cdot d\vec{r} = \phi(\vec{r}(b)) - \phi(\vec{r}(a))\]

No: Use formula

\[\int_C \vec{F} \cdot d\vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt\]

Today: Compute $\oint_C \vec{F} \cdot \vec{T} \, ds$ when $C$ is a simple closed curve using Greens Theorem.

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Review Example:

Is $\vec{F} = \langle \sin y, x \cos y + \cos z, -y \sin z \rangle$ conservative?

To determine if the vector field is conservative, let $\vec{F} = \langle f, g, h \rangle$. We check the equality of mixed partial derivatives:

$f_y = g_x \implies \begin{cases} f_y = \cos y \\ g_x = \cos y \end{cases}$
$f_z = h_x \implies \begin{cases} f_z = 0 \\ h_x = 0 \end{cases}$
$g_z = h_y \implies \begin{cases} g_z = -\sin z \\ h_y = -\sin z \end{cases}$

Because all equalities hold, $\vec{F}$ is conservative.

* Find $\phi$ such that $\vec{F} = \vec{\nabla}\phi = \left\langle \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}, \frac{\partial\phi}{\partial z} \right\rangle$, by solving:

$\frac{\partial\phi}{\partial x} = \sin y$
$\frac{\partial\phi}{\partial y} = x \cos y + \cos z$
$\frac{\partial\phi}{\partial z} = -y \sin z$

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Example: Finding a Potential Function

Solve the following system of partial differential equations:

(1) $\frac{\partial \phi}{\partial x} = \sin y$
(2) $\frac{\partial \phi}{\partial y} = x \cos y + \cos z$
(3) $\frac{\partial \phi}{\partial z} = -y \sin z$

Starting with equation (1): \[ \frac{\partial \phi}{\partial x} = \sin y \] Integrate with respect to $x$, treating $y$ and $z$ as constants: \[ \phi(x, y, z) = x \sin y + k_1(y, z) \]

Substitute this expression into equation (2): \[ x \cos y + \cos z = \frac{\partial \phi}{\partial y} = x \cos y + \frac{\partial k_1}{\partial y} \] Simplifying, we find: \[ \frac{\partial k_1}{\partial y} = \cos z \]

Integrate with respect to $y$, treating $z$ as a constant: \[ k_1(y, z) = y \cos z + k_2(z) \] Substitute this result back into the expression for $\phi$: \[ \phi(x, y, z) = x \sin y + y \cos z + k_2(z) \]

Now, substitute this updated expression into equation (3): \[ -y \sin z = \frac{\partial \phi}{\partial z} = 0 - y \sin z + k_2'(z) \] This simplifies to: \[ \Rightarrow k_2'(z) = 0 \Rightarrow k_2(z) = C \]

Therefore, the complete solution for the potential function is: \[ \phi(x, y, z) = x \sin y + y \cos z + C \]

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Curl & Divergence (2D)

Given $\vec{F} = \langle f(x,y), g(x,y) \rangle$

Scalar (or 2D) Curl of $\vec{F} = g_x - f_y$

Divergence of $\vec{F} = f_x + g_y$

Think!

$\vec{F} = \langle f(x,y), g(x,y), 0 \rangle$

$\vec{\nabla} = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle \to$ "Del" operator

\[ \text{Curl} = \vec{\nabla} \times \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f & g & 0 \end{vmatrix} = \langle 0, 0, \underbrace{g_x - f_y}_{\text{Scalar or 2D Curl}} \rangle \] \[ \text{Divergence} = \vec{\nabla} \cdot \vec{F} = \underbrace{f_x + g_y}_{\text{Divergence}} \]

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Greens theorem

Suppose $\vec{F} = \langle f(x,y), g(x,y) \rangle$.
$C$ is a simple (no intersections) closed curve (in loop, same start & end points).
$R$ is the region enclosed by $C$.

Visual Description: A 2D coordinate system diagram illustrating Green's Theorem. It features a horizontal x-axis and a vertical y-axis. An irregular, bean-shaped closed curve, labeled C in green, is plotted across the axes. An arrow on the curve indicates a counter-clockwise orientation. The region inside the curve is filled with blue diagonal hatch marks and is labeled with a blue capital letter R.

\[ \oint_C \vec{F} \cdot \vec{T} ds = \oint_C \vec{F} \cdot d\vec{r} = \oint_C f dx + g dy \xrightarrow{\text{Greens}} \iint_R (g_x - f_y) dA \]

Note: $(g_x - f_y)$ represents the 2D curl of $\vec{F}$.

\[ \text{Flux} = \oint_C \vec{F} \cdot \vec{N} ds = \oint_C g dx - f dy \xrightarrow{\text{Greens}} \iint_R (f_x + g_y) dA \]

Note: $(f_x + g_y)$ represents the Div $\vec{F}$ (Divergence of $\vec{F}$).

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Theme: integral "cancels" derivative

Line diagram showing a horizontal line segment representing a 1D interval. The left endpoint is marked with a point labeled 'a' and the right endpoint is marked with a point labeled 'b'. This illustrates the boundary points of the interval [a, b].

Visual Description: Line diagram showing a horizontal line segment representing a 1D interval. The left endpoint is marked with a point labeled 'a' and the right endpoint is marked with a point labeled 'b'. This illustrates the boundary points of the interval [a, b].

\[\int_a^b f'(t) dt = f(b) - f(a) \quad \text{FTC}\]

$x=a, x=b$ are boundary points of interval $[a, b]$

$Diagram showing a smooth, directed curve C in space. The starting point is labeled $\vec{r}(a)$ and the endpoint is labeled $\vec{r}(b)$. An arrow indicates the orientation of the curve. A general position vector on the curve is labeled $\vec{r}(t)$. This represents the path for a line integral and its boundary points.$

Visual Description: Diagram showing a smooth, directed curve C in space. The starting point is labeled $\vec{r}(a)$ and the endpoint is labeled $\vec{r}(b)$. An arrow indicates the orientation of the curve. A general position vector on the curve is labeled $\vec{r}(t)$. This represents the path for a line integral and its boundary points.

\[\int_C \vec{\nabla}\phi \cdot d\vec{r} = \phi(\vec{r}(b)) - \phi(\vec{r}(a)) \quad \text{ FTCLI}\]

$\vec{r}(a)$ and $\vec{r}(b)$ are boundary points of curve $C$

Diagram of a two-dimensional region R enclosed by a closed curve C. The region R is indicated by diagonal blue hatching lines. The boundary curve defining the perimeter of the region is labeled C. This illustrates the domain for a double integral and its boundary for a line integral.

Visual Description: Diagram of a two-dimensional region R enclosed by a closed curve C. The region R is indicated by diagonal blue hatching lines. The boundary curve defining the perimeter of the region is labeled C. This illustrates the domain for a double integral and its boundary for a line integral.

\[\iint_R (g_x - f_y) dA = \oint_C \vec{F} \cdot \vec{T} ds \quad \text{Green's Theorem}\]

Curve $C$ is the boundary of Region $R$

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Example: Evaluation of a Line Integral

Evaluate the following line integral:

\[ \oint_{C} xy^2 \, dx + (3x + x^2y) \, dy \]

Here, the vector field is defined as $\vec{F} = \langle f, g \rangle = \langle xy^2, 3x + x^2y \rangle$, where $f = xy^2$ and $g = 3x + x^2y$.

A coordinate geometry diagram on a grid background showing a right-angled triangle in the first quadrant. The triangle's vertices are at the origin (0,0), point (1,0) on the x-axis, and point (0,1) on the y-axis. The interior region is shaded with blue diagonal lines and labeled 'R'. The boundary curve 'C' consists of three oriented segments: C1 is on the x-axis from (0,0) to (1,0) with a rightward arrow; C2 is a diagonal line from (1,0) to (0,1) with an arrow pointing toward (0,1); and C3 is on the y-axis from (0,1) back to the origin (0,0) with a downward arrow. This indicates a counter-clockwise orientation of the boundary C.

Visual Description: A coordinate geometry diagram on a grid background showing a right-angled triangle in the first quadrant. The triangle's vertices are at the origin (0,0), point (1,0) on the x-axis, and point (0,1) on the y-axis. The interior region is shaded with blue diagonal lines and labeled 'R'. The boundary curve 'C' consists of three oriented segments: C1 is on the x-axis from (0,0) to (1,0) with a rightward arrow; C2 is a diagonal line from (1,0) to (0,1) with an arrow pointing toward (0,1); and C3 is on the y-axis from (0,1) back to the origin (0,0) with a downward arrow. This indicates a counter-clockwise orientation of the boundary C.

Method 1: Using Parametrization

The total line integral over the closed curve $C$ is the sum of the line integrals over each individual segment:

\[ \oint_{C} \vec{F} \cdot d\vec{r} = \int_{C_1} \vec{F} \cdot d\vec{r} + \int_{C_2} \vec{F} \cdot d\vec{r} + \int_{C_3} \vec{F} \cdot d\vec{r} \]

Method 2: Using Green's Theorem

Applying Green's Theorem for a region $R$ with boundary $C$:

\[ \oint_{C} xy^2 \, dx + (3x + x^2y) \, dy = \oint_{C} \vec{F} \cdot d\vec{r} = \iint_{R} (g_x - f_y) \, dA \]

In this context, $g_x - f_y$ is the 2D curl of $\vec{F}$.

\[ = \iint_{R} \left( \frac{\partial}{\partial x}(3x + x^2y) - \frac{\partial}{\partial y}(xy^2) \right) \, dA \] \[ = \iint_{R} (3 + 2xy - 2xy) \, dA \] \[ = \iint_{R} 3 \, dA \]

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Application of Green's Theorem

2D Cartesian coordinate graph showing a triangular region R in the first quadrant, shaded with blue diagonal lines. The region R is bounded by the x-axis, the y-axis, and the diagonal line segment x + y = 1 (also labeled as y = 1 - x). The vertices of the triangle are at coordinates (0,0), (1,0), and (0,1). The boundary curve C is composed of three segments oriented counter-clockwise: C1 along the x-axis from (0,0) to (1,0), C2 along the hypotenuse from (1,0) to (0,1), and C3 along the y-axis from (0,1) back to (0,0). Green arrows along these segments indicate the direction of traversal. A vertical line segment is drawn within the region at a general position x to illustrate vertical slicing for integration.

Visual Description: 2D Cartesian coordinate graph showing a triangular region R in the first quadrant, shaded with blue diagonal lines. The region R is bounded by the x-axis, the y-axis, and the diagonal line segment x + y = 1 (also labeled as y = 1 - x). The vertices of the triangle are at coordinates (0,0), (1,0), and (0,1). The boundary curve C is composed of three segments oriented counter-clockwise: C1 along the x-axis from (0,0) to (1,0), C2 along the hypotenuse from (1,0) to (0,1), and C3 along the y-axis from (0,1) back to (0,0). Green arrows along these segments indicate the direction of traversal. A vertical line segment is drawn within the region at a general position x to illustrate vertical slicing for integration.

\[\oint_{C} xy^2 dx + (3x + x^2y) dy = \iint_{R} 3 dA = 3 \iint_{R} 1 dA\] \[= 3 (\text{Area of } R) = 3 \times \frac{1}{2}(1 \times 1) = \frac{3}{2}\]

OR, compute by finding bounds of $R$:

\[\iint_{R} 3 dA = \int_{0}^{1} \int_{0}^{1-x} 3 dy dx = \int_{0}^{1} 3(1-x) dx\] \[= 3 \left(x - \frac{x^2}{2}\right) \bigg|_{0}^{1} = \frac{3}{2}\]

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Example:

Find work done in $\vec{F} = \langle 2xy^3, 4x^2y^2 \rangle$ moving along the line segments from $(0,0)$ to $(1,0)$ and $(1,0)$ to $(1,1)$ and then moving back to $(0,0)$ along $y = x^3$.

A Cartesian coordinate system graph illustrating the path of integration C and the enclosed region R. The path C is a closed loop with counter-clockwise orientation, consisting of three segments: 1. A horizontal segment along the x-axis from (0,0) to (1,0). 2. A vertical segment from (1,0) up to the point (1,1). 3. A curved segment following the function y = x³ that returns from (1,1) to the origin (0,0). The region R enclosed by these segments is shaded with horizontal blue lines. Labels on the graph include the origin (0,0), points (1,0) and (1,1), the curve equation y = x³, the path label C, and the region label R.

Visual Description: A Cartesian coordinate system graph illustrating the path of integration C and the enclosed region R. The path C is a closed loop with counter-clockwise orientation, consisting of three segments: 1. A horizontal segment along the x-axis from (0,0) to (1,0). 2. A vertical segment from (1,0) up to the point (1,1). 3. A curved segment following the function y = x³ that returns from (1,1) to the origin (0,0). The region R enclosed by these segments is shaded with horizontal blue lines. Labels on the graph include the origin (0,0), points (1,0) and (1,1), the curve equation y = x³, the path label C, and the region label R.

The work done is given by the line integral: \[W = \oint_C \vec{F} \cdot d\vec{r} = \oint_C f \, dx + g \, dy\] where $f = 2xy^3$ and $g = 4x^2y^2$.

Applying Green's Theorem: \[\stackrel{\text{Green's}}{=} \iint_R (g_x - f_y) \, dA\] \[= \iint_R \left[ \frac{\partial}{\partial x}(4x^2y^2) - \frac{\partial}{\partial y}(2xy^3) \right] dA\] \[= \iint_R (8xy^2 - 6xy^2) \, dA\] \[= \iint_R 2xy^2 \, dA\]

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Example: Calculation of Work via Green's Theorem

A Cartesian coordinate graph on a grid illustrating a region R in the first quadrant. The region is bounded by the x-axis, the vertical line segment from (1,0) to (1,1), and the curve y = x^3. The interior of the region is labeled R and shaded with horizontal blue hatching lines. The boundary path C is indicated with a counter-clockwise orientation: an arrow points right along the x-axis (labeled C), another points up along the vertical line at x=1, and a third points down-left along the curve y = x^3 toward the origin. Points (0,0), (1,0), and (1,1) are labeled. A representative vertical slice is shown starting at coordinate x on the x-axis and extending up to a point on the curve y = x^3.

Visual Description: A Cartesian coordinate graph on a grid illustrating a region R in the first quadrant. The region is bounded by the x-axis, the vertical line segment from (1,0) to (1,1), and the curve y = x^3. The interior of the region is labeled R and shaded with horizontal blue hatching lines. The boundary path C is indicated with a counter-clockwise orientation: an arrow points right along the x-axis (labeled C), another points up along the vertical line at x=1, and a third points down-left along the curve y = x^3 toward the origin. Points (0,0), (1,0), and (1,1) are labeled. A representative vertical slice is shown starting at coordinate x on the x-axis and extending up to a point on the curve y = x^3.

Using Green's Theorem, the work $W$ is given by:

\[ W = \oint_{C} \vec{F} \cdot d\vec{r} = \iint_{R} (g_x - f_y) \, dA \] \[ = \iint_{R} 2xy^2 \, dA \]

We evaluate the double integral over the region $R$ by setting up iterated integrals with respect to $y$ then $x$:

\[ \iint_{R} 2xy^2 \, dA = \int_{0}^{1} \int_{0}^{x^3} 2xy^2 \, dy \, dx \] \[ = \int_{0}^{1} \left[ 2x \frac{y^3}{3} \right]_{0}^{x^3} \, dx \] \[ = \int_{0}^{1} 2 \frac{x^{10}}{3} \, dx \] \[ = \left. \frac{2 x^{11}}{33} \right|_{0}^{1} \] \[ = \frac{2}{33} \]

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Example:

A Cartesian coordinate system showing a shaded region R and its boundary curve C. The region R is bounded by the x-axis from x=0 to x=1, a vertical line at x=1 from y=0 to y=1, and a curved line representing y=x^3 from (1,1) back to the origin (0,0). The region is filled with blue horizontal shading lines. Key coordinates are labeled: (0,0) at the origin, (1,0) on the x-axis, and (1,1) where the curve meets the vertical boundary. The boundary curve C is oriented counter-clockwise, indicated by directional arrows. The equation y=x^3 is written next to the upper curved segment of the boundary.

Visual Description: A Cartesian coordinate system showing a shaded region R and its boundary curve C. The region R is bounded by the x-axis from x=0 to x=1, a vertical line at x=1 from y=0 to y=1, and a curved line representing y=x^3 from (1,1) back to the origin (0,0). The region is filled with blue horizontal shading lines. Key coordinates are labeled: (0,0) at the origin, (1,0) on the x-axis, and (1,1) where the curve meets the vertical boundary. The boundary curve C is oriented counter-clockwise, indicated by directional arrows. The equation y=x^3 is written next to the upper curved segment of the boundary.

\[ \vec{F} = \langle 2xy^3, 4x^2y^2 \rangle \] Find flux over the curve $C$. \[ \text{Flux} = \oint_{C} \vec{F} \cdot \vec{N} \, ds \] Applying **Green's Theorem**: \[ = \iint_{R} (2\text{D Divergence of } \vec{F}) \, dA \] Where the divergence components are calculated as: \[ f_x + g_y = 2y^3 + 8x^2y \] Substituting this into the double integral: \[ = \iint_{R} 2y^3 + 8x^2y \, dA \] Based on the region shown in the diagram, we set up the iterated integral with $x$ from 0 to 1 and $y$ from 0 to $x^3$: \[ = \int_{0}^{1} \int_{0}^{x^3} 2y^3 + 8x^2y \, dy \, dx \] Integrating with respect to $y$: \[ = \int_{0}^{1} \left[ \frac{y^4}{2} + 4x^2y^2 \right]_{y=0}^{y=x^3} \, dx \] Substituting the upper limit $y = x^3$: \[ = \int_{0}^{1} \frac{x^{12}}{2} + 4x^{8} \, dx \] Integrating with respect to $x$: \[ = \left[ \frac{x^{13}}{26} + \frac{4x^9}{9} \right]_{x=0}^{x=1} \] Evaluating at the limits: \[ = \frac{1}{26} + \frac{4}{9} \]