Lesson 30: Curl and Divergence (17.5)

Accessible transcription generated on 4/3/2026

Original Notes

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Lesson 30: Curl and Divergence (17.5)

Review: Computing \(\int_C \vec{F} \cdot d\vec{r} = \int_C \vec{F} \cdot \vec{T} \, ds = \int_C f \, dx + g \, dy\)

The method to compute the line integral depends on a primary question: Is \(C\) a Simple Closed Curve in 2D?

Case: Yes

If the curve \(C\) is a simple closed curve, we apply Green's Theorem to relate the line integral to a double integral over the region \(R\) enclosed by \(C\).

A hand-drawn diagram illustrating the geometric domain for Green's Theorem. It shows an irregularly shaped region labeled R, shaded with purple diagonal parallel lines. The region R is enclosed by a simple closed curve labeled C. An arrow on curve C indicates a counter-clockwise (positive) orientation.
Visual Description: A hand-drawn diagram illustrating the geometric domain for Green's Theorem. It shows an irregularly shaped region labeled R, shaded with purple diagonal parallel lines. The region R is enclosed by a simple closed curve labeled C. An arrow on curve C indicates a counter-clockwise (positive) orientation.

Tangential form of Green's Theorem:

\[ \oint_C \vec{F} \cdot \vec{T} \, ds = \iint_R (g_x - f_y) \, dA \]

The integrand \((g_x - f_y)\) is known as the Scalar Curl.

Normal form of Green's Theorem:

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

The integrand \((f_x + g_y)\) is known as div \(\vec{F}\) (Divergence of \(\vec{F}\)).

Case: No

If the curve \(C\) is not a simple closed curve, the next question is: Is the vector field \(\vec{F}\) conservative? (i.e., does there exist a potential function \(\phi\) such that \(\vec{F} = \nabla \phi\)?)

  • If Yes: Apply FTCLI (Fundamental Theorem of Calculus for Line Integrals) \[ \int_C \nabla \phi \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) \]
  • If No: Use the Definition of a line integral \[ \int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \]

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Worked Example: Line Integrals and Green's Theorem

Given the vector field \(\vec{F} = \langle \underbrace{-y^3 + 9}_{f}, \underbrace{x^3 + 3y^2 + 1}_{g} \rangle\) and the curve \(C\) defined by \(\vec{r}(t) = \langle \cos t, \sin t \rangle\) for \(0 \le t \le 2\pi\), evaluate the line integrals \(\oint_C \vec{F} \cdot \vec{T} \, ds\) and \(\oint_C \vec{F} \cdot \vec{N} \, ds\).

Note that \(C\) is a circle of radius 1 and is a simple closed curve.

A 2D plot on a Cartesian coordinate system showing a unit circle centered at the origin. The boundary of the circle is labeled C and has a counter-clockwise orientation arrow. The interior of the circle, representing the region R, is shaded with diagonal purple lines.
Visual Description: A 2D plot on a Cartesian coordinate system showing a unit circle centered at the origin. The boundary of the circle is labeled C and has a counter-clockwise orientation arrow. The interior of the circle, representing the region R, is shaded with diagonal purple lines.
1. Evaluating \(\oint_C \vec{F} \cdot \vec{T} \, ds\) (Circulation form of Green's Theorem)

Using Green's Theorem for circulation, we have: \[\oint_C \vec{F} \cdot \vec{T} \, ds = \iint_R (g_x - f_y) \, dA\]

First, calculate the partial derivatives: \[g_x = \frac{\partial}{\partial x}(x^3 + 3y^2 + 1) = 3x^2\] \[f_y = \frac{\partial}{\partial y}(-y^3 + 9) = -3y^2\]

Substituting these into the double integral: \[\iint_R (3x^2 - (-3y^2)) \, dA = \iint_R 3(x^2 + y^2) \, dA\]

Converting to polar coordinates for the region \(R\) (where \(0 \le r \le 1\) and \(0 \le \theta \le 2\pi\)): \[= \int_0^{2\pi} \int_0^1 3(r^2) \cdot r \, dr \, d\theta\] \[= \left. \frac{3r^4}{4} \right|_0^1 \cdot \left. \theta \right|_0^{2\pi}\] \[= \frac{3}{4} \cdot 2\pi = \frac{3\pi}{2}\]

2. Evaluating \(\oint_C \vec{F} \cdot \vec{N} \, ds\) (Flux form of Green's Theorem)

Using Green's Theorem for flux, we have: \[\oint_C \vec{F} \cdot \vec{N} \, ds = \iint_R (f_x + g_y) \, dA\]

Calculate the partial derivatives: \[f_x = \frac{\partial}{\partial x}(-y^3 + 9) = 0\] \[g_y = \frac{\partial}{\partial y}(x^3 + 3y^2 + 1) = 6y\]

Substituting these into the double integral: \[\iint_R (0 + 6y) \, dA = \iint_R 6y \, dA\]

Converting to polar coordinates (where \(y = r \sin \theta\)): \[= \int_0^{2\pi} \int_0^1 (6r \sin \theta) \cdot r \, dr \, d\theta\] \[= \left. 2r^3 \right|_0^1 \cdot \left. -\cos \theta \right|_0^{2\pi}\] \[= 2(1 - 0) \cdot (-(\cos 2\pi - \cos 0))\] \[= 2 \cdot -(1 - 1) = 0\]

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Curl and Divergence of a Vector Field \(\vec{F} = \langle f, g, h \rangle\)

Notation: \(\vec{\nabla} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle \rightarrow\) "del" is an operator.

\[\text{grad}(\phi) = \vec{\nabla}\phi = \left\langle \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}, \frac{\partial\phi}{\partial z} \right\rangle\]

In this operation, \(\phi\) is a scalar input, and the resulting gradient is a vector.

\[\text{Curl}(\vec{F}) = \vec{\nabla} \times \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f & g & h \end{vmatrix} = \langle h_y - g_z, -(h_x - f_z), g_x - f_y \rangle\]

Key Notes:

  • Takes a vector input (\(\vec{F}\)).
  • Often described as the "vector derivative of \(\vec{F}\)" or "Circulation Density."
  • The output is a vector.

\[\text{div}(\vec{F}) = \vec{\nabla} \cdot \vec{F} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle \cdot \langle f, g, h \rangle = f_x + g_y + h_z\]

Key Notes:

  • Often described as the "Scalar derivative of \(\vec{F}\)" or "Flux density."
  • The result is a Scalar output.

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Example

\[ \vec{F} = \langle -y, x, 0 \rangle \]

A vector field plot on a 2D Cartesian coordinate system representing \(\vec{F} = \langle -y, x, 0 \rangle\). The diagram shows blue arrows indicating the direction and magnitude of the field: in the upper-left quadrant, arrows point left; in the lower-left, they point down; in the lower-right, they point right; and in the upper-right, they point up, overall creating a counter-clockwise circulation around the origin. A red rectangular object, labeled as a 'Log', is placed diagonally across the origin with internal hatching. Below the plot, the text notes 'Log Rotates' and 'Does not Drift', signifying that the vector field causes rotation but has no net diverging or converging flow at the center.
Visual Description: A vector field plot on a 2D Cartesian coordinate system representing \(\vec{F} = \langle -y, x, 0 \rangle\). The diagram shows blue arrows indicating the direction and magnitude of the field: in the upper-left quadrant, arrows point left; in the lower-left, they point down; in the lower-right, they point right; and in the upper-right, they point up, overall creating a counter-clockwise circulation around the origin. A red rectangular object, labeled as a 'Log', is placed diagonally across the origin with internal hatching. Below the plot, the text notes 'Log Rotates' and 'Does not Drift', signifying that the vector field causes rotation but has no net diverging or converging flow at the center.

Log Rotates
Does not Drift

\[ \text{curl } \vec{F} = \nabla \times \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & 0 \end{vmatrix} \] \[ = \langle 0, 0, 2 \rangle \]

\(\text{curl } \vec{F} \neq \langle 0, 0, 0 \rangle \rightsquigarrow\) Rotating

\[ \text{div } \vec{F} = \nabla \cdot \vec{F} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle \cdot \langle -y, x, 0 \rangle \] \[ = 0 \]

\(\text{div } \vec{F} = 0 \rightsquigarrow\) Source free, incompressible, divergence free

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Example: Analyzing a Vector Field

Consider the vector field defined by:

\[ \vec{F} = \langle x, y, 0 \rangle \]
A vector field plot on a Cartesian coordinate system representing \(\vec{F} = \langle x, y, 0 \rangle\). The diagram features blue arrows radiating outward from the origin in all directions. Specific arrows are drawn along the x and y axes and the diagonal lines \(y=x\) and \(y=-x\). A red rectangle, representing a 'log', is placed diagonally across the origin along the line \(y=x\). The rectangle contains red cross-hatching to indicate its physical extent. The radial nature of the arrows suggests an expanding field with no swirling or circular motion around the origin.
Visual Description: A vector field plot on a Cartesian coordinate system representing \(\vec{F} = \langle x, y, 0 \rangle\). The diagram features blue arrows radiating outward from the origin in all directions. Specific arrows are drawn along the x and y axes and the diagonal lines \(y=x\) and \(y=-x\). A red rectangle, representing a 'log', is placed diagonally across the origin along the line \(y=x\). The rectangle contains red cross-hatching to indicate its physical extent. The radial nature of the arrows suggests an expanding field with no swirling or circular motion around the origin.

To determine the rotational properties of the field, we calculate the curl:

\[ \text{curl } \vec{F} = \vec{\nabla} \times \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & 0 \end{vmatrix} = \langle 0, 0, 0 \rangle \]
Conclusion: Since \(\text{curl } \vec{F} = \vec{0}\), the vector field is irrotational or curl free.

Next, we calculate the divergence of the field:

\[ \text{div } \vec{F} = \vec{\nabla} \cdot \vec{F} = \frac{\partial}{\partial x} x + \frac{\partial}{\partial y} y + \frac{\partial}{\partial z} 0 = 1 + 1 + 0 = 2 \]

"Log Drifts"

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Example: Vector Operations

Given a vector field \(\vec{F}\), which of the following does not make sense?

  1. \(\text{grad } \vec{F} = \vec{\nabla} \vec{F}\) (Does not make sense)
  2. \(\text{curl } \vec{F} = \vec{\nabla} \times \vec{F}\) (Makes sense)
  3. \(\text{div } \vec{F} = \vec{\nabla} \cdot \vec{F}\) (Makes sense)
  4. \(\text{grad}(\text{curl } \vec{F})\) (Does not make sense)
  5. \(\text{curl}(\text{curl } \vec{F})\) (Makes sense)
  6. \(\text{curl}(\text{div } \vec{F})\) (Does not make sense)
  7. \(\text{div}(\text{curl } \vec{F})\) (Makes sense)
  8. \(\text{div}(\text{div } \vec{F})\) (Does not make sense)
  9. \(\text{grad}(\text{div } \vec{F})\) (Makes sense)

Note: Operation input and output types:

\(\text{grad } [\text{scalar}] = \text{vector}\)

\(\text{curl } [\text{vector}] = \text{vector}\)

\(\text{div } [\text{vector}] = \text{scalar}\)

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Vector Calculus Operations Example

Example: \(\vec{F} = \langle x^2y, ze^x, xyz \rangle\)

Compute \(\text{curl } \vec{F}\), \(\text{div } \vec{F}\), and \(\text{grad}(\text{div } \vec{F})\)

To compute \(\text{curl } \vec{F}\):

\[\text{curl } \vec{F} = \vec{\nabla} \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ x^2y & ze^x & xyz \end{vmatrix}\] \[= \left\langle \frac{\partial}{\partial y}(xyz) - \frac{\partial}{\partial z}(ze^x), -\left( \frac{\partial}{\partial x}(xyz) - \frac{\partial}{\partial z}(x^2y) \right), \frac{\partial}{\partial x}(ze^x) - \frac{\partial}{\partial y}(x^2y) \right\rangle\] \[= \langle xz - e^x, -yz, ze^x - x^2 \rangle\]

To compute \(\text{div } \vec{F}\):

\[\text{div } \vec{F} = \vec{\nabla} \cdot \vec{F} = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial y}(ze^x) + \frac{\partial}{\partial z}(xyz) = 2xy + 0 + xy = 3xy\]

To compute \(\text{grad}(\text{div } \vec{F})\):

\[\text{grad}(\text{div } \vec{F}) = \vec{\nabla}(3xy) = \left\langle \frac{\partial}{\partial x} 3xy, \frac{\partial}{\partial y} 3xy, \frac{\partial}{\partial z} 3xy \right\rangle = \langle 3y, 3x, 0 \rangle\]

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Vector Calculus Identities

\[ \text{div}(\text{curl } \vec{F}) = 0 \]

Consider the specific vector field:

\[ \vec{F} = \langle x^2 y, z e^x, x y z \rangle \]

First, calculate the curl of \( \vec{F} \):

\[ \text{curl } \vec{F} = \langle x z - e^x, -y z, z e^x - x^2 \rangle \]

Then, calculate the divergence of that curl:

\[ \text{div}(\text{curl } \vec{F}) = \vec{\nabla} \cdot (\text{curl } \vec{F}) = \vec{\nabla} \cdot (\vec{\nabla} \times \vec{F}) \] \[ = \frac{\partial}{\partial x}(x z - e^x) + \frac{\partial}{\partial y}(-y z) + \frac{\partial}{\partial z}(z e^x - x^2) \] \[ = z - e^x + (-z) + e^x = 0 \]

General Derivation

To prove this identity for any smooth vector field, let:

\[ \vec{F} = \langle f, g, h \rangle \]

The curl of \( \vec{F} \) is expressed as:

\[ \vec{\nabla} \times \vec{F} = \text{curl } \vec{F} = \langle h_y - g_z, -(h_x - f_z), g_x - f_y \rangle \]

Next, we take the divergence of the resulting vector:

\[ \text{div}(\text{curl } \vec{F}) = \frac{\partial}{\partial x}(h_y - g_z) + \frac{\partial}{\partial y}(-h_x + f_z) + \frac{\partial}{\partial z}(g_x - f_y) \] \[ = h_{yx} - g_{zx} - h_{xy} + f_{zy} + g_{xz} - f_{yz} \]

By Clairaut's Theorem, for functions with continuous second partial derivatives, the mixed partials are equal (e.g., \( h_{yx} = h_{xy} \)). Applying this to the terms above leads to total cancellation:

\[ = 0 \]

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Example: Vector Operations on Scalar Functions

Given a scalar function \(\phi\), which of these makes sense?

1. \(\text{grad } \phi\) ✓

2. \(\text{div } \phi\) ✗

3. \(\text{curl } \phi\) ✗

4. \(\text{curl}(\text{grad } \phi)\) ✓

5. \(\text{div}(\text{curl } \phi)\) ✗

6. \(\text{div}(\text{grad } \phi)\) ✓

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Example

\(\phi = x^2 + yz^3\), find \(grad \, \phi\), \(div \, \phi\), \(curl \, \phi\)

Note: In the original handwritten notes, \(div \, \phi\) and \(curl \, \phi\) are marked with red "X"s, as these operations are not defined for scalar fields.

\(grad \, \phi = \vec{\nabla}\phi = \langle 2x, z^3, 3yz^2 \rangle\)

\(div(grad \, \phi) = \vec{\nabla} \cdot \vec{\nabla}\phi = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle \cdot \langle 2x, z^3, 3yz^2 \rangle\)
\(= 2 + 0 + 6yz\)

\(div(grad \, \phi) = \vec{\nabla} \cdot \vec{\nabla}\phi = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle \cdot \langle \phi_x, \phi_y, \phi_z \rangle\)
\(= \phi_{xx} + \phi_{yy} + \phi_{zz}\)

Notation:

\[div(grad \, \phi) = \vec{\nabla} \cdot \vec{\nabla}\phi = \Delta \phi\] Laplacian

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Curl of a Gradient

Given a specific gradient vector field:

\[ \text{grad } \phi = \vec{\nabla} \phi = \langle 2x, z^3, 3yz^2 \rangle \]

The curl of this gradient is calculated using the cross product of the del operator and the gradient vector:

\[ \text{curl}(\text{grad } \phi) = \vec{\nabla} \times \vec{\nabla} \phi = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2x & z^3 & 3yz^2 \end{vmatrix} \] \[ = \langle 3z^2 - 3z^2, -(0 - 0), 0 - 0 \rangle \] \[ = \langle 0, 0, 0 \rangle \]
\[ \text{curl}(\text{grad } \phi) = \langle 0, 0, 0 \rangle \]

General Case Derivation:

For any scalar field \(\phi\) with continuous second-order partial derivatives, the curl of the gradient is always the zero vector. Using subscript notation for partial derivatives:

\[ \vec{\nabla} \times \text{grad } \phi = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \phi_x & \phi_y & \phi_z \end{vmatrix} = \langle \phi_{zy} - \phi_{yz}, -(\phi_{zx} - \phi_{xz}), \phi_{yx} - \phi_{xy} \rangle \]

By Clairaut's Theorem (equality of mixed partials), each component simplifies to zero:

\[ = \langle 0, 0, 0 \rangle \]