In our lecture on Green's Theorem, we interpreted the quantity $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ as the "micro-swirl" or 2D Curl of a vector field at a specific point. We saw that summing up these micro-swirls inside a region gave us the total circulation around the boundary.
Today, we ask: what happens if the field is three-dimensional? A vector field in space can rotate around any axis, not just the vertical one. To describe this, we need to upgrade our scalar "2D Curl" into a powerful vector operator: the 3D Curl.
We will also introduce a second machine, called Divergence, which measures how a field expands or contracts.
Throughout Calculus III, we have used the symbol $\nabla$ (del) to compute gradients. But $\nabla$ is not just for gradients. It is a vector operator:
$$ \nabla = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle $$
Think of $\nabla$ as a machine waiting for a command. It doesn't mean anything by itself. It needs to operate on a function or a vector field.
| Operator Name | Symbol | Input Type | Output Type |
|---|---|---|---|
| Gradient | $\nabla f$ | Scalar Function | Vector Field |
| Curl | $\nabla \times \mathbf{F}$ | Vector Field | Vector Field |
| Divergence | $\nabla \cdot \mathbf{F}$ | Vector Field | Scalar Function |
If $\mathbf{F} = \langle P, Q, R \rangle$ is a vector field, the Curl of $\mathbf{F}$ is the vector product of $\nabla$ and $\mathbf{F}$.
$$ \text{curl } \mathbf{F} = \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix} $$
When computing this determinant, students frequently miss the negative sign on the $\mathbf{j}$ component. Be careful!
$$ = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\mathbf{i} \color{red}{- \left(\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z}\right)\mathbf{j}} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\mathbf{k} $$
What does Curl actually measure? Imagine dropping a microscopic paddle wheel into a fluid flowing according to $\mathbf{F}$.
Interactive: animate slider s to see the vector field, slider t to see the paddle wheel "curl" and you can move the point $(x_1,y_1)$ to see the magnitude of the curl change
Look at the $\mathbf{k}$-component of the curl formula above: $\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\mathbf{k}$.
Does that look familiar? It is exactly the "2D Curl" scalar we used in Green's Theorem last week!
$$ \text{Green's Integrand} = (\text{curl } \mathbf{F}) \cdot \mathbf{k} $$
This tells us that Green's Theorem is really just a special case of 3D Curl where we only care about rotation around the vertical $z$-axis (the $\mathbf{k}$ direction).
Find $\text{curl } \mathbf{F}$ for $\mathbf{F}(x,y,z) = \langle xz, xyz, -y^2 \rangle$.
$$ \text{curl } \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ xz & xyz & -y^2 \end{vmatrix} $$
i-component: $\frac{\partial}{\partial y}(-y^2) - \frac{\partial}{\partial z}(xyz) = -2y - xy$.
j-component: $\frac{\partial}{\partial x}(-y^2) - \frac{\partial}{\partial z}(xz) = 0 - x = -x$. (Don't forget the extra minus! $\to +x$).
k-component: $\frac{\partial}{\partial x}(xyz) - \frac{\partial}{\partial y}(xz) = yz - 0 = yz$.
$$ \text{curl } \mathbf{F} = \langle -2y - xy, x, yz \rangle $$
Let's see what happens at the specific point $(1, 1, 1)$. Substituting $x=1, y=1, z=1$ into our result:
$$ \text{curl } \mathbf{F}(1,1,1) = \langle -2(1) - (1)(1), 1, (1)(1) \rangle = \langle -3, 1, 1 \rangle $$
What this means: If you placed a paddle wheel at $(1,1,1)$, the fluid would try to spin it. The axis of rotation would point in the direction of the vector $\langle -3, 1, 1 \rangle$.
Find $\text{curl } \mathbf{F}$ for $\mathbf{F}(x,y,z) = \langle \sin y, \cos z, \tan x \rangle$.
We set up the determinant:
$$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ \sin y & \cos z & \tan x \end{vmatrix} $$
$$ \text{curl } \mathbf{F} = \langle \sin z, -\sec^2 x, -\cos y \rangle $$
Calculate $\text{curl } \mathbf{F}$ for the field $\mathbf{F} = \langle 0, z, -y \rangle$.
$$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ 0 & z & -y \end{vmatrix} $$
Result: $\mathbf{\langle -2, 0, 0 \rangle}$. The field rotates purely around the x-axis.
The Divergence of $\mathbf{F}$ is the dot product of $\nabla$ and $\mathbf{F}$.
$$ \text{div } \mathbf{F} = \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} $$
Notice that there are no $\mathbf{i}, \mathbf{j}, \mathbf{k}$ vectors in the result. Divergence is a SCALAR function. It gives you a single number at every point, not a vector.
Imagine $\mathbf{F}$ represents the flow of a gas.
Find $\text{div } \mathbf{F}$ for the same field $\mathbf{F}(x,y,z) = \langle xz, xyz, -y^2 \rangle$.
$$ \frac{\partial}{\partial x}(xz) + \frac{\partial}{\partial y}(xyz) + \frac{\partial}{\partial z}(-y^2) $$
$$ = z + xz + 0 = z + xz $$
Let's evaluate this at the point $(1,1,1)$.
$$ \text{div } \mathbf{F}(1,1,1) = (1) + (1)(1) = \color{red}{2} $$
What this means: Since the result is positive ($2 > 0$), the fluid is expanding at this location. It acts like a "source," meaning more fluid is flowing out of a small region around this point than is flowing in.
Find the divergence of the magnetic field $\mathbf{B} = \langle y^2, z^2, x^2 \rangle$.
$$ \text{div } \mathbf{B} = \frac{\partial}{\partial x}(y^2) + \frac{\partial}{\partial y}(z^2) + \frac{\partial}{\partial z}(x^2) $$
$$ = 0 + 0 + 0 = \color{red}{0} $$
Interpretation: This field is incompressible.
Suppose a vector field $\mathbf{F}$ represents water flow. At point $P$, you calculate $\text{div } \mathbf{F}(P) = -5$. Is water being created (a source) or destroyed (a sink) at this point?
It is a Sink.
Because the divergence is negative ($-5 < 0$), the fluid is compressing or draining away at that point.
Just like $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ and $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0$, we have special identities for our differential operators.
Think of a gradient field as a landscape of hills. If you walk along the path of steepest ascent (the gradient vectors), you are always going up. It is impossible to walk in a closed loop (ending where you started) while constantly going uphill—unless you are in an M.C. Escher drawing! Therefore, a gradient field cannot have "spin" or circulation.
Let's verify these theorems with specific examples.
Let $f(x,y,z) = x^2 y z$. First, we compute the gradient:
$$ \nabla f = \langle 2xyz, x^2z, x^2y \rangle $$
Now, we take the curl of this gradient vector:
$$ \text{curl}(\nabla f) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ 2xyz & x^2z & x^2y \end{vmatrix} $$
Result: $\langle 0, 0, 0 \rangle = \mathbf{0}$. Verified.
Let $\mathbf{F} = \langle z, x, y \rangle$. First, compute the curl:
$$ \text{curl } \mathbf{F} = \langle (1-0), -(0-1), (1-0) \rangle = \langle 1, 1, 1 \rangle $$
Now, take the divergence of this result:
$$ \text{div}(\langle 1, 1, 1 \rangle) = \frac{\partial}{\partial x}(1) + \frac{\partial}{\partial y}(1) + \frac{\partial}{\partial z}(1) = 0 + 0 + 0 = 0 $$
Result: $0$. Verified.
Is the expression $\text{curl}(\text{div } \mathbf{F})$ valid? Why or why not?
No, it is Invalid.
$\text{div } \mathbf{F}$ results in a Scalar function. However, the Curl operator ($\nabla \times$) requires a Vector Field as an input. You cannot cross a vector with a scalar!
In Section 16.3, we learned that a 2D field $\mathbf{F} = \langle P, Q \rangle$ is conservative if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. We can now state the 3D version of this test using Curl.
If $\mathbf{F} = \langle P, Q, R \rangle$ is defined on a simply-connected region and $\text{curl } \mathbf{F} = \mathbf{0}$, then $\mathbf{F}$ is conservative. This requires all three components of the curl to be zero:
1. Curl about the x-axis ($\mathbf{i}$): $$ \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial z} $$
2. Curl about the y-axis ($\mathbf{j}$): $$ \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} $$
3. Curl about the z-axis ($\mathbf{k}$): $$ \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} $$
Notice that condition #3 is exactly the test we used for 2D fields!
Is $\mathbf{F} = \langle y^2 z^3, 2xyz^3, 3xy^2 z^2 \rangle$ a conservative vector field?
We compute the curl:
$$ \text{curl } \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ y^2 z^3 & 2xyz^3 & 3xy^2 z^2 \end{vmatrix} $$
Since $\text{curl } \mathbf{F} = \langle 0,0,0 \rangle$, the field is conservative.
Is $\mathbf{F} = \langle z, x, y \rangle$ conservative?
$$ \text{curl } \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ z & x & y \end{vmatrix} $$
We can stop right here. Since the first component is $1 \neq 0$, the curl cannot be the zero vector $\mathbf{0}$. Therefore, $\mathbf{F}$ is not conservative.
It is important to understand that Curl (Rotation) and Divergence (Expansion) are independent properties. A field can have one without the other.
| Incompressible (Div = 0) |
Compressible (Div $\neq$ 0) |
|
|---|---|---|
| Irrotational (Curl = 0) Conservative |
1. Uniform Flow $\mathbf{F} = \langle 1, 0, 0 \rangle$ No spin, no expansion. |
2. Pure Expansion $\mathbf{F} = \langle x, y, z \rangle$ Explodes outward, but particles don't rotate. |
| Rotational (Curl $\neq$ 0) Not Conservative |
3. Pure Rotation $\mathbf{F} = \langle -y, x, 0 \rangle$ Spins like a whirlpool, but water doesn't expand. |
4. Spiraling Expansion $\mathbf{F} = \langle -y+x, x+y, 0 \rangle$ Particles spin AND move outward (like a tornado). |
There is one final operator that is crucial for physics and engineering students: the Laplacian. It is formed by taking the "del dot del" of a function.
$$ \nabla^2 f = \nabla \cdot \nabla f = \text{div}(\nabla f) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} $$
While the Gradient measures slope and Curl measures rotation, the Laplacian measures how the value of a function at a specific point compares to the average value of its neighbors.
If $f(x,y,z) = x^3 + y^3 + z^3$, find $\nabla^2 f$.
First find the second derivatives:
$f_{xx} = \frac{\partial}{\partial x}(3x^2) = 6x$
$f_{yy} = \frac{\partial}{\partial y}(3y^2) = 6y$
$f_{zz} = \frac{\partial}{\partial z}(3z^2) = 6z$
$$ \nabla^2 f = 6x + 6y + 6z $$
A function is called harmonic if its Laplacian is zero. Show that $f(x,y,z) = x^2 + y^2 - 2z^2$ is harmonic.
$$ \nabla^2 f = \frac{\partial^2}{\partial x^2}(x^2) + \frac{\partial^2}{\partial y^2}(y^2) + \frac{\partial^2}{\partial z^2}(-2z^2) $$
$$ = 2 + 2 - 4 = 0 $$
Since $\nabla^2 f = 0$, the function is harmonic.
Finally, we can rewrite Green's Theorem using our new notation. This is excellent practice for identifying the physical meaning of Curl and Divergence.
Recall that the Line Integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ represents Work (or Circulation). It sums the component of the field that pushes along the curve (tangent to the boundary). Green's Theorem relates this "push along the edge" to the "curl inside the region."
However, there is another type of line integral called Flux. Instead of measuring how much fluid flows along the curve, Flux measures how much fluid flows across the curve (normal to the boundary).
We use the symbol $\mathbf{n}$ for the outward normal vector. The flux integral is $\oint \mathbf{F} \cdot \mathbf{n} \, ds$. Green's Theorem has a form for this too!
This "Flux Form" is physically intuitive: The total amount of fluid leaving the boundary equals the sum of all the "sources" (divergence) inside the region.
Look at the vector field $\mathbf{F} = \langle -y, x \rangle$ (a whirlpool spinning counter-clockwise). Without calculating, what can you say about its Divergence and Curl?
Divergence is 0: Nothing is being created or destroyed; the water just spins. It doesn't expand.
Curl is Positive (k): A paddle wheel placed in this fluid would spin counter-clockwise. Specifically, $\text{curl } \mathbf{F} = (1 - (-1))\mathbf{k} = 2\mathbf{k}$.
Which of the following operations makes sense?
1. Invalid. $\text{div } \mathbf{F}$ is a scalar. You cannot take the curl of a scalar.
2. Valid. $\text{div } \mathbf{F}$ is a scalar. You CAN take the gradient of a scalar (resulting in a vector).
If $\text{curl } \mathbf{F} = \mathbf{0}$ everywhere, what is the value of $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $C$ is the unit circle?
The integral is 0.
If $\text{curl } \mathbf{F} = \mathbf{0}$, the field is Conservative. The line integral of a conservative field over any closed path is always zero.
Calculate the divergence of $\mathbf{F} = \langle x, y, z \rangle$ at the point $(2, 2, 2)$. Is the fluid expanding or contracting?
$$ \text{div } \mathbf{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 1 + 1 + 1 = \color{red}{3} $$
Interpretation: Since $3 > 0$, the fluid is expanding.