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Divergence and Curl

Determine divergence from the formula for a given vector field.

Divergence

Divergence is an operation on a vector field that tells us how the field behaves toward or away from a point. Locally, the divergence of a vector field F in \({ℝ}^{2}\) or \({ℝ}^{3}\) at a particular point P is a measure of the “outflowing-ness” of the vector field at P. If F represents the velocity of a fluid, then the divergence of F at P measures the net rate of change with respect to time of the amount of fluid flowing away from P (the tendency of the fluid to flow “out of” P). In particular, if the amount of fluid flowing into P is the same as the amount flowing out, then the divergence at P is zero.

Note the divergence of a vector field is not a vector field, but a scalar function. In terms of the gradient operator \(∇=〈\frac{∂}{∂x},\frac{∂}{∂y},\frac{∂}{∂z}〉,\) divergence can be written symbolically as the dot product

\[\text{div}\ \text{F}=∇\cdot \text{F}.\]

Note this is merely helpful notation, because the dot product of a vector of operators and a vector of functions is not meaningfully defined given our current definition of dot product.

If \(\text{F}=〈P,Q〉\) is a vector field in \({ℝ}^{2},\) and \({P}_{x}\) and \({Q}_{y}\) both exist, then the divergence of F is defined similarly as

\[\text{div}\ \text{F}={P}_{x}+{Q}_{y}=\frac{∂P}{∂x}+\frac{∂Q}{∂y}=∇\cdot \text{F}.\]

To illustrate this point, consider the two vector fields in . At any particular point, the amount flowing in is the same as the amount flowing out, so at every point the “outflowing-ness” of the field is zero. Therefore, we expect the divergence of both fields to be zero, and this is indeed the case, as

\[\text{div}(〈1,2〉)=\frac{∂}{∂x}(1)+\frac{∂}{∂y}(2)=0\ \text{and}\ \text{div}(〈\text{-}y,x〉)=\frac{∂}{∂x}(\text{-}y)+\frac{∂}{∂y}(x)=0.\]

By contrast, consider radial vector field \(\text{R}(x,y)=〈\text{-}x,\text{-}y〉\) in . At any given point, more fluid is flowing in than is flowing out, and therefore the “outgoingness” of the field is negative. We expect the divergence of this field to be negative, and this is indeed the case, as \(\text{div}(\text{R})=\frac{∂}{∂x}(\text{-}x)+\frac{∂}{∂y}(\text{-}y)=-2.\)

Condensed — the full section is in OpenStax Calculus Volume 3.

Curl

The second operation on a vector field that we examine is the curl, which measures the extent of rotation of the field about a point. Suppose that F represents the velocity field of a fluid. Then, the curl of F at point P is a vector that measures the tendency of particles near P to rotate about the axis that points in the direction of this vector. The magnitude of the curl vector at P measures how quickly the particles rotate around this axis. In other words, the curl at a point is a measure of the vector field’s “spin” at that point. Visually, imagine placing a paddlewheel into a fluid at P, with the axis of the paddlewheel aligned with the curl vector (). The curl measures the tendency of the paddlewheel to rotate.

Consider the vector fields in . In part (a), the vector field is constant and there is no spin at any point. Therefore, we expect the curl of the field to be zero, and this is indeed the case. Part (b) shows a rotational field, so the field has spin. In particular, if you place a paddlewheel into a field at any point so that the axis of the wheel is perpendicular to a plane, the wheel rotates counterclockwise. Therefore, we expect the curl of the field to be nonzero, and this is indeed the case (the curl is \(2\text{k}).\)

To see what curl is measuring globally, imagine dropping a leaf into the fluid. As the leaf moves along with the fluid flow, the curl measures the tendency of the leaf to rotate. If the curl is zero, then the leaf doesn’t rotate as it moves through the fluid.

The definition of curl can be difficult to remember. To help with remembering, we use the notation \(∇\ \times \ \text{F}\) to stand for a “determinant” that gives the curl formula:

\[|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z} \\ P & Q & R\end{array}|.\]

The determinant of this matrix is

\[({R}_{y}-{Q}_{z})\text{i}-({R}_{x}-{P}_{z})\text{j}+({Q}_{x}-{P}_{y})\text{k}=({R}_{y}-{Q}_{z})\text{i}+({P}_{z}-{R}_{x})\text{j}+({Q}_{x}-{P}_{y})\text{k}=\text{curl}\ \text{F}.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Using Divergence and Curl

Now that we understand the basic concepts of divergence and curl, we can discuss their properties and establish relationships between them and conservative vector fields.

If F is a vector field in \({ℝ}^{3},\) then the curl of F is also a vector field in \({ℝ}^{3}.\) Therefore, we can take the divergence of a curl. The next theorem says that the result is always zero. This result is useful because it gives us a way to show that some vector fields are not the curl of any other field. To give this result a physical interpretation, recall that divergence of a velocity field v at point P measures the tendency of the corresponding fluid to flow out of P. Since \(\text{div}\ \text{curl}\ (\text{v})=0,\) the net rate of flow in vector field curl(v) at any point is zero. Taking the curl of vector field F eliminates whatever divergence was present in F.

By the definitions of divergence and curl, and by Clairaut’s theorem,

\[\begin{array}{ll}\text{div curl}\ \text{F} & =\text{div}[({R}_{y}-{Q}_{z})\text{i}+({P}_{z}-{R}_{x})\text{j}+({Q}_{x}-{P}_{y})\text{k}] \\ & ={R}_{yx}-{Q}_{xz}+{P}_{yz}-{R}_{yx}+{Q}_{zx}-{P}_{zy} \\ & =0.\end{array}\]

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Example

Try it.

Show that \(\text{F}(x,y,z)={e}^{x}\text{i}+yz\text{j}+x{z}^{2}\text{k}\) is not the curl of another vector field. That is, show that there is no other vector G with \(\text{curl}\ \text{G}=\text{F}.\)

Solution

Notice that the domain of F is all of \({ℝ}^{3}\) and the second-order partials of F are all continuous. Therefore, we can apply the previous theorem to F.

The divergence of F is \({e}^{x}+z+2xz.\) If F were the curl of vector field G, then \(\text{div}\ \text{F}=\text{div curl}\ \text{G}=0.\) But, the divergence of F is not zero, and therefore F is not the curl of any other vector field.

With the next two theorems, we show that if F is a conservative vector field then its curl is zero, and if the domain of F is simply connected then the converse is also true. This gives us another way to test whether a vector field is conservative.

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

The divergence of a vector field is a scalar function. Divergence measures the “outflowing-ness” of a vector field. If v is the velocity field of a fluid, then the divergence of v at a point is the outflow of the fluid less the inflow at the point.
The curl of a vector field is a vector field. The curl of a vector field at point P measures the tendency of particles at P to rotate about the axis that points in the direction of the curl at P.
A vector field with a simply connected domain is conservative if and only if its curl is zero.

Key Equations

Curl	\(∇\ \times \ \text{F}=({R}_{y}-{Q}_{z})\text{i}+({P}_{z}-{R}_{x})\text{j}+({Q}_{x}-{P}_{y})\text{k}\)
Divergence	\(∇\cdot \text{F}={P}_{x}+{Q}_{y}+{R}_{z}\)
Divergence of curl is zero	\(∇\cdot (∇\ \times \ \text{F})=0\)
Curl of a gradient is the zero vector	\(∇\ \times \ (\text{∇}f)=0\)

Divergence and Curl

For the following exercises, determine whether the statement is true or false.

For the following exercises, find the curl of F.

For the following exercises, find the divergence of F.

For the following exercises, determine whether each of the given scalar functions is harmonic.

For the following exercises, use \(r=||\text{r}||\) and \(\text{r}=\left\.\)

For the following exercises, use a computer algebra system to find the curl of the given vector fields.

For the following exercises, find the divergence of F at the given point.

Condensed — the full section is in OpenStax Calculus Volume 3.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

If \(\text{F}(x,y,z)={e}^{x}\text{i}+yz\text{j}-y{z}^{2}\text{k},\) then find the divergence of F at \((0,2,-1).\)

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The divergence of F is
\[\frac{∂}{∂x}({e}^{x})+\frac{∂}{∂y}(yz)-\frac{∂}{∂z}(y{z}^{2})={e}^{x}+z-2yz.\]
Therefore, the divergence at \((0,2,-1)\) is \({e}^{0}-1+4=4.\) If F represents the velocity of a fluid, then more fluid is flowing out than flowing in at point \((0,2,-1).\)
Find \(\text{div}\ \text{F}\) for \(\text{F}(x,y,z)=〈xy,5-{z}^{2}y,{x}^{2}+{y}^{2}〉.\)

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\(y-{z}^{2}\)
Is it possible for \(\text{F}(x,y)=〈{x}^{2}y,y-x{y}^{2}〉\) to be a magnetic field?

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If F were magnetic, then its divergence would be zero. The divergence of F is
\[\frac{∂}{∂x}({x}^{2}y)+\frac{∂}{∂y}(y-x{y}^{2})=2xy+1-2xy=1\]
and therefore F cannot model a magnetic field ().
Is field \(\text{F}(x,y)=〈{x}^{2}y,5-x{y}^{2}〉\) source free?

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Note the domain of F is \({ℝ}^{2},\) which is simply connected. Furthermore, F is continuous with differentiable component functions. Therefore, we can use to analyze F. The divergence of F is
\[\frac{∂}{∂x}({x}^{2}y)+\frac{∂}{∂y}(5-x{y}^{2})=2xy-2xy=0.\]
Therefore, F is source free by .
Let \(\text{F}(x,y)=〈\text{-}ay,bx〉\) be a rotational field where a and b are positive constants. Is F source free?

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Yes
Suppose \(\text{v}(x,y)=〈\text{-}xy,y〉,y>0\) models the flow of a fluid. Is more fluid flowing into point \((1,4)\) than flowing out?

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To determine whether more fluid is flowing into \((1,4)\) than is flowing out, we calculate the divergence of v at \((1,4)\text{:}\)
\[\text{div}(\text{v})=\frac{∂}{∂x}(\text{-}xy)+\frac{∂}{∂y}(y)=\text{-}y+1.\]
To find the divergence at \((1,4),\) substitute the point into the divergence: \(-4+1=-3.\) Since the divergence of v at \((1,4)\) is negative, more fluid is flowing in than flowing out ().
For vector field \(\text{v}(x,y)=〈\text{-}xy,y〉,y>0,\) find all points P such that the amount of fluid flowing in to P equals the amount of fluid flowing out of P.

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All points on line \(y=1.\)
Find the curl of \(\text{F}(P,Q,R)=〈{x}^{2}z,{e}^{y}+xz,xyz〉.\)

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The curl is
\[\begin{array}{llllllll}\text{curl}\ \text{F} & =∇\ \times \ \text{F} \\ & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ ∂\text{/}∂x & ∂\text{/}∂y & ∂\text{/}∂z \\ P & Q & R\end{array}| \\ & =({R}_{y}-{Q}_{z})\text{i}+({P}_{z}-{R}_{x})\text{j}+({Q}_{x}-{P}_{y})\text{k} \\ & =(xz-x)\text{i}+({x}^{2}-yz)\text{j}+z\text{k}.\end{array}\]
Find the curl of \(\text{F}=〈\text{sin}\ x\ \text{cos}\ z,\text{sin}\ y\ \text{sin}\ z,\text{cos}\ x\ \text{cos}\ y〉\) at point \((0,\frac{\pi }{2},\frac{\pi }{2}).\)

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\(\text{-}\text{i}\)
Find the curl of \(\text{F}=〈P,Q〉=〈y,0〉.\)

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Notice that this vector field consists of vectors that are all parallel. In fact, each vector in the field is parallel to the x-axis. This fact might lead us to the conclusion that the field has no spin and that the curl is zero. To test this theory, note that
\[\text{curl}\ \text{F}=({Q}_{x}-{P}_{y})\text{k}=\text{-}\text{k}\ne 0.\]
Therefore, this vector field does have spin. To see why, imagine placing a paddlewheel at any point in the first quadrant (). The larger magnitudes of the vectors at the top of the wheel cause the wheel to rotate. The wheel rotates in the clockwise (negative) direction, causing the coefficient of the curl to be negative.
Show that a gravitational field has no spin.

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To show that F has no spin, we calculate its curl. Let \(P(x,y,z)=\frac{x}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}},\) \(Q(x,y,z)=\frac{y}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}},\) and \(R(x,y,z)=\frac{z}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}}.\) Then,
\[\begin{array}{ll}\text{curl}\ \text{F} & =\text{-}G{m}_{1}{m}_{2}[({R}_{y}-{Q}_{z})\text{i}+({P}_{z}-{R}_{x})\text{j}+({Q}_{x}-{P}_{y})\text{k}] \\ & =\text{-}G{m}_{1}{m}_{2}[\begin{array}{l}(\frac{-3yz}{{({x}^{2}+{y}^{2}+{z}^{2})}^{5\text{/}2}}-(\frac{-3yz}{{({x}^{2}+{y}^{2}+{z}^{2})}^{5\text{/}2}}))\text{i} \\ +(\frac{-3xz}{{({x}^{2}+{y}^{2}+{z}^{2})}^{5\text{/}2}}-(\frac{-3xz}{{({x}^{2}+{y}^{2}+{z}^{2})}^{5\text{/}2}}))\text{j} \\ +(\frac{-3xy}{{({x}^{2}+{y}^{2}+{z}^{2})}^{5\text{/}2}}-(\frac{-3xy}{{({x}^{2}+{y}^{2}+{z}^{2})}^{5\text{/}2}}))\text{k}\end{array}] \\ & =0.\end{array}\]
Since the curl of the gravitational field is zero, the field has no spin.
Field \(\text{v}(x,y)=〈-\frac{y}{{x}^{2}+{y}^{2}},\frac{x}{{x}^{2}+{y}^{2}}〉\) models the flow of a fluid. Show that if you drop a leaf into this fluid, as the leaf moves over time, the leaf does not rotate.

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\(\text{curl}\ \text{v}=0\)
Show that \(\text{F}(x,y,z)={e}^{x}\text{i}+yz\text{j}+x{z}^{2}\text{k}\) is not the curl of another vector field. That is, show that there is no other vector G with \(\text{curl}\ \text{G}=\text{F}.\)

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Notice that the domain of F is all of \({ℝ}^{3}\) and the second-order partials of F are all continuous. Therefore, we can apply the previous theorem to F.
The divergence of F is \({e}^{x}+z+2xz.\) If F were the curl of vector field G, then \(\text{div}\ \text{F}=\text{div curl}\ \text{G}=0.\) But, the divergence of F is not zero, and therefore F is not the curl of any other vector field.
Is it possible for \(\text{G}(x,y,z)=〈\text{sin}\ x,\text{cos}\ y,\text{sin}(xyz)〉\) to be the curl of a vector field?

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No
Use the curl to determine whether \(\text{F}(x,y,z)=〈yz,xz,xy〉\) is conservative.

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Note that the domain of F is all of \({ℝ}^{3},\) which is simply connected (). Therefore, we can test whether F is conservative by calculating its curl.
The curl of F is
\[(\frac{∂}{∂y}xy-\frac{∂}{∂z}xz)\text{i}+(\frac{∂}{∂y}yz-\frac{∂}{∂z}xy)\text{j}+(\frac{∂}{∂y}xz-\frac{∂}{∂z}yz)\text{k}=(x-x)\text{i}+(y-y)\text{j}+(z-z)\text{k}=0.\]
Thus, F is conservative.
Is it possible for \(f(x,y)={x}^{2}+x-y\) to be the potential function of an electrostatic field that is located in a region of \({ℝ}^{2}\) free of static charge?

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If \(f\) were such a potential function, then \(f\) would be harmonic. Note that \({f}_{xx}=2\) and \({f}_{yy}=0,\) and so \({f}_{xx}+{f}_{yy}\ne 0.\) Therefore, \(f\) is not harmonic and \(f\) cannot represent an electrostatic potential.
Is it possible for function \(f(x,y)={x}^{2}-{y}^{2}+x\) to be the potential function of an electrostatic field located in a region of \({ℝ}^{2}\) free of static charge?

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Yes
If the coordinate functions of \(\text{F}:{ℝ}^{3}\to {ℝ}^{3}\) have continuous second partial derivatives, then \(\text{curl}\ (\text{div}(\text{F}))\) equals zero.
\(∇\cdot (x\text{i}+y\text{j}+z\text{k})=1.\)

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False
All vector fields of the form \(\text{F}(x,y,z)=f(x)\text{i}+g(y)\text{j}+h(z)\text{k}\) are conservative.
If \(\text{curl}\ \text{F}=0,\) then F is conservative.

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True
If F is a constant vector field then \(\text{div}\ \text{F}=0.\)
If F is a constant vector field then \(\text{curl}\ \text{F}=0.\)

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True
\(\text{F}(x,y,z)=x{y}^{2}{z}^{4}\text{i}+(2{x}^{2}y+z)\text{j}+{y}^{3}{z}^{2}\text{k}\)
\(\text{F}(x,y,z)={x}^{2}z\text{i}+{y}^{2}x\text{j}+(y+2z)\text{k}\)

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\(\text{curl}\ \text{F}=\text{i}+{x}^{2}\text{j}+{y}^{2}\text{k}\)
\(\text{F}(x,y,z)=3xy{z}^{2}\text{i}+{y}^{2}\text{sin}\ z\text{j}+x{e}^{2z}\text{k}\)
\(\text{F}(x,y,z)={x}^{2}yz\text{i}+x{y}^{2}z\text{j}+xy{z}^{2}\text{k}\)

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\(\text{curl}\ \text{F}=(x{z}^{2}-x{y}^{2})\text{i}+({x}^{2}y-y{z}^{2})\text{j}+({y}^{2}z-{x}^{2}z)\text{k}\)
\(\text{F}(x,y,z)=(x\ \text{cos}\ y)\text{i}+x{y}^{2}\text{j}\)
\(\text{F}(x,y,z)=(x-y)\text{i}+(y-z)\text{j}+(z-x)\text{k}\)

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\(\text{curl}\ \text{F}=\text{i}+\text{j}+\text{k}\)
\(\text{F}(x,y,z)=xyz\text{i}+{x}^{2}{y}^{2}{z}^{2}\text{j}+{y}^{2}{z}^{3}\text{k}\)
\(\text{F}(x,y,z)=xy\text{i}+yz\text{j}+xz\text{k}\)

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\(\text{curl}\ \text{F}=\text{-}y\text{i}-z\text{j}-x\text{k}\)
\(\text{F}(x,y,z)={x}^{2}\text{i}+{y}^{2}\text{j}+{z}^{2}\text{k}\)
\(\text{F}(x,y,z)=ax\text{i}+by\text{j}+c\text{k}\) for constants a, b, c

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\(\text{curl}\ \text{F}=0\)
\(\text{F}(x,y,z)={x}^{2}z\text{i}+{y}^{2}x\text{j}+(y+2z)\text{k}\)
\(\text{F}(x,y,z)=3xy{z}^{2}\text{i}+{y}^{2}\text{sin}\ z\text{j}+x{e}^{2z}\text{k}\)

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\(\text{div}\ \text{F}=3y{z}^{2}+2y\ \text{sin}\ z+2x{e}^{2z}\)
\(\text{F}(x,y)=(\text{sin}\ x)\text{i}+(\text{cos}\ y)\text{j}\)
\(\text{F}(x,y,z)={x}^{2}\text{i}+{y}^{2}\text{j}+{z}^{2}\text{k}\)

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\(\text{div}\ \text{F}=2(x+y+z)\)
\(\text{F}(x,y,z)=(x-y)\text{i}+(y-z)\text{j}+(z-x)\text{k}\)
\(\text{F}(x,y)=\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\text{i}+\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}\text{j}\)

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\(\text{div}\ \text{F}=\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\)
\(\text{F}(x,y)=x\text{i}-y\text{j}\)

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

imaginary unit
i² = −1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Divergence and Curl

Determine divergence from the formula for a given vector field.
Determine curl from the formula for a given vector field.
Use the properties of curl and divergence to determine whether a vector field is conservative.
The divergence of a vector field is a scalar function. Divergence measures the “outflowing-ness” of a vector field. If
The curl of a vector field is a vector field. The curl of a vector field at point
A vector field with a simply connected domain is conservative if and only if its curl is zero.

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Versuch es selbst.

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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