Calculus – Vector calculus – Curl and divergence

The curl of a vector field

Definition: If \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}+R\,\mathbf{k}\) is a vector field on \(\mathbb{R}^3\) and the partial derivatives of \(P\), \(Q\) and \(R\) all exist, then the curl of \(\mathbf{F}\) is the vector field on \(\mathbb{R}^3\) defined by

\[\textrm{curl}\,\mathbf{F}=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\,\mathbf{i} +\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\,\mathbf{j} +\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,\mathbf{k}.\]

The vector differential operator \(\nabla\)

Definition: The vector differential operator \(\nabla\) ("del") is defined by

\[\nabla=\frac{\partial}{\partial x}\,\mathbf{i}+\frac{\partial}{\partial y}\,\mathbf{j}+\frac{\partial}{\partial z}\,\mathbf{k}.\]

Applied to a scalar function \(f\) of three variables it produces the gradient (vector) of \(f\):

\[\nabla f=\frac{\partial f}{\partial x}\,\mathbf{i}+\frac{\partial f}{\partial y}\,\mathbf{j}+\frac{\partial f}{\partial z}\,\mathbf{k}.\]

For the curl of a vector field \(\mathbf{F}\) we now have: \(\textrm{curl}\,\mathbf{F}=\nabla\times\mathbf{F}\).

Stewart §16.5, Example 1
If \(\mathbf{F}(x,y,z)=xz\,\mathbf{i}+xyz\,\mathbf{j}-y^2\,\mathbf{k}\), find \(\textrm{curl}\,\mathbf{F}\).

Solution: Using the definition we obtain

\begin{align*} \textrm{curl}\,\mathbf{F}=\nabla\times\mathbf{F} &=\left(\frac{\partial}{\partial y}(-y^2)-\frac{\partial}{\partial z}(xyz)\right)\,\mathbf{i} -\left(\frac{\partial}{\partial x}(-y^2)-\frac{\partial}{\partial z}(xz)\right)\,\mathbf{j} +\left(\frac{\partial}{\partial x}(xyz)-\frac{\partial}{\partial y}(xz)\right)\,\mathbf{k}\\[2.5mm] &=(-2y-xy)\,\mathbf{i}-(0-x)\,\mathbf{j}+(yz-0)\,\mathbf{k}=-y(2+x)\,\mathbf{i}+x\,\mathbf{j}+yz\,\mathbf{k}. \end{align*}

The curl of a conservative vector field

Theorem: If \(f\) is a function of three variables that has continuous second-order partial derivatives, then: \(\textrm{curl}\left(\nabla f\right)=\mathbf{0}\).

Proof: We have

\begin{align*} \textrm{curl}\left(\nabla f\right)=\nabla\times\left(\nabla f\right) &=\left(\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial z}\right)-\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial y}\right)\right)\,\mathbf{i} -\left(\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial z}\right)-\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial x}\right)\right)\,\mathbf{j} +\left(\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)-\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)\right)\,\mathbf{k}\\[2.5mm] &=\left(\frac{\partial^2f}{\partial y\,\partial x}-\frac{\partial^2f}{\partial x\,\partial y}\right)\,\mathbf{i} -\left(\frac{\partial^2f}{\partial x\,\partial z}-\frac{\partial^2f}{\partial z\,\partial x}\right)\,\mathbf{j} +\left(\frac{\partial^2f}{\partial x\,\partial y}-\frac{\partial^2f}{\partial y\,\partial x}\right)\,\mathbf{k} =0\,\mathbf{i}-0\,\mathbf{j}+0\,\mathbf{k}=\mathbf{0} \end{align*}

by Clairaut's theorem.

A vector field \(\mathbf{F}\) is called conservative if there exists a potential function \(f\) such that \(\mathbf{F}=\nabla f\).

Hence we have: if \(\mathbf{F}\) is conservative, then \(\textrm{curl}\,\mathbf{F}=\mathbf{0}\).

This implies: if \(\textrm{curl}\,\mathbf{F}\neq\mathbf{0}\), then \(\mathbf{F}\) cannot be conservative.

Stewart §16.5, Example 2
Show that the vector field \(\mathbf{F}(x,y,z)=xz\,\mathbf{i}+xyz\,\mathbf{j}-y^2\,\mathbf{k}\) is not conservative.

Solution: In the previous example we showed that

\[\textrm{curl}\,\mathbf{F}=-y(2+x)\,\mathbf{i}+x\,\mathbf{j}+yz\,\mathbf{k}.\]

This shows that \(\textrm{curl}\,\mathbf{F}\neq\mathbf{0}\), which implies that \(\mathbf{F}\) is not conservative.

Theorem: If \(\mathbf{F}\) is a vector field defined on all of \(\mathbb{R}^3\) whose component functions have continuous partial derivatives and \(\textrm{curl}\,\mathbf{F}=\mathbf{0}\), then \(\mathbf{F}\) is a conservative vector field.

Stewart §16.5, Example 3
Show that

\[\mathbf{F}(x,y,z)=y^2z^3\,\mathbf{i}+2xyz^3\,\mathbf{j}+3xy^2z^2\,\mathbf{k}\]

is a conservative vector field and find a potential function \(f\) such that \(\mathbf{F}=\nabla f\).

Solution: Note that

\[\textrm{curl}\,\mathbf{F}=\nabla\times\mathbf{F} =(6xyz^2-6xyz^2)\,\mathbf{i}-(3y^2z^2-3y^2z^2)\,\mathbf{j}+(2yz^3-2yz^3)\,\mathbf{k}=\mathbf{0}.\]

Since \(\textrm{curl}\,\mathbf{F}=\mathbf{0}\) and the domain of \(\mathbf{F}\) is \(\mathbb{R}^3\), the vector field \(\mathbf{F}\) is conservative. For a potential function \(f\) we obtain

\[\left\{\begin{array}{lcl}f_x(x,y,z)=y^2z^3&\Longrightarrow&f(x,y,z)=xy^2z^3+g(y,z)\\[2.5mm] f_y(x,y,z)=2xyz^3&\Longrightarrow&f(x,y,z)=xy^2z^3+h(x,z)\\[2.5mm] f_z(x,y,z)=3xy^2z^2&\Longrightarrow&f(x,y,z)=xy^2z^3+k(x,y).\end{array}\right.\]

We conclude that \(f(x,y,z)=xy^2z^3+K\) for any constant \(K\).

A vector field \(\mathbf{F}\) on \(\mathbb{R}^3\) with \(\textrm{curl}\,\mathbf{F}=\mathbf{0}\) at a point \(P\) is called irrotational at \(P\).

The divergence of a vector field

Definition: If \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}+R\,\mathbf{k}\) is a vector field on \(\mathbb{R}^3\) and \(\displaystyle\frac{\partial P}{\partial x}\), \(\displaystyle\frac{\partial Q}{\partial y}\) and \(\displaystyle\frac{\partial R}{\partial z}\) exist, then the divergence of \(\mathbf{F}\) is the (scalar) function of three variables defined by

\[\textrm{div}\,\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.\]

Note that \(\textrm{div}\,\mathbf{F}=\nabla\cdot\mathbf{F}\).

Stewart §16.5, Example 4
If \(\mathbf{F}(x,y,z)=xz\,\mathbf{i}+xyz\,\mathbf{j}-y^2\,\mathbf{k}\), find \(\textrm{div}\,\mathbf{F}\).

Solution: Using the definition we obtain

\[\textrm{div}\,\mathbf{F}=\nabla\cdot\mathbf{F}=\frac{\partial}{\partial x}(xz)+\frac{\partial}{\partial y}(xyz)+\frac{\partial}{\partial z}(-y^2) =z+xz+0=z(1+x).\]

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The divergence of the curl of a vector field

Theorem: If \(\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}+R\,\mathbf{k}\) is a vector field on \(\mathbb{R}^3\) and \(P\), \(Q\) and \(R\) have continuous second-order partial derivatives, then \(\textrm{div}\,\textrm{curl}\,\mathbf{F}=0\).

Proof: Using the definitions of divergence and curl, we obtain

\begin{align*} \textrm{div}\,\textrm{curl}\,\mathbf{F}=\nabla\cdot\left(\nabla\times\mathbf{F}\right) &=\frac{\partial}{\partial x}\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) +\frac{\partial}{\partial y}\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) +\frac{\partial}{\partial z}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\\[2.5mm] &=\frac{\partial^2R}{\partial x\,\partial y}-\frac{\partial^2Q}{\partial x\,\partial z}+\frac{\partial^2P}{\partial y\,\partial z} -\frac{\partial^2R}{\partial y\,\partial x}+\frac{\partial^2Q}{\partial z\,\partial x}-\frac{\partial^2P}{\partial z\,\partial y}=0 \end{align*}

by Clairaut's theorem.

Hence: if \(\textrm{div}\,\mathbf{F}\neq0\), then \(\mathbf{F}\) cannot be the curl of a vector field.

If \(\textrm{div}\,\mathbf{F}=0\), then the vector field \(\mathbf{F}\) is said to be incompressible.

Stewart §16.5, Example 5
Show that the vector field \(\mathbf{F}(x,y,z)=xz\,\mathbf{i}+xyz\,\mathbf{j}-y^2\,\mathbf{k}\) cannot be written as the curl of a vector field, that is \(\mathbf{F}\neq\textrm{curl}\,\mathbf{G}\).

Solution: In the previous example we showed that \(\textrm{div}\,\mathbf{F}=z(1+x)\neq0\). Therefore \(\mathbf{F}\) is not the curl of another vector field.

The Laplace operator

Another differential operator occurs when we compute the divergence of a gradient vector field \(\nabla f\). If \(f\) is a function of three variables, then we have:

\[\textrm{div}\left(\nabla f\right)=\nabla\cdot\left(\nabla f\right)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}.\]

The operator \(\nabla^2=\nabla\cdot\nabla\) is called the Laplace operator and

\[\nabla^2f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}=0\]

is called Laplace's (differential) equation. See the course on differential equations.

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Last modified on October 14, 2021