The Divergence and Curl of a Vector Field In Two Dimensions

From The Divergence of a Vector Field and The Curl of a Vector Field pages we gave formulas for the divergence and for the curl of a vector field $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ on $\mathbb{R}^3$ given by the following formulas:

(1)

\begin{align} \quad \mathrm{div}( \mathbf{F}) = \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \end{align}

(2)

\begin{align} \quad \mathrm{curl} ( \mathbf{F} ) = \nabla \times \mathbf{F} = \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \vec{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right ) \vec{j} + \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k} \end{align}

Now suppose that $\mathbf{F}(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a vector field in $\mathbb{R}^2$. Then we define the divergence and curl of $\mathbb{F}$ as follows:

Definition: If $\mathbf{F}(x, y) = P(x, y)\vec{i} + Q(x, y) \vec{j}$ and $\frac{\partial P}{\partial x}$ and $\frac{\partial Q}{\partial y}$ both exist then the Divergence of $\mathbf{F}$ is the scalar field given by $\mathrm{div} (\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$.

Definition: If $\mathbf{F}(x, y) = P(x,y)\vec{i} + Q(x, y) \vec{j}$ and $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$ both existence then the Curl of $\mathbf{F}$ is the vector field given by $\mathrm{curl} (\mathbf{F}) = \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k}$.

It is important to note that the curl of $\mathbf{F}$ exists in three dimensional space despite $\mathbf{F}$ be a vector field on $\mathbb{R}^2$.

Example 1

Find the divergence of the vector field $\mathbf{F}(x, y) = 2xy \vec{i} + 3 \cos y \vec{j}$.

We can apply the formula above directly to get that:

(3)

\begin{align} \quad \mathrm{div} (\mathbf{F}) = \frac{\partial}{\partial x} (2xy) + \frac{\partial}{\partial y} (3 \cos y) = 2y - 3 \sin y \end{align}

Example 2

Find the divergence of the vector field $\mathbf{F}(x, y) = e^x y^2 \vec{i} + (x + 2y) \vec{j}$.

We can apply the formula above directly to get that:

(4)

\begin{align} \quad \mathrm{div} (\mathbf{F}) = \frac{\partial}{\partial x} (e^x y^2) + \frac{\partial}{\partial y} (x + 2y) = e^x y^2 + 2 \end{align}

Example 3

Find the curl of the vector field $\mathbf{F}(x, y) = 2xy \vec{i} + 3 \cos y \vec{j}$.

We can apply the formula above directly to get that:

(5)

\begin{align} \quad \mathrm{curl} (\mathbf{F}) = \left ( \frac{\partial}{\partial x} (3 \cos y) - \frac{\partial}{\partial y} (2xy) \right ) \vec{k} \\ \quad \mathrm{curl} (\mathbf{F}) = -2x \vec{k} \end{align}

Example 4

Find the curl of the vector field $\mathbf{F}(x, y) = e^x y^2 \vec{i} + (x + 2y) \vec{j}$.

We can apply the formula above directly to get that:

(6)

\begin{align} \quad \mathrm{curl} (\mathbf{F}) = \left ( \frac{\partial}{\partial x} (x + 2y) - \frac{\partial}{\partial y} (e^x y^2) \right ) \vec{k} \\ \quad \mathrm{curl} (\mathbf{F}) = (1 - 2e^xy) \vec{k} \end{align}