The Curl of a Vector Field Examples 1

# The Curl of a Vector Field Examples 1

Recall from The Curl of a Vector Field page that if $\mathbf{F} (x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a vector field on $\mathbb{R}^3$ and the appropriate partial derivatives of $P$, $Q$, and $R$ exist then the curl of $\mathbf{F}$ is given by the following formula:

(1)
\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}

We will now look at some examples of computing the curl of a vector field.

## Example 1

Compute the curl of the vector field $\mathbf{F} (1 + y + z^2) \vec{i} + e^{xyz} \vec{j} + (xyz) \vec{k}$.

Applying the formula for the curl of $\mathbf{F}$ directly and we have that:

(2)
\begin{align} \quad \quad \mathrm{curl} ( \mathbf{F}) = \nabla \times \mathbf{F} = \left ( \frac{\partial}{\partial y} (xyz) - \frac{\partial}{\partial z} (e^{xyz}) \right ) \vec{i} + \left ( \frac{\partial}{\partial z} (1 + y + z^2) - \frac{\partial}{\partial x} (xyz) \right ) \vec{j} + \left ( \frac{\partial}{\partial x} (e^{xyz}) - \frac{\partial}{\partial y} (1 + y + z^2) \right ) \vec{k} \\ \quad \quad \mathrm{curl} ( \mathbf{F} ) = \left ( xz -xye^{xyz} \right ) \vec{i} + \left (2z - yz \right ) \vec{k} + \left ( yze^{xyz} - 1 \right ) \vec{k} \end{align}

## Example 2

Compute the curl of the vector field $\mathbf{F}(x, y, z) = \frac{2xy}{z} \vec{i} + xe^{xy} \vec{j} + \cos (xy^2) \vec{k}$.

Applying the formula for the curl of $\mathbf{F}$ directly and we have that:

(3)
\begin{align} \quad \quad \mathrm{curl} ( \mathbf{F}) = \nabla \times \mathbf{F} = \left ( \frac{\partial}{\partial y} (\cos (xy^2) ) - \frac{\partial}{\partial z} (xe^{xy}) \right ) \vec{i} + \left ( \frac{\partial}{\partial z} \left (\frac{2xy}{z} \right ) - \frac{\partial}{\partial x} (\cos (xy^2)) \right ) \vec{j} + \left ( \frac{\partial}{\partial x} (xe^{xy}) - \frac{\partial}{\partial y} \left (\frac{2xy}{z} \right ) \right ) \vec{k} \\ \quad \quad \mathrm{curl} (\mathbf{F}) = \left ( -2xy \sin (xy^2) - 0 \right ) \vec{i} + \left (-\frac{2xy}{z^2} + y^2 \sin (xy^2) \right ) \vec{j} + \left (xye^{xy} + e^{xy} - \frac{2x}{z} \right ) \vec{k} \end{align}

## Example 3

Compute the curl of the vector field $\mathbf{F} (x, y, z) = \log (xy) \vec{i} + \sec (xy) \vec{j} + \frac{3}{x^2} \vec{k}$. Are there any points in $\mathbb{R}^3$ for which $\mathbf{F}$ irrotational?

Applying the formula for the curl of $\mathbf{F}$ directly and we have that:

(4)
\begin{align} \quad \quad \mathrm{curl} ( \mathbf{F}) = \nabla \times \mathbf{F} = \left ( \frac{\partial}{\partial y} \left ( \frac{3}{x^2} \right ) - \frac{\partial}{\partial z} (\sec (xy)) \right ) \vec{i} + \left ( \frac{\partial}{\partial z} (\log (xy)) - \frac{\partial}{\partial x} \left ( \frac{3}{x^2} \right) \right ) \vec{j} + \left ( \frac{\partial}{\partial x} (\sec (xy)) - \frac{\partial}{\partial y} \log (xy) \right ) \vec{k} \\ \quad \quad \mathrm{curl} (\mathbf{F}) = \left ( 0 - 0 \right ) \vec{i} + \left ( 0 + \frac{6}{x^3} \right ) \vec{j} + \left ( y \sec (xy) \tan (xy) - \frac{1}{y \ln (10)} \right ) \vec{k} = 0 \vec{i} + \frac{6}{x^3} \vec{k} + \left ( y \sec (xy) \tan (xy) - \frac{1}{y \ln (10)} \right ) \vec{k} \end{align}

Now $\mathbf{F}$ is irrotational if $\mathrm{curl} (\mathbf{F}) = \vec{0}$ which implies that we must have:

(5)
\begin{align} \left\{\begin{matrix} 0 = 0\\ \frac{6}{x^3} = 0\\ y \sec (xy) \tan (xy) - \frac{1}{y \ln (10)} =0 \end{matrix}\right. \end{align}

However, note that $\frac{6}{x^3} \neq 0$ for all $(x, y, z) \in \mathbb{R}^3$, so $\mathbf{F}$ is not irrotational anywhere.