Test the vector field F to determine if it is conservative.
-$$\mathbf { F } = - \csc ^ { 2 } x \csc y \mathbf { i } - \cot x \cot y \csc y \mathbf { j } - \cos x \mathbf { k }$$
A) Conservative
B) Not conservative

Question

Quizplus · Accepted Answer

To determine if a vector field is conservative, we must check if it is curl-free, i.e. if $
abla 	imes \mathbf{F} = \mathbf{0}$. Computing the curl of $\mathbf{F}$ gives us:

\begin{align*}

abla 	imes \mathbf{F} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ -\csc^2 x \csc y & -\cot x \cot y \csc y & -\cos x \end{vmatrix} \
&= (-\cos x \csc y - 0) \mathbf{i} - (-\sin x \csc^2 y - 0) \mathbf{j} + (\csc y \cot x \sin x - 0) \mathbf{k} \
&= -\cos x \csc y \mathbf{i} + \sin x \csc^2 y \mathbf{j} + \csc y \cot x \sin x \mathbf{k}
\end{align*}

Since $
abla 	imes \mathbf{F}$ is not equal to $\mathbf{0}$ (unless we're at a point where $\csc y = 0$, which is not defined), the vector field $\mathbf{F}$ is not conservative.

Test the Vector Field F to Determine If It Is F=−csc⁡2xcsc⁡yi−cot⁡xcot⁡ycsc⁡yj−cos⁡xk\mathbf { F } = - \csc ^ { 2 } x \csc y \mathbf { i } - \cot x \cot y \csc y \mathbf { j } - \cos x \mathbf { k }F=−csc2xcscyi−cotxcotycscyj−cosxk

Test the Vector Field F to Determine If It Is $\mathbf { F } = - \csc ^ { 2 } x \csc y \mathbf { i } - \cot x \cot y \csc y \mathbf { j } - \cos x \mathbf { k }$