\(\rm \vec v\) = yz î + 3xz ĵ + z k̂, then Curl \(\rm \vec v\) is:

Concept:

Curl of a Vector:

Let \(\rm \vec{v} = v_1 \rm \vec{i} + v_2 \rm \vec{j}+ v_3 \rm \vec{k}\)

Then the curl of the vector function v is given by:

\(\rm curl \rm \;\vec v =∇ \times {\rm{\;\vec v\;}}=\begin{vmatrix} \rm \vec i & \rm \vec j & \rm \vec k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \rm v_1 & \rm v_2 & \rm v_3 \end{vmatrix}\)

Where ∇ = \(\rm \frac{{\partial }}{{\partial x}} \hat{i}+ \frac{{\partial }}{{\partial y }} \hat{j} +\frac{{\partial }}{{\partial z }} \hat{k}\)

Calculation:

Given: \(\rm \vec v\) = yz î + 3xz ĵ + z k̂

\(\rm curl \rm \;\vec v =∇ \times {\rm{\;\vec v\;}}=\begin{vmatrix} \rm \vec i & \rm \vec j & \rm \vec k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \rm yz & \rm 3xz & \rm z \end{vmatrix}\)

\(= \rm \vec i (\frac{\partial z}{\partial y} - \frac{\partial (3xz)}{\partial z})-\rm \vec j (\frac{\partial z}{\partial x} - \frac{\partial (yz)}{\partial z})+\rm \vec k (\frac{\partial (3xz)}{\partial x} - \frac{\partial (yz)}{\partial y})\)

\(= \rm \vec i (0-3x)- \rm \vec j (0-y)+\rm \vec k (3z-z)\\= -3x\rm \vec i +y \rm \vec j +2z\rm \vec k \)