Curl Operator on Vector Space is Cross Product of Del Operator

Theorem
Let $R$ be a region of Cartesian $3$-space $\R^3$.

Let $\map {\mathbf V} {x, y, z}$ be a vector field acting over $R$.

Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.

Then
 * $\curl \mathbf V = \nabla \times \mathbf V$

where:
 * $\curl \mathbf V $ denotes the curl of $\mathbf V$
 * $\nabla$ denotes the del operator.

Proof
We have by definition of curl of $\mathbf V$:


 * $\curl \mathbf V = \paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k$

Now: