Show that \(\nabla \times \mathbf{V}=0\) when \(\mathbf{V}\) is the gradient of a scalar function of position, \(\mathbf{V}=\nabla \phi\).

The curl of a vector field (\(\nabla \times \mathbf{V}\)) is a vector that represents the "curliness" or "vorticity" of the field. Mathematically, it is defined as the cross product of the gradient operator \(\nabla\) with the vector field \(\mathbf{V}\). In Cartesian coordinates, the curl is given by: $$ \nabla \times \mathbf{V} = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \times \left(V_x, V_y, V_z\right) $$

Since we are given that \(\mathbf{V}=\nabla \phi\), we can write the curl of \(\mathbf{V}\) as the curl of \(\nabla \phi\). In Cartesian coordinates, \(\nabla \phi\) is given by: $$ \nabla \phi = \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) $$ So the curl now looks like: $$ \nabla \times \mathbf{V} = \nabla \times \nabla \phi = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \times \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) $$

To calculate the cross product, we can use the determinant of a matrix containing the unit vectors, the gradient operator, and the gradient of \(\phi\). The curl becomes: $$ \nabla \times \mathbf{V} = \nabla \times \nabla \phi = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{vmatrix} $$ Expanding this determinant, we obtain: $$ \nabla \times \mathbf{V} = \nabla \times \nabla \phi = \left(\frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y}\right)\hat{i} - \left(\frac{\partial^2 \phi}{\partial x \partial z} - \frac{\partial^2 \phi}{\partial z \partial x}\right)\hat{j} + \left(\frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x}\right)\hat{k} $$

Notice that for the second-order partial derivatives, the order of differentiation does not matter (according to Schwarz's theorem, assuming \(\phi\) has continuous second derivatives on the domain). Therefore, we have: $$ \frac{\partial^2 \phi}{\partial y \partial z} = \frac{\partial^2 \phi}{\partial z \partial y},\quad \frac{\partial^2 \phi}{\partial x \partial z} = \frac{\partial^2 \phi}{\partial z \partial x},\quad \text{and} \quad \frac{\partial^2 \phi}{\partial x \partial y} = \frac{\partial^2 \phi}{\partial y \partial x} $$ Substituting these equalities back into the expression for the curl, we get: $$ \nabla \times \mathbf{V} = \nabla \times \nabla \phi = (0)\hat{i} - (0)\hat{j} + (0)\hat{k} = \mathbf{0} $$

We showed that the curl of the vector field \(\mathbf{V}=\nabla \phi\) is equal to the zero vector (i.e., \(\nabla \times \mathbf{V}=0\)), thus completing the exercise.