Solutions

1. \( \mathrm{curl}(\vec{F}) = \vec{\nabla} \times \vec{F} = x\hat{i} - y\hat{j} + (y-2xy)\hat{k} \)
2. \( \mathrm{curl}(\vec{E}) = \vec{\nabla} \times \vec{E} = -z^2\sin (yz^2) \hat{i} + (y \cos (xy) - xe^{xy})\hat{k} \)

3. \( \vec{\nabla} \circ (\vec{\nabla} \times \vec{V}) = \big(\frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}\big) \circ \bigg[\bigg(\frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}\bigg)\hat{i} + \bigg(\frac{\partial V_x}{\partial z} - \frac{\partial V_z}{\partial x}\bigg)\hat{j} + \bigg(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\bigg)\hat{k}\bigg] \)

   \( = \frac{\partial}{\partial x}\bigg(\frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}\bigg) + \frac{\partial}{\partial y}\bigg(\frac{\partial V_x}{\partial z} - \frac{\partial V_z}{\partial x}\bigg) + \frac{\partial}{\partial z}\bigg(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\bigg)\hat{k} \)

   \( = \frac{\partial ^2 V_z}{\partial x \partial y} - \frac{\partial ^2 V_y}{\partial x \partial z} + \frac{\partial ^2 V_x}{\partial y \partial z} - \frac{\partial ^2 V_z}{\partial y \partial x} + \frac{\partial^2 V_y}{\partial z \partial x} - \frac{\partial ^2 V_x}{\partial z \partial y} = 0 \).

   This result applies to any vector field.

4. Faraday's Law in differential form states \( \frac{\partial \vec{B}}{\partial t} = -\vec{\nabla} \times \vec{E} \), so \( \frac{\partial \vec{B}}{\partial t} = - 2E_0 \bigg[ \bigg(\frac{y}{y_0^2}\bigg)\hat{i} + \bigg( \frac{z}{z_0^2} \bigg)\hat{j} + \bigg( \frac{x}{x_0^2} \bigg)\hat{k} \bigg]\).

5. Consider an arbitrary vector field \( \vec{E} \). Conservative vector fields can be expressed in terms of a potential function like so \( \vec{E}=-\vec{\nabla}V = -\bigg(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z} \hat{k} \bigg) \).

   \( \vec{\nabla} \times \vec{E} = - \vec{\nabla} \times (\vec{\nabla}V) = - \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial V}{\partial x} & \frac{\partial V}{\partial y} & \frac{\partial V}{\partial z} \end{vmatrix} \)

   \( = - \bigg(\frac{\partial ^2 V}{\partial y \partial z} - \frac{ \partial ^2 V}{\partial z \partial y} \bigg)\hat{i} - \bigg(\frac{\partial ^2 V}{\partial z \partial x} - \frac{ \partial ^2 V}{\partial x \partial z} \bigg)\hat{j} - \bigg(\frac{\partial ^2 V}{\partial x \partial y} - \frac{ \partial ^2 V}{\partial y \partial x} \bigg)\hat{k} = 0 \)

   A less painful route to the proof that conservative fields are irrotational is the following. If a field is irrotational, the circulation around any path is always zero, so if we look at our definition of curl, \( (\vec{\nabla} \times \vec{E})\circ \hat{n} \equiv \lim_{\Delta A \rightarrow 0} \frac{\mathrm{circulation}}{\Delta A} \), when the circulation is zero, the curl is zero.