ধরা যাক, একটি ভেক্টর ক্ষেত্র $\vec{V} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}$। ভেক্টরটির কার্ল, $\vec{\nabla} \times \vec{V}$ নির্ণয় করি: $$ \begin{aligned} \vec{\nabla} \times \vec{V} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ V_x & V_y & V_z \end{vmatrix} \\ &= \hat{i} \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z} \right) - \hat{j} \left( \frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z} \right) + \hat{k} \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \end{aligned} $$ এখন, এই কার্লের ডাইভারজেন্স, $\vec{\nabla} \cdot (\vec{\nabla} \times \vec{V})$ নির্ণয় করি: $$ \begin{aligned} \vec{\nabla} \cdot (\vec{\nabla} \times \vec{V}) &= \left( \frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k} \right) \cdot \left[ \hat{i} \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z} \right) - \hat{j} \left( \frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z} \right) + \hat{k} \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \right] \\ &= \frac{\partial}{\partial x} \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z} \right) - \frac{\partial}{\partial y} \left( \frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z} \right) + \frac{\partial}{\partial z} \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \\ &= \frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_y}{\partial x \partial z} - \frac{\partial^2 V_z}{\partial y \partial x} + \frac{\partial^2 V_x}{\partial y \partial z} + \frac{\partial^2 V_y}{\partial z \partial x} - \frac{\partial^2 V_x}{\partial z \partial y} \end{aligned} $$ যেহেতু আংশিক অন্তরীকরণের ক্রম পরিবর্তনযোগ্য (অর্থাৎ $\frac{\partial^2}{\partial x \partial y} = \frac{\partial^2}{\partial y \partial x}$), তাই পদগুলো কাটাকাটি গিয়ে শূন্য হয়: $$ \begin{aligned} \vec{\nabla} \cdot (\vec{\nabla} \times \vec{V}) &= 0 \end{aligned} $$ অতএব, যেকোনো ভেক্টরের কার্লের ডাইভারজেন্স সর্বদাই শূন্য। প্রমাণিত।