দেওয়া আছে, ভেক্টর ক্ষেত্র, $$ \vec{A} = x^2 \hat{i} - 2x^3 yz \hat{j} + 3yz^3 \hat{k} $$ কার্ল বা $\vec{\nabla} \times \vec{A}$ নির্ণয়ের সূত্রটি হলো, $$ \begin{aligned} \vec{\nabla} \times \vec{A} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 & -2x^3 yz & 3yz^3 \end{vmatrix} \\ &= \hat{i} \left( \frac{\partial}{\partial y}(3yz^3) - \frac{\partial}{\partial z}(-2x^3 yz) \right) - \hat{j} \left( \frac{\partial}{\partial x}(3yz^3) - \frac{\partial}{\partial z}(x^2) \right) + \hat{k} \left( \frac{\partial}{\partial x}(-2x^3 yz) - \frac{\partial}{\partial y}(x^2) \right) \\ &= \hat{i} (3z^3 - (-2x^3 y)) - \hat{j} (0 - 0) + \hat{k} (-6x^2 yz - 0) \\ &= (3z^3 + 2x^3 y) \hat{i} - 6x^2 yz \hat{k} \end{aligned} $$ এখন, $(1, 1, -1)$ বিন্দুতে মান বসিয়ে পাই, $$ \begin{aligned} \vec{\nabla} \times \vec{A} &= (3(-1)^3 + 2(1)^3(1)) \hat{i} - 6(1)^2(1)(-1) \hat{k} \\ &= (-3 + 2) \hat{i} + 6 \hat{k} \\ &= -\hat{i} + 6 \hat{k} \end{aligned} $$ নির্ণেয় কার্ল হলো $-\hat{i} + 6 \hat{k}$।