## Thomas' Calculus 13th Edition

A vector field is said to be conservative when $curl F=\nabla \times F=0$ Now, $$curl F= \nabla \times F \\=(\dfrac{\partial z}{\partial y}-\dfrac{\partial y}{\partial z}) \space i+(\dfrac{\partial x}{\partial z}-\dfrac{\partial z}{\partial x}) \space j+(\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{\partial y}) \space k$$ From the given equation, we have $curl F=(0-ye^z)i+(0-ze^x)j+(0-xe^y)k \ne 0$ Therefore, the given field is not conservative.