Answer
a) $ze^xi+(xye^z-yze^x)j-xe^zk$
b) $y(e^z+e^x)$
Work Step by Step
a) Here, we have
$curl F=[ze^x-0]i+[xye^z-yze^x]j+[0-xe^z]k=ze^xi+(xye^z-yze^x)j-xe^zk$
b) $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
Thus, we have $div F=\dfrac{\partial (xye^z)}{\partial x}+\dfrac{\partial (0)}{\partial y}+\dfrac{\partial (yze^x)}{\partial z}$
or, $=y(e^z+e^x)$
