\text { (a) }\left\{\begin{array}{c}-\frac{\partial W_{z}}{\partial x}=F_{y} \Rightarrow W_{z}(x, y, z)=-\int_{0}^{x} F_{y}\left(x^{\prime}, y, z\right) d x^{\prime}+C_{1}(y, z) \\\frac{\partial W_{y}}{\partial x}=F_{z} \Rightarrow W_{y}(x, y, z)=+\int_{0}^{x}F_{z}\left(x^{\prime}, y, z\right) d x^{\prime}+C_{2}(y, z)\end{array}\right\}
These satisfy (ii) and (iii), for any C_{1} \text { and } C_{2}; it remains to choose these functions so as to satisfy (i):
-\int_{0}^{x} \frac{\partial F_{y}\left(x^{\prime}, y, z\right)}{\partial y} d x^{\prime}+\frac{\partial C_{1}}{\partial y}-\int_{0}^{x} \frac{\partial F_{z}\left(x^{\prime}, y, z\right)}{\partial z} d x^{\prime}-\frac{\partial C_{2}}{\partial z}=F_{x}(x, y, z) . \text { But } \frac{\partial F_{x}}{\partial x}+\frac{\partial F_{y}}{\partial y}+\frac{\partial F_{z}}{\partial z}=0 , so
\int_{0}^{x} \frac{\partial F_{x}\left(x^{\prime}, y, z\right)}{\partial x^{\prime}} d x^{\prime}+\frac{\partial C_{1}}{\partial y}-\frac{\partial C_{2}}{\partial z}=F_{x}(x, y, z) . \quad \text { Now } \int_{0}^{x} \frac{\partial F_{x}\left(x^{\prime}, y, z\right)}{\partial x^{\prime}} d x^{\prime}=F_{x}(x, y, z)-F_{x}(0, y, z) , so
\frac{\partial C_{1}}{\partial y}-\frac{\partial C_{2}}{\partial z}=F_{x}(0, y, z) \text {. We may as well pick } C_{2}=0, C_{1}(y, z)=\int_{0}^{y} F_{x}\left(0, y^{\prime}, z\right) d y^{\prime} \text {, and we're done, with }
W_{x}=0 ; \quad W_{y}=\int_{0}^{x} F_{z}\left(x^{\prime}, y, z\right) d x^{\prime} ; \quad W_{z}=\int_{0}^{y} F_{x}\left(0, y^{\prime}, z\right) d y^{\prime}-\int_{0}^{x} F_{y}\left(x^{\prime}, y, z\right) d x^{\prime} .
\text { (b) } \nabla \times W =\left(\frac{\partial W_{z}}{\partial y}-\frac{\partial W_{y}}{\partial z}\right) \hat{ x }+\left(\frac{\partial W_{x}}{\partial z}-\frac{\partial W_{z}}{\partial x}\right) \hat{ y }+\left(\frac{\partial W_{y}}{\partial x}-\frac{\partial W_{x}}{\partial y}\right) \hat{ z }
=\left[F_{x}(0, y, z)-\int_{0}^{x} \frac{\partial F_{y}\left(x^{\prime}, y, z\right)}{\partial y} d x^{\prime}-\int_{0}^{x} \frac{\partial F_{z}\left(x^{\prime}, y, z\right)}{\partial z} d x^{\prime}\right] \hat{ x }+\left[0+F_{y}(x, y, z)\right] \hat{ y }+\left[F_{z}(x, y, z)-0\right] \hat{ z } .
\text { But } \nabla \cdot F =0, \text { so the } \hat{ x } \text { term is }\left[F_{x}(0, y, z)+\int_{0}^{x} \frac{\partial F_{x}\left(x^{\prime}, y, z\right)}{\partial x^{\prime}} d x^{\prime}\right]=F_{x}(0, y, z)+F_{x}(x, y, z)-F_{x}(0, y, z) ,
\text { so } \nabla \times W = F .
\nabla \cdot W =\frac{\partial W_{x}}{\partial x}+\frac{\partial W_{y}}{\partial y}+\frac{\partial W_{z}}{\partial z}=0+\int_{0}^{x} \frac{\partial F_{z}\left(x^{\prime}, y, z\right)}{\partial y} d x^{\prime}+\int_{0}^{y} \frac{\partial F_{x}\left(0, y^{\prime}, z\right)}{\partial z} d y^{\prime}-\int_{0}^{x} \frac{\partial F_{y}\left(x^{\prime}, y, z\right)}{\partial z} d x^{\prime} \neq 0 ,
in general.
\text { (c) } W_{y}=\int_{0}^{x} x^{\prime} d x^{\prime}=\frac{x^{2}}{2} ; W_{z}=\int_{0}^{y} y^{\prime} d y^{\prime}-\int_{0}^{x} z d x^{\prime}=\frac{y^{2}}{2}-z x .
W =\frac{x^{2}}{2} \hat{ y }+\left(\frac{y^{2}}{2}-z x\right) \hat{ z } . \nabla \times W =\left|\begin{array}{ccc}\hat{ x } & \hat{ y } & \hat{ z } \\\partial / \partial x & \partial / \partial y & \partial / \partial z \\0 & x^{2} / 2 & \left(y^{2} / 2-z x\right)\end{array}\right|=y \hat{ x }+z \hat{ y }+x \hat{ z }= F .