Answer
See proof
Work Step by Step
Let $\vec{F} = x^3\hat{i} - y^3\hat{j} + y^2\hat{k}$ be a vector field. The divergence is given by: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x^3) - \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(y^2) = 3x^2 - 3y^2 + 2y. \] Since $\nabla \cdot \vec{F} > 0$, it follows that $3x^2 - 3y^2 + 2y > 0$, which implies $y - x > 0$, or $y > x$. Therefore, at all points $(x_0, y_0, z_0)$ such that $z_0 > 0$ and $y_0 > x_0$, the vector field has sources. Analogously, when $\nabla \cdot \vec{F} < 0$, it implies $y - x < 0$, or $y < x$. Therefore, at all points $(x_0, y_0, z_0)$ such that $z_0 < 0$ and $y_0 < x_0$, the vector field has sinks. Finally, in the plane $z = 0$ (or $xy$-plane), where $y = x$, the given vector field doesn't have sources or sinks.
