Answer
a) $P_1$ is a source and $P_2$ is sink.
b) $P_1$ is a source and $P_2$ is sink.
Work Step by Step
a) Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$
We found that at the point $P_1$ the vector end near that point are shorter than the vectors that start near that point. This implies that the net flow is outwards and this yields $P_1$ is a source and the vectors at the point $P_2$ that the vector end near that point are greater than the vectors that start near that point , this implies that the net flow is inwards and $P_2$ is a sink.
This gives us that $P_1$ is a source and $P_2$ is sink.
b) Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\dfrac{\partial x}{\partial x}+\dfrac{\partial y^2}{\partial y}=1+2y$
We found that the y-value of $P_1$ is positive that is, div F $\gt 0$ and so $P_1$ is a source. Also, the y-value of $P_2$ is less than $-1$, that is, div F $\lt 0$ and thus, $P_2$ is a sink.
This gives us that $P_1$ is a source and $P_2$ is sink.
