Let $V$ be a volume in $\mathbb{R}^3$ bounded by a simple closed piecewise-smooth surface $S$ with outward pointing normal vector $\mathbf{n}$. For which one of the following vector fields is the surface integral $\iint_S f\cdot\mathbf{n} \space dS$ equal to the volume of $V$?
A: $\mathbf{f}(\mathbf{r})=(1,1,1)$
B: $\mathbf{f}(\mathbf{r})=(2x,-y^2,2yz-z)$
C: $\mathbf{f}(\mathbf{r})=(z^2,y^2,1-2yz)$
D: $\mathbf{f}(\mathbf{r})=\frac{1}{2}(x,y,z)$
Hint: For the surface integral for outward pointing normal vector to be equal to volume, $ div F$ (divergence of the vector field) should be positive constant $1$ independent of $x, y, z$.
$\frac{\partial (2x)}{\partial x} + \frac{\partial (-y^2)}{\partial y} + \frac{\partial (2yz-z)}{\partial z}$ is the one that gives $1$.
$\\\int_V div F dv$ becomes $\int_V 1 dv = V$