A flow field is given by:

$V(r)=\left[\begin{array}{c}u\\ v\\ w\end{array}\right]=\left[\begin{array}{c}\frac{y}{x+1}\\ -\frac{x}{y+1}-\frac{y}{x+1}\\ 1\end{array}\right]$

where $r={\left[\begin{array}{ccc}x& y& z\end{array}\right]}^T$ is a position vector. The velocity field this describes is shown here:

$($the file generating this figure is on blackboard if you'd like to zoom or rotate it around to help

you understand the problem$)$ The fluid has a uniform density of $1kg{m}^{-3}.$ Consider a volume in

this flow field bounded by with side length of $4$ meters whose center is at the point:

$r_{\text{c}\text{u}\text{b}\text{e}}={\left[\begin{array}{ccc}5& 5& 5\end{array}\right]}^T($this is the cube depicted in the figure above$).$

Use a surface integral over the cube to find the rate of change of fluid mass within the cube.

$3$

Compute the divergence of the velocity field from problem $2$

$4$

Recompute the rate of change of fluid in the cube by invoking the divergence theorem to integrate

over the interior of the cube from problem $2.$