Divergence Theorem Questions Assistance (elliptical parabolid?)

Question

Compute the flux of water through parabolic cylinder $S:y=x^2$, from $0\le x\le 2$, $0\le z\le 3$, if the velocity vector is $F(x,y,z)=3z^2i+6j+6xzk$. enter image description here Here's my set up right now, it doesn't look right... my main confusion is how to set up the bound such that only the bounds of x and z stated are included

Answer 1

Given the vector field $\mathbf{F} : \mathbb{R}^3 \to \mathbb{R}^3$ of law:

$$ \mathbf{F}(x,y,z) = \left(3z^2,\,6,\,6xz\right) $$

and given the surface $\mathbf{r} : [0,2]\times[0,3] \to \mathbb{R}^3$ of law:

$$ \mathbf{r}(u,v) = (u,\,u^2,\,v) $$

with support $\Sigma$, by definition the flux of $\mathbf{F}$ through $\Sigma$ is equal to:

$$ \begin{aligned} \Phi_{\Sigma}(\mathbf{F}) & := \iint\limits_{\Sigma} \mathbf{F}\cdot\mathbf{n}_{\Sigma}\,\text{d}S \\ & = \iint\limits_{[0,2]\times[0,3]} \mathbf{F}(\mathbf{r}(u,v))\cdot\left(\mathbf{r}_u(u,v) \land \mathbf{r}_v(u,v)\right)\text{d}u\,\text{d}v \\ & = \iint\limits_{[0,2]\times[0,3]} \left(3v^2,\,6,\,6uv\right)\cdot\left(2u,\,-1,\,0\right)\text{d}u\,\text{d}v \\ & = \int_0^2 \text{d}u \int_0^3 6\left(u\,v^2-1\right)\text{d}v \\ & = \int_0^2 18\,(3\,u-1)\,\text{d}u \\ & = 72\,. \end{aligned} $$

In particular, if $\mathbf{F}$ is the velocity field of a fluid, then $\Phi$ is the flow rate through $\Sigma$:

$\quad\quad\quad\quad\quad$

where, being an open surface, the application of the divergence theorem is awkward.