Day 11: Hot Chocolate!

Note: This is the second question I have posted in this set about 3D flux integrals. Sorry for the repetition, but I really like these problems!!

Let's start by visualizing this situation. We know that the hot chocolate maker is a cube with the bottom on the $xy$ -plane. For the purpose of this question, we are specifically interested in the side that lies on the side that does not lie on the $xz$ -plane, but lies perpendicular to the $y$ axis. This surface can be modelled with the equation $y \ = \ 6$ , and looks somewhat like this:

We know that the nozzle is a square with side-length $2$ , which lies on this surface, so we will integrate over these bounds over the surface $y \ = \ 6$ . We are measuring flow rate, which is the flux through this surface bounded by the nozzle area. The general form of a 3D flux integral is:

$\displaystyle\int \displaystyle\int_{C} \ \vec{F} \ \cdot \ \nabla f \ \text{dA}$

Where $f$ is a function that is equal to $z \ - \ g(x, \ y) \ = \ 0$ , where $z \ = \ g(x, \ y)$ represents the surface through which we are measuring the flux. In our case, $z \ = \ 6$ , so $f \ = \ z \ - \ 6$ . We are given the vector field, so this integral becomes:

$\displaystyle\int \displaystyle\int_{C} \ \begin{bmatrix} 3x^{3}z^{2} \\ 2y^{2}z^{2} \\ 3z \end{bmatrix} \ \cdot \ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \ \text{dA} \ = \ \displaystyle\int \displaystyle\int_{C} \ 2y^{2}z^{2} \ \text{dA}$

Now, all we have to do is add in the bounds of our surface. The nozzle is symmetric across the $y$ -axis, so $-1 \ \leq \ x \ \leq \ 1$ . We also know that the nozzle has a height of $2$ , so its bounds will be $a \ \leq \ z \ \leq \ a \ + \ 2$ . It is stated that the flow rate through the nozzle is $1248$ , so we get:

$\displaystyle\int_{a}^{a \ + \ 2} \displaystyle\int_{-1}^{1} \ 2y^{2}z^{2} \ \text{dx} \ \text{dz}$

We can substitute in $y \ = \ 6$ :

$\displaystyle\int_{a}^{a \ + \ 2} \displaystyle\int_{-1}^{1} \ 72z^{2} \ \text{dx} \ \text{dz} \ \Rightarrow\ \displaystyle\int_{a}^{a \ + \ 2} \ 144z^{2} \ \text{dz} \ \Rightarrow \ 48(a \ + \ 2)^3 \ - \ 48a^3 \ = \ 48(6a^2 \ + \ 12a \ + \ 8)$

Setting this equal to $1248$ :

$48(6a^2 \ + \ 12a \ + \ 8) \ = \ 1248 \ \Rightarrow \ 6a^2 \ + \ 12a \ + \ 8 \ = \ 26 \ \Rightarrow \ 6a^2 \ + \ 12a \ - \ 18 \ = \ 0 \ \Rightarrow \ a^2 \ + \ 2a \ - \ 3 \ = \ 0$

Solving this equation, we get two solutions of $a \ = \ -3$ and $a \ = \ 1$ . Since the $xy$ -plane represents the bottom of the hot chocolate machine, the amount that we shift the nozzle up from the bottom can't be negative, so we find that $a \ = \ 1$ , therefore the nozzle is shifted up from the bottom of the hot chocolate machine by $1$ unit.