## Related questions with answers

A pile of wet sand having total volume $5 \pi$ covers the disk $x^{2}+y^{2} \leq 1, z=0$. The momentum of water vapour is given by $\mathbf{F}=\operatorname{grad} \phi+\mu \mathrm{curl} \mathbf{G}$, where $\phi=x^{2}-y^{2}+z^{2}$ is the water concentration, $\mathbf{G}=\frac{1}{3}\left(-y^{3} \mathbf{i}+x^{3} \mathbf{j}+z^{3} \mathbf{k}\right)$, and $\mu$ is a constant. Find the flux of F upward through the top surface of the sand pile.

Solution

VerifiedThe region is enclosed by the upper surface $\gamma$ of the sand pile and the disk:

$\begin{equation} \delta: x^{2}+y^{2} \leq 1 \end{equation}$

in the $xy$ plane $(z=0)$. By using the Divergence Theorem, we have that

$\begin{equation} \phi_{T}=\int\!\!\!\!\!\int_{\gamma} \vec{F} \cdot \hat{N} d S+\int\!\!\!\!\!\int_{\delta} \vec{F} \cdot \hat{N} d S=\int\!\!\!\!\!\int\!\!\!\!\!\int_{D} \vec{\nabla} \cdot \vec{F} d V \end{equation}$

where $D$ is the region enclosed by the surfaces $\delta$ and $\gamma$. The divergence of the vector field is

$\begin{equation} \begin{aligned} \vec{\nabla} \cdot \vec{F} &=\vec{\nabla} \cdot(\vec{\nabla} \phi+\mu \vec{\nabla} \times \vec{G})=\vec{\nabla} \cdot(\vec{\nabla} \phi)+\mu(0) \\ &=\vec{\nabla} \cdot(2 x \hat{\imath}-2 y \hat{\jmath}+2 z \hat{k})=2-2+2=2 \end{aligned} \end{equation}$

because $\vec{\nabla}\cdot(\vec{\nabla}\times \vec{G})=0$ for all smooth vector field $\vec{G}$. Therefore,

$\begin{equation} \phi_{T}=2 \int\!\!\!\!\!\int\!\!\!\!\!\int_{D} d V=2 V=2(5 \pi)=10 \pi \end{equation}$

Now, we need to evaluate the flux across the disk $\delta: x^{2}+y^{2} \leq 1$. The vector field is

$\begin{equation} \begin{aligned} \vec{F} &=\vec{\nabla} \phi+\mu \vec{\nabla} \times \vec{F} \\ &=2 x \hat{\imath}-2 y \hat{\jmath}+2 z \hat{k}+\mu\left(3 x^{2}+3 y^{2}\right) \hat{k} \\ &=2 x \hat{\imath}-2 y \hat{\jmath}+\left(2 z+3 \mu\left(x^{2}+y^{2}\right)\right) \hat{k} \end{aligned} \end{equation}$

and the outward normal vector to the disk is $\hat{N}=\hat{k}$, hence

$\begin{equation} \vec{F} \cdot \hat{N}=\vec{F} \cdot \hat{k}=2 z+3 \mu\left(x^{2}+y^{2}\right)=3 \mu\left(x^{2}+y^{2}\right) \end{equation}$

because $z=0$ on the disk. So that:

$\begin{equation} \int\!\!\!\!\!\int_{\delta} \vec{F} \cdot \hat{N} d S=\int\!\!\!\!\!\int_{\gamma} 3 \mu\left(x^{2}+y^{2}\right) d S \end{equation}$

In polar coordinates:

$\begin{equation} \begin{aligned} x^{2}+y^{2} &=r^{2} \\ d S &=r d r d \theta \end{aligned} \end{equation}$

where $0 \leq r \leq 1\ \text { and }\ 0 \leq \theta \leq 2 \pi$. Thus

$\begin{equation} \int\!\!\!\!\!\int_{\delta} \vec{F} \cdot \hat{N} d S=3 \mu \int_{0}^{2 \pi} \int_{0}^{1} r^{3} d r d \theta=\frac{3}{2} \pi\mu \end{equation}$

Finally, the flux across the top of the surface is:

$\begin{equation} \phi=\frac{3}{2} \mu \pi+10 \pi=\left(\frac{3}{2} \mu+10\right) \pi \end{equation}$

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