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A pile of wet sand having total volume 5π5 \pi covers the disk x2+y21,z=0x^{2}+y^{2} \leq 1, z=0. The momentum of water vapour is given by F=gradϕ+μcurlG\mathbf{F}=\operatorname{grad} \phi+\mu \mathrm{curl} \mathbf{G}, where ϕ=x2y2+z2\phi=x^{2}-y^{2}+z^{2} is the water concentration, G=13(y3i+x3j+z3k)\mathbf{G}=\frac{1}{3}\left(-y^{3} \mathbf{i}+x^{3} \mathbf{j}+z^{3} \mathbf{k}\right), and μ\mu is a constant. Find the flux of F upward through the top surface of the sand pile.


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Answered 1 month ago
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The region is enclosed by the upper surface γ\gamma of the sand pile and the disk:

δ:x2+y21\begin{equation} \delta: x^{2}+y^{2} \leq 1 \end{equation}

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

ϕT= ⁣ ⁣ ⁣ ⁣ ⁣γFN^dS+ ⁣ ⁣ ⁣ ⁣ ⁣δFN^dS= ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣DFdV\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 DD is the region enclosed by the surfaces δ\delta and γ\gamma. The divergence of the vector field is

F=(ϕ+μ×G)=(ϕ)+μ(0)=(2xı^2yȷ^+2zk^)=22+2=2\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 (×G)=0\vec{\nabla}\cdot(\vec{\nabla}\times \vec{G})=0 for all smooth vector field G\vec{G}. Therefore,

ϕT=2 ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣DdV=2V=2(5π)=10π\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 δ:x2+y21\delta: x^{2}+y^{2} \leq 1. The vector field is

F=ϕ+μ×F=2xı^2yȷ^+2zk^+μ(3x2+3y2)k^=2xı^2yȷ^+(2z+3μ(x2+y2))k^\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 N^=k^\hat{N}=\hat{k}, hence

FN^=Fk^=2z+3μ(x2+y2)=3μ(x2+y2)\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=0z=0 on the disk. So that:

 ⁣ ⁣ ⁣ ⁣ ⁣δFN^dS= ⁣ ⁣ ⁣ ⁣ ⁣γ3μ(x2+y2)dS\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:

x2+y2=r2dS=rdrdθ\begin{equation} \begin{aligned} x^{2}+y^{2} &=r^{2} \\ d S &=r d r d \theta \end{aligned} \end{equation}

where 0r1  and  0θ2π0 \leq r \leq 1\ \text { and }\ 0 \leq \theta \leq 2 \pi. Thus

 ⁣ ⁣ ⁣ ⁣ ⁣δFN^dS=3μ02π01r3drdθ=32πμ\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:

ϕ=32μπ+10π=(32μ+10)π\begin{equation} \phi=\frac{3}{2} \mu \pi+10 \pi=\left(\frac{3}{2} \mu+10\right) \pi \end{equation}

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