Exercise 1: Flux through an Open and a Closed Surface
A function $\,h:\reel^2 \rightarrow \reel\,$ is given by the expression
$$\,h(x,y)=1-x^3\,.$$
We consider a rectangle in the $(x,y)$-plane that is determined by $\,0\leq x\leq 1\,$ and $\,-\frac{\pi}2\leq y\leq \frac{\pi}2\,.$ Let the surface $\,\mathcal F\,$ be the part of the graph for $\,h\,$ that lies vertically above the rectangle.
A
Determine a parametric representation for $\mathcal F\,.$
hint
$\mathcal F$ is a so-called graph surface.
answer
$\mathbf s(u,v)=(u,v,1-u^3)\,$ where $u\in \left[0,1\right]$ and $v\in \left[-\pi/2,\pi/2\right]\,.$
Now let $\Omega$ denote the solid spatial field the lies vertically between the rectangle in the $(x,y)$-plane and $\mathcal F\,.$
C
Determine a parametric representation for $\Omega\,.$
hint
Probably you can use the parametric representation you made for $\mathcal F\,,$ you should just adjust the third coordinate a bit.
answer
$\mr(u,v,w)=(u,v,w(1-u^3))\,$ where $u\in \left[0,1\right]\,,$$v\in \left[-\pi/2,\pi/2\right]\,$ and $w\in \left[0,1\right]\,.$
D
Use Gauss’ Theorem to determine the flux of $\mV$ out through the surface of $\Omega\,.$
hint
Actually the surface of $\Omega$ consist of 5 parts. The good thing about Gauss’ Theorem is that we only need one integral, but it is a space integral. Which?
hint
You must find the space integral of the divergence of $\mV$ over $\Omega\,.$ Find the integrand, that is
$$\textrm{Div}\mV(\mr(u,v,w))\mathrm{Jacobi}_{\mathbf r }(u,v,w)\,$$
and determine the tripple integral, possibly using Maple.
answer
The integrand is
$$\,u^6w-2u^3w+w+u^4\sin(v)-u\sin(v)\,$$
and the answer is $\,\displaystyle{\frac{9}{28}}\,\pi\,.$
Exercise 2: 12 Fluxes in Fields with Constant Divergence
A spatial field $\Omega_1$ is a solid unit sphere with centre at the origin, and a spatial field $\Omega_2$ is given by the parametric representation.
The surfaces of the two fields $\partial \Omega_1$ og $\partial \Omega_2$ are oriented with outward pointing unit normal vectors.
In addition we are given the vector fields
\begin{align}
\mV_1(x,y,z)&=(1,2,3)\
\mV_2(x,y,z)&=(-x,\frac y2,-\frac z3)\
\mV_3(x,y,z)&=(x-yz,-2y+xz^2,3z+yx^3)\
\mV_4(x,y,z)&=(k_1,k_2,k_3)\
\mV_5(x,y,z)&=(y-x^3,3x^2y,25+10z)\
\mV_6(x,y,z)&=(2xz-2xy-z,z^3+y^2,-z^2)
\end{align}
Determine the volume of $\Omega_1$ and $\Omega_2\,$ and then (by mental computation!) the divergence of each of the six vector fields.
answer
Vol($\Omega_1$) is – as is well known – $\frac43\pi\,$ while Vol($\Omega_2$)=$\frac {\pi}2\,.$ Then it follows from Gauss’ Theorem that the wanted fluxes are found by multiplication of volume by divergence.
Exercise 3: Untitled, But with the Use of Gauss
A parametrized spatial field $\Omega_{\mathbf r}$ in the $(x,y,z)$-space has the parametric representation
together with a hemi-spherical surface $F,$ that is given by
$$x^2+y^2+z^2-4z=0\,\,\mathrm{og}\,\,z\leq 2\,.$$
A
Draw a sketch of $F$ using pen and paper.
hint
Make a square completion. That’s the southern half part of a spherical shell with centre in $(0,0,2)$ and radius 2.
$F$ is thought to be oriented with a unit normal vector field with negative $z$-coordinate. We wish to determine the flux though $F,$ but it turns out to be rather difficult to integrate over the surface $F\,,$ since the vector field is a bit complicated. On the other hand it is not difficult to find Div$(\mV)(x,y,z)\,,$ therefore we will tune the problem, so it can be solved using Gauss’ Theorem. We start by integrating the divergence of $\mV$ over the solid hemisphere $\Omega$ that fills $F$.
B
Compute the flux of $\mV$ out through the surface $\partial \Omega$ of $\Omega$, by computing the flux as
Here you shall of course need the ordinary parametric representation for a solid sphere (remember the centre). And its Jacobi function.
hint
$\mr(u,v,w)=(u\sin(v)\cos(w), u\sin(v)\sin(w), u\cos(v)+2)\,.$
Jacobi is of course $\,u^2\sin(v)\,.$
Now state the integrand, and readily use Maple to compute the tripple integral.
But, the hemi-spherical surface is open on the top side, but we have computed the flux through the closed surface!
C
Find a parametric representation of the circular disc that can cover the top side of the hemisphere.
We shall find the flux through the surface without the circular disc…
We have already found the flux through the surface of the solid hemisphere.
answer
The flux through the hemi-spherical surface is found as the flux through the surface of the solid hemisphere minus the flux through the circular disc.
Flux($\mV$,hemi-spherical surface)$=-\frac{64\pi}{15}\,.$
