1
$\begingroup$

A net is dipped in a river. Determine the flow rate of water across the net if the velocity vector field for the river is given by $\vec{v}=(x-y)\vec{i}+(z+y)\vec{j}+z^2\vec{k}$ and the net is given by $y=\sqrt{1-x^2-z^2}, y\geq 0$, oriented in the positive $y$-direction.

My attempt:

We need to find $\iint_{S}\vec{v}\,d\vec{S}$

The surface describes a hemisphere whose bottom is the disk $x^2+z^2\leq 1$ in the $xz$-plane. This sphere can be parameterized as $\vec{r}(\theta,\phi)=\cos\theta \sin\phi\vec{i}+\sin\theta \sin\phi\vec{j}+\cos\phi\vec{k}$ with $0\leq\theta\leq \pi$ and $0\leq\phi\leq \pi$.

Then I calculated $\vec{r}_{\theta}\times\vec{r}_{\phi}$ which turns out to be $-\cos\theta \sin^2\phi\vec{i}-\sin\theta \sin^2\phi\vec{j}-\sin\phi \cos\phi\vec{k}$.

Since we are interested in the positive $y$-direction, our normal vector $\vec{n}=-\vec{r}_{\theta}\times\vec{r}_{\phi}=\cos\theta \sin^2\phi\vec{i}+\sin\theta \sin^2\phi\vec{j}+\sin\phi \cos\phi\vec{k}$ because the $y$-component, which is $\sin\theta \sin^2\phi$ is always positive in the interval $0\leq\theta\leq \pi$.

Rewriting $\vec{v}$ in terms of our parameters we get, $\vec{v}=(\cos\theta \sin\phi-\sin\theta \sin\phi)\vec{i}+(\cos\phi+\sin\theta \sin\phi)\vec{j}+\cos^2\phi\vec{k}$

So we can express what we are seeking as the double integral $\int^\pi_{0}\int^\pi_{0}\vec{v}\cdot\vec{n}\,d\phi\,d\theta$ which equals $\int^\pi_{0}\int^\pi_{0}((\cos\theta \sin\phi-\sin\theta \sin\phi)\vec{i}+(\cos\phi+\sin\theta \sin\phi)\vec{j}+\cos^2\phi\vec{k})\cdot(\cos\theta \sin^2\phi\vec{i}+\sin\theta \sin^2\phi\vec{j}+\sin\phi \cos\phi\vec{k})\,d\phi\,d\theta$

This looks extremely tedious to deal with. I expanded the brackets but could not see any similar terms that could be simplifed. Am I miscalculating something or is my approach not the best one? This is a freshman year multivariable calculus problem so I don't know any other fancy methods to solve this.

$\endgroup$
1
$\begingroup$

The divergence theorem works well here, provided you close the surface beforehand with the disk $S_1$ of equation $x^2+z^2=1$ in the $xz$ plane.

Your flux equals $$ \iint_S \vec{v}\cdot d\vec{S}+\iint_{S_1} \vec{v}\cdot d\vec{S}-\iint_{S_1} \vec{v}\cdot d\vec{S}\\ =\iint_{S\cup S_1} \vec{v}\cdot d\vec{S}-\iint_{S_1} \vec{v}\cdot d\vec{S}\\ =\iiint_E \nabla \cdot\vec{v} \;dV-\iint_{S_1} \vec{v}\cdot d\vec{S}\\ =\iiint_E (1+1+2z)\; dV + \iint_{S_1}z \;dS $$

Computations are easier once you are there.

$\endgroup$
1
$\begingroup$

Great News: your calculus is correct top-to-bottom. And yes, this integration is complicated, but in the end, you get a cute answer.

Given: \begin{align} Q &= \iint\limits_{S}\vec{v} \cdot d\vec{S}\\ \end{align}

The flow through the net is, after solving that equation using Mathematica,...

SPOILER ALLERT

$$Q = \dfrac{4\pi}{3}$$

$\endgroup$
  • 1
    $\begingroup$ We usually use a mid-dot for a scalar product, not a full-stop dot. A middot has its own command \cdot in $\LaTeX$ – x \cdot y is rendered as $x \cdot y$. $\endgroup$ – CiaPan Dec 16 '15 at 15:10
  • $\begingroup$ Thank you, @CiaPan. In fact, I'm rather ok with the usage of $\LaTeX$, but as I used the question code something like this could have passed. $\endgroup$ – Guilherme Thompson Dec 16 '15 at 15:13
0
$\begingroup$

I suppose you might try to integrate over the disk, which is the base of your hemisphere...

Take the disc element $dx\,dz$ at coordinates $x,z$ – the corresponding element of the hemisphere is $dS = \frac{dx\,dz}y$ at $(x,y,z)$. The vector $d\vec S$ is parallel to vector $[x, y, z]$, so it is $$d\vec S = x\,dS\,\vec i + y\,dS\,\vec j + z\,dS\,\vec k$$ because $(x,y,z)$ is on a unit hemisphere, so $x^2+y^2+z^2=1$ is unit length.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.