0
$\begingroup$

Considering a fluid whose velocity field is

$\vec{v}(x,y,z)= (y^{3}e^{-z^{2}} + x)\vec{i} + (ze^{x} + y^{2})\vec{j} + (cos(x^{2}+ y^{2}) +2z)\vec{k}$

Calculate the flow of fluid through the closed surface S comprises the border of the area bounded by the half-spheres $z = -\sqrt{4-x^{2}-y^{2}}$ and $z = \sqrt{9-x^{2}-y^{2}}$ and by the plan $z = 0$.

For the moment, I calculated the divergence of $\vec{v}$ to apply the divergence theorem in the volume of the cap between the half-spheres using cylindrical coordinates and I found a answer. But I'm not sure if I need to do something special in reason of the minus sign in the first half-sphere.

I'm afraid of doing something really wrong because I don't understand the divergence theorem really well. So any tip will be helpful.

Thanks in advance

$\endgroup$
  • $\begingroup$ the spheres dont fit together, is it a non smooth surface ? $\endgroup$ – Rene Schipperus Aug 10 '14 at 15:50
  • $\begingroup$ Probably not the best question to do if your new to divergence theorem, but have you calculated the divergence ? $\endgroup$ – Rene Schipperus Aug 10 '14 at 15:51
  • $\begingroup$ Yes I did @Rene Schipperus. I found $ div \ \vec{v} = 2z + 3$ $\endgroup$ – user78723 Aug 10 '14 at 16:00
  • $\begingroup$ I think its $2y+3$, but ok just integrate this over the two half spheres. $\endgroup$ – Rene Schipperus Aug 10 '14 at 16:04
  • $\begingroup$ @Rene Schipperus. I really don't know why I wrote 2z instead of the 2y :/ I found 38$\pi$. $\endgroup$ – user78723 Aug 10 '14 at 16:12
0
$\begingroup$

The Volume of Integration is created by the overlapping of two semi-spheres centered at the origin ($S_1$ being the upper sphere, and $S_2$ being the lower one).

$\nabla\cdot\vec{u}=1+2y+2=3+2y$

$\begin{eqnarray} \displaystyle\iint\limits_{\partial S_1\mathop\cup\partial S_2}\vec{v}\cdot\vec{n}\,\mathrm dS&=&\iiint\limits_{S_1\mathop\cup S_2}\nabla\cdot\vec{v}\,\mathrm dV\\ &=&\iiint\limits_{S_1}\nabla\cdot\vec{v}\,\mathrm dV+\iiint\limits_{S_2}\nabla\cdot\vec{v}\,\mathrm dV \end{eqnarray}$

SPOILER!

$\displaystyle\int\limits_0^3\int\limits_0^{\pi\over2}\int\limits_0^{2\pi}3\rho^2\sin(\phi)+2\rho^3\sin^2(\phi)\sin(\theta)\,\mathrm d\theta\,\mathrm d\phi\,\mathrm d\rho+\int\limits_0^2\int\limits_{\pi\over2}^\pi\int\limits_0^{2\pi}3\rho^2\sin(\phi)+2\rho^3\sin^2(\phi)\sin(\theta)\,\mathrm d\theta\,\mathrm d\phi\,\mathrm d\rho=70\pi$

$\endgroup$
  • $\begingroup$ Do you see the two half-spheres overlapping each other above the $z = 0$ plane? $\endgroup$ – user78723 Aug 10 '14 at 17:17
  • $\begingroup$ I guess that term is wrong, i meant that one half-sphere is above the other one, so it looks like a cartoon-ish spaceship. how-to-draw-funny-cartoons.com/image-files/… $\endgroup$ – Daniel Charry Aug 10 '14 at 17:26
  • $\begingroup$ Ah, ok. I did one half-sphere above another like a half plum and I found $38\pi$ dreamstime.com/… $\endgroup$ – user78723 Aug 10 '14 at 18:18
  • $\begingroup$ Well, your mistake is probably because you didn't see the minus sign on $z=-\sqrt{4-x^2-y^2}$ ... $\endgroup$ – Daniel Charry Aug 10 '14 at 18:25
  • 1
    $\begingroup$ Not at all, but consider using the plain definition of a surface integral in this exercise. It would require the parameterizations of 3 different surfaces. $\endgroup$ – Daniel Charry Aug 10 '14 at 19:26

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.