1
$\begingroup$

Calculate the flux $$\iint\limits_{S}\mathbf{F}\cdot\mathrm{d}\mathbf{S}$$

when

$$\mathbf{F}(x,y,z) = {x\,{\bf i}+y\,{\bf j}+z^4\,{\bf k}}$$

and the surface S is given by

$$\mathbf{r}(u,v) = {3\,\sin \left( u \right)\,\cos \left( v \right)\,{\bf i}+3\,\sin \left( u \right)\,\sin \left( v \right)\,{\bf j}+3\,\cos \left( u \right)\,{\bf k}}$$

with ${0}\le u\le{\frac{\pi}{2}}$ and ${0}\le v\le{2\,\pi}$

Done:

$$\frac{\delta r}{\delta u}= 3\cos(u)\cos(v)\mathbf i+3\cos(u)\sin(v)\mathbf j-3\sin(u)\mathbf k$$

and

$$\frac{\delta r}{\delta v}= -3\sin(v)\sin(u)\mathbf i+3\cos(v)\sin(u)\mathbf j+0\mathbf k$$

then $d\mathbf S=r_u\times r_v= 9\sin^2(u)\cos(v)\mathbf i+9\sin^2(u)\sin(v)\mathbf j+9\sin^2(u)\cos(v)\mathbf k$

then

$$\iint\limits_{S}\mathbf{F}\cdot\mathrm{d}\mathbf{S}$$

$$\int_0^{\frac{\pi}{2}}\int_0^{2\pi}({x\,{\bf i}+y\,{\bf j}+z^4\,{\bf k}})\cdot (9\sin^2(u)\cos(v)\mathbf i+9\sin^2(u)\sin(v)\mathbf j+9\sin^2(u)\cos(v)\mathbf k)\,dudv$$

this gives me answer $18\pi + \frac{19683\pi ^2}{128}$ which is wrong. I don't know where I have done mistake.

$\endgroup$
  • $\begingroup$ Well is it just $F(r(u,v))$ $\endgroup$ – engineerstudent Oct 13 at 15:46
  • $\begingroup$ @MarkViola Please don't be obtuse. The function $\mathbf{r}$ engineerstudent has written is what denotes $x$, $y$, and $z$ $\endgroup$ – Ninad Munshi Oct 13 at 15:51
  • $\begingroup$ @engineerstudent given that this is the parametrization of a sphere, the $dS$ is supposed to be $9\sin^2u \cdot \mathbf{r}(u,v)$ but that is not what you got, so your mistake is there. $\endgroup$ – Ninad Munshi Oct 13 at 15:53
  • $\begingroup$ @MarkViola Apologies, it seems I misinterpreted your initial comments to OP as being overly hostile. As something to note, though, in modern textbooks the notation that appears frequently $\mathbf{r}(u,v) = (x(u,v),y(u,v),z(u,v))$. OP did their due diligence in providing information and I did not appreciate the attitude you seemed to be taking with them. $\endgroup$ – Ninad Munshi Oct 13 at 15:56
  • $\begingroup$ sorry I meant $9\sin u \cdot \mathbf{r}(u,v)/|\mathbf{r}(u,v)|$, but I cannot edit my comment anymore. To explain further, I use the heuristic that the Jacobian for a sphere should be the position vector $r$ times $\rho^2\sin\phi$ or $r^2\sin\theta$, whichever variable convention your book chooses. $\endgroup$ – Ninad Munshi Oct 13 at 16:02
0
$\begingroup$

First, note that the differential surface element is given by

$$\begin{align} \frac{\partial \vec r}{\partial u}\times \frac{\partial \vec r}{\partial v}&=9\left(\hat i \cos(u) \cos(v)+\hat j \cos(u)\sin(v)-\hat k \sin(u)\right)\times\left(-\hat i \sin(u) \sin(v)+\hat j \sin(u)\cos(v)\right)\\\\&=9\sin(u) \left(\hat i \sin(u)\cos(v)+\hat j \sin(u) \sin(v)+\hat k \cos(u)\right) \end{align}$$

Then, the flux integral is

$$\int_0^{2\pi } \int_0^{\pi/2} \left(27 \sin^3(u)\cos^2(v)+729 \cos^5(u)\sin(u)\right)\,du\,dv$$

Can you finish now?

$\endgroup$
  • $\begingroup$ Please feel free to up vote an answer as you see fit. $\endgroup$ – Mark Viola Nov 11 at 23:27
1
$\begingroup$

The last term in $d\mathbf S$ is wrong. The term in $\mathbf k$ is supposed to be $$(3\cos u\cos v )(3\sin u\cos v)-(3\cos u\sin v)(-3\sin u\sin v)=\\9\cos u\sin u(\cos^2 v+\sin^2 v)=\\9\cos u\sin u$$

To confirm, $d\mathbf S$ is supposed to be the area element on the upper part of the sphere of radius $R=3$. If you write in polar coordinates, it should be $$R^2\sin\theta d\theta d\phi\hat r=9\sin\theta(\sin \theta\cos\phi\hat i+\sin\theta\sin\phi\hat j+\cos\theta \hat k)d\theta d\phi$$ Now just use $\theta = u$ and $\phi= v$

$\endgroup$

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.