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


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

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 ${0}\le v\le{2\,\pi}$


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


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

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



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

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

  • $\begingroup$ 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

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?

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

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 $\phi= v$


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