Flux in the surface 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.

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 $\phi= v$
