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.
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?