Finding a Surface Integral for the Vector Field: $F=\langle xz,x^2+y^2,y \rangle$ I need help calculating $$\iint_S F\cdot ds$$
where $$F=\langle xz,x^2+y^2,y \rangle$$ and $$S=\left\{(x,y,z)\mid x^2+y^2+z^2=25 ,y\ge0\right\}$$ oriented in the positive $y$ direction.
My Thoughts:
Maybe we can do the following:
$$\iint_S F\cdot ds=\pm \iint_D F(r(u,v))\cdot [r_u(u,v)\times r_v(u,v)]\: dA(u,v)$$
where the sign is chose so that $\pm r_u\times r_v$ points in the direction we want.
Is this what we need to do here, or is there a much easier way. The easier approach may be to use the divergence theorem, but I need help in doing so.
 A: We parameterize the vector $\vec r$ that spans the surface $S$ using the parameterization $x=5 \sin u \cos v$, $y=5\sin u \sin v$, and $z=5\cos u $, $u\in [0,\pi]$, $v\in [0,\pi]$.  Then, the position vector $\vec r$ in Cartesian coordinates is given by
$$\vec r(u,v)=\hat x 5 \sin u \cos v+\hat y 5\sin u \sin v+ \hat z 5\cos u$$
The vector surface differential $\vec n\,dS$ is then given by
$$\begin{align}
\hat n dS&=\frac{\partial \vec r}{\partial u}\times \frac{\partial \vec r}{\partial v}\,du\,dv\\\\
&=\left(\hat x \sin u\cos v+\hat y \sin u \sin v+\hat z  \cos v\right)\,25\,\sin u\,du\,dv
\end{align}$$
Now, to see whether $\hat n$ has the correct orientation, observe that the $y$ component of $\hat n$, $\hat n\cdot \hat y =\sin u \sin v$.  Inasmuch as $\hat n \cdot \hat y$ is positive for $u\in [0,\pi]$ and $v\in [0,\pi]$, the surface is appropriately oriented.


NOTE 1: 
Had we attempted to define the vector surface differential as $\frac{\partial \vec r}{\partial v}\times \frac{\partial \vec r}{\partial u}\,du\,dv$, then the resulting component of $\hat n$ would have pointed in the negative $y$ direction.  We would then simply reorient by multiplying by $-1$.


Finally, we compute the surface integral of interest as 
$$\iint_S \vec F\cdot \hat n\,dS=\int_0^{\pi}\,\int_0^{\pi}\,\left(\hat x(25\sin u \cos u \cos v)+\hat y(25\sin^2 u)+\hat z(5\sin u \sin v)\right)\cdot \left(\left(\hat x \sin u\cos v+\hat y \sin u \sin v+\hat z  \cos v\right)\right)\,25\,\sin u\,du\,dv
$$
We leave the computation of the integral as an exercise for the reader.


NOTE 2: 
The evaluation of this integral is significantly facilitated by using the Divergence Theorem.  If $V$ is the volume of the hemi-sphere $y>0$, $r=5$, and $S'$ is the surface enclosing $V$, then  
$$\begin{align}
\int_V \nabla \cdot \vec F\,dV&=\oint_{S'}\vec F\cdot \hat n\,dS\\\\
&=\iint_{S}\vec F\cdot \hat n\,dS+\int_0^5\int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}\vec F(x,y=0,z)(-\hat y)\,dz\,dx\\\\
&=\iint_{S}\vec F\cdot \hat n\,dS
\end{align}$$
Therefore, the integral of interest is given by 
$$\begin{align}
\iint_{S}\vec F\cdot \hat n\,dS&=\int_V \nabla \cdot \vec F\,dV\\\\
&=\int_0^{\pi}\int_0^{\pi}\int_0^5\,(r\cos u+2r\sin u \sin v)r^2\sin u \,dr\,du\,dv
\end{align}$$
which again is left as an exercise, albeit perhaps an easier exercise, to the reader.

