Problem. Let $\textbf{F}=\langle 0,0,z^2\rangle$ and let $S$ be the upper hemisphere for $x^2+y^2+z^2=4$. Find $$ \iint\limits_S\textbf{F}\cdot\textbf{n}\;dS, $$ where $\textbf{n}$ is the outward normal vector.
Attempted Solution. We will use the formula $$ \iint\limits_S\textbf{F}\cdot\textbf{n}\;dS=\iint\limits_R\textbf{F}\cdot(\textbf{t}_u\times\textbf{t}_v)\;dA $$ where $\textbf{t}=\textbf{t}(u,v)$ is a parametrization of $S$. To find the parametrization, we use spherical coordinates as follows: $$ \textbf{t}=\langle 2\sin v\cos u, 2\sin v\sin u,2\cos v\rangle,\;\;\;0\leq u\leq 2\pi,\;\;\;0\leq v\leq\frac{\pi}{2} $$ Due to the fact that the $x$- and $y$-coordinates of $\textbf{F}$ are zero, we only need the $z$-coordinate of $\textbf{t}_u\times\textbf{t}_v$, which is given by $$ (-2\sin v\sin u)(2\cos v\sin u)-(2\sin v\cos u)(2\cos v\cos u)=-4\sin v\cos v $$ Now $\textbf{F}=\langle 0,0,4\cos^2v \rangle$, which means $$ \textbf{F}\cdot(\textbf{t}_u\times\textbf{t}_v)=-16\sin v\cos^3v. $$ We can now compute $$ \iint\limits_R\textbf{F}\cdot(\textbf{t}_u\times\textbf{t}_v)\;dA =\int_0^{2\pi}\int_0^{\pi/2}(-16\sin v\cos^3v)\;dv\;du=-8\pi. $$ However, the Divergence Theorem yields a different answer: $+8\pi$.
I seem to be missing a negative somewhere, and I strongly suspect it has to do with the orientation of $S$. If I take $v$ to range from $\pi/2$ to $0$ (instead of from $0$ to $\pi/2$) then it works out. But, how would I know to orient it that way? Or is there something else I'm missing?
Thanks!