Determine the flux of the vector field $$\mathbf{A}=(x^2,2y,z)$$ through the sphere $|r|=R$. Use the parametrization $\mathbf{r} = R(\sin u \cos v,\sin u \sin v,\cos u)$.
Okay so this seems simple enough. We get $\mathbf{A}(\mathbf{r}) = (R^2 \sin^2 u \cos^2 v, ...)$ and $d\mathbf{S}$ = $R^2 \sin u \: dudv\: \hat{e}_r$ where $u: 0 \to \pi, v: 0\to 2\pi$. This yields $$\iint_{r=R} \mathbf{A} \cdot d\mathbf{S} = \iint R^4 \sin^3 u \cos^2 v \:dudv = R^4 \dfrac{4\pi}{3}$$ which is wrong (answer is $4\pi R^3$). Can anyone spot my mistake(s)?
$d\mathbf S$ needs to be a vector element of surface area (i.e., $\mathbf n\,dS$). But, by the way, $dS = R^2\sin u\,du\,dv$. So you should have $$d\mathbf S = R^2\sin u\big(\sin u\cos v,\sin u\sin v,\cos u\big)du\,dv.$$ I'm actually not sure how you managed to get a scalar integral from what you were doing.