$\def\hl#1#2{\bbox[#1,1px]{#2}} \def\box#1#2#3#4#5{\color{#2}{\bbox[0px, border: 2px solid #2]{\hl{#3}{\color{white}{\color{#3}{\boxed{\underline{\large\color{#1}{\text{#4}}}\\\color{#1}{#5}\\}}}}}}} \def\verts#1{\left\vert#1\right\vert} \def\Verts#1{\left\Vert#1\right\Vert} \def\pra#1{\left(#1\right)} \def\R{\mathbb{R}} \def\N{\mathbb{N}} \def\Z{\mathbb{Z}}$ $\box{black}{black}{} {Question} {\text{Let F$(x,y,z)=(xy,yz,xz)$ and $S=\{(x,y,z):x^2+y^2\le1,0\le z\le1\}$}\\ \text{$(a)$ Compute $\iint_{\partial S}\text{F}\cdot \text{n}dA$, where $\partial S$ is oriented outwards, without using }\\ \text{$\hspace{3.5ex}$any theorems.}\\ \text{$(b)$ Compute $\iint_{\partial S}\text{F}\cdot \text{n}dA$ using divergence theorem}}$
My attempts
$(a)$ Consider union of $3$ surfaces such that $S_{\text{top}}\cup S_{\text{side}}\cup S_{\text{bottom}}=\partial S$ i.e. \begin{align} &S_{\text{top}}=\{(x,y,z):x^2+y^2\le1,z=1\}\\ &S_{\text{side}}=\{(x,y,z):x^2+y^2=1,0\le z\le1\}\\ &S_{\text{bottom}}=\{(x,y,z):x^2+y^2\le1,z=0\}\\ \end{align} Say $S_{\text{top}}$ parametrized by a function $G:T\to S_{\text{top}}$ where $T=\{(r,\theta):0\le r\le1,0\le \theta\le2\pi\}$, such that \begin{align} G(r,\theta)=&(r\cos(\theta),r\sin(\theta),1)\\ \partial_1G=&(\cos(\theta),\sin(\theta),0)\\ \partial_2G=&(-\sin(\theta),\cos(\theta),0)\\ \partial_1G\times\partial_2G=&(0,0,1) \end{align} Therefore \begin{align} \iint_{S_{\text{top}}}\text{F}\cdot\text{n}dA=&\iint_T \text{F$(G(r,\theta))$}\cdot(\partial_1G\times\partial_2G)dA\\ =&\int_0^{2\pi}\int_0^1 \cos(\theta)r^2drd\theta\\ =&0 \end{align} For the similar reason also have \begin{align} \iint_{S_{\text{bottom}}}\text{F}\cdot\text{n}dA =&0 \end{align} Then Integral over $S_{\text{side}}$, consider $G:T\to S_{\text{side}}$, where $T=\{(\theta,h):0\le \theta\le2\pi,0\le h\le1\}$ such that \begin{align} G(\theta,h)=&(\cos(\theta),\sin(\theta),h)\\ \partial_1G=&(-\sin(\theta),\cos(\theta),0)\\ \partial_2G=&(0,0,1)\\ \partial_1G\times\partial_2G=&(\cos(\theta),\sin(\theta),0) \end{align} Therefore \begin{align} \iint_{S_{\text{side}}}\text{F}\cdot\text{n}dA =&\iint_T \text{F$(G(\theta,h))$}\cdot(\partial_1G\times\partial_2G)dhd\theta\\ =&\int_0^{2\pi}\int_0^1 \cos^2(\theta)\sin(\theta)+\sin^2(\theta)h~dhd\theta=\frac{\pi}{2} \end{align}
Have $$\iint_{\partial S}\text{F}\cdot \text{n}dA=0+0+\frac{\pi}{2}=\frac{\pi}{2}$$
$(b)$ Apply divergence theorem that \begin{align} \iint_{\partial S}\text{F}\cdot\text{n}dA=&\iiint_S\text{div F}dV\\ =&\iiint_Sx+y+z~dV\\ =&\int_0^1\int_0^{2\pi}\int_0^1(r\cos(\theta)+r\sin(\theta)+z)r~drd\theta dz=\frac{\pi}{2} \end{align}
Is my solution correct, I know if $\partial S$ oriented inwards, we can mulitply this by $-1$. However, how do I know if it has correct orientation, could someone expain this to me in details?