A little help integrating this torus? 
Let $\mathbf{F}\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be given by
  $$\mathbf{F}(x,y,z)=(x,y,z).$$
  Evaluate $$\iint\limits_S \mathbf{F}\cdot dS$$ where $S$ is the surface of the torus given by
  $$\begin{align*}
x&=(R+\cos(\phi))\cdot \cos(\theta)\\  
y&=(R+\cos(\phi))\cdot \sin(\theta)\\
z&=\sin(\phi)
\end{align*}$$
  and$$0\leq{\theta}\leq{2\pi},\qquad 0\leq{\theta}\leq{2\pi}.$$
  Assume $S$ is oriented outward using the outward unit normal.

My Take
Ok, so I know I have to start by taking the cross product of the partial derivatives of theta and phi, but when it says using the outward unit normal, which vector has to be positive in order for it to be the outward normal? Is it the $i$, $j$ or $k$ vector? How do I know which vector it is? What will my next step be after I find the cross product?
 A: I don't know if you wanna compute the surface integral the hard way, which is, computing the surface outward normal pointwisely and then integrating. For me this problem looks rather like a standard "applying the divergence theorem" exercise.
Denote the interior of the torus as $T$, and $\partial T = S$, then by Divergence theorem:
$$
\iint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_T \nabla \cdot \mathbf{F} \,dV  = 3 |T|,
$$
where $|T|$ is the volume of $T$. Now we compute the volume using the parametrization you gave: based on the parametrization, if we assume $R>1$, then the torus has $R$ as the distance from the center of the tube to its center, the radius of the tube is 1. Thus the interior's parametrization is:
$$
\begin{aligned}
&x =  (R + r \cos \phi) \cos{\theta} 
\\
&y=  (R + r \cos \phi) \sin{\theta} 
\\
&z=  r \sin \phi
\end{aligned}
$$
for $(r,\phi,\theta)\in [0,1]\times[0,2\pi]\times[0,2\pi]$. Compute the Jacobian:
$$
\left|\frac{\partial (x,y,z) }{\partial (r,\phi,\theta)} \right|= r(R+r\cos\phi).
$$
Hence the volume $|T|$ is:
$$
|T| = \int^1_0 \int^{2\pi}_0 \int^{2\pi}_0 r(R+r\cos\phi) d\phi d\theta dr = 2\pi^2 R,
$$
and
$$
\iint_S \mathbf{F}\cdot d\mathbf{S} = 6\pi^2 R.
$$
Wikipedia's entry of Torus uses Pappus's centroid theorem to compute the volume, which is much simpler, but I guess we are supposed to use the integral to compute for we are given a parametrization here.
A: A related problem. You can use the identity
$$ \iint\limits_S {F}\cdot dS = \iint\limits_D {F}\cdot ({r}_\theta \times {r}_{\phi})\,{dA} = \iint\limits_D {F}\cdot ({r}_\theta \times {r}_{\phi}) {d \phi d\theta}, $$
where
$$ r(\theta,\phi) = (R+\cos(\phi)) \cos(\theta) {i} + ( R +\cos(\phi))  \cos(\theta)  {j} + \sin(\phi) {k},  $$
and you can write ${F}$ as 
$$ {F} (x,y,z)={F}( {r}(\theta,\phi) )= (R+\cos(\phi)) \cos(\theta) {i} + ( R +\cos(\phi))  \cos(\theta)  {j} + \sin(\phi) {k}.$$
I leave it here for you to finish the task.
