# Gauss & Stokes, direction of normal vector

I've decided to finish my education through completing my last exam (I've been working for 5 years). The exam is in multivariable calculus and I took the classes 6 years ago so I am very rusty. Will ask a bunch of questions over the following weeks and I love you all for helping me.

Translating the question from Swedish, sorry if I'm not using the right terminology.

Question:

Let $S$ be the surface $x^2+y^2=1 ; 0 \le z \le 2;$ oriented so that the normal points from the z-axis. Calculate:

$$\iint\limits_S F\cdot dS;F(x,y,z)=xy^2i+x^2yj+zk;$$

The teacher starts out by adding a bottom $B$ and top $T$ to the cylinder, closing the surface.

$$\iint\limits_{S+B+T} F\cdot dS = \iiint\limits_V \nabla F dV=\iint\limits_{x^2+y^2\le1}\int_0^2(y^2+x^2+1) dzdxdy=[polar]=...=3\pi$$

Now it's time to subtract $B$ and $T$ again. The teacher puts up the following two equations:

$$\iint\limits_B F\cdot dS=\iint\limits_{x^2+y^2\le1}F(x,y,0)\cdot (0,0,-1)dxdy= 0$$ and $$\iint\limits_T F\cdot dS=\iint\limits_{x^2+y^2\le1}F(x,y,2)\cdot (0,0,1)dxdy= 2\pi$$

I guess he's using the general Stokes (edit: this is wrong) $$\int\limits_CF\cdot dS=\iint\limits_S \nabla F\cdot NdS$$

My questions are: If he is using that formula, where's the $\nabla$ disappearing? Why are the normal vectors chosen in that particular way? How do I know which way is which?

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"I guess he's using the general Stokes" Not sure what you mean by this. The two integrals that are evaluated before this statement are just surface integrals. You've also written "Stokes theorem" in a confusing way; no one would ever write $dS$ in a boundary integral on one side and also use $dS$ in the non-boundary integral on the other side. –  Muphrid Aug 17 '13 at 2:03
Yeah sorry, I'm not very good at the peculiars of this. Appreciate all constructive feedback you have. –  Mattis Aug 17 '13 at 2:06

Since you're doing a surface integral over a hollow cylinder from $z = 0$ to $z = 2$ he's decided to pretend the cylinder is a closed surface by adding two 'caps' to the cylinder. Then he can use the Gauss Divergence theorem (a special case of the general Stokes theorem & the reason why the gradient disappears) to turn the surface integral over the closed surface to a volume integral of what's flowing out of the surface.

$$\smallint_S \vec{F} \cdot d \vec{S} + \smallint_B \vec{F} \cdot d \vec{S} + \smallint_T \vec{F} \cdot d \vec{S} = \smallint_{S+B+T} \vec{F} \cdot d \vec{S} = \smallint_V \vec{\nabla} \cdot \vec{F} dV$$

implies

$$\smallint_S \vec{F} \cdot d \vec{S} = \smallint_V \vec{\nabla} \cdot \vec{F} dV - \smallint_B \vec{F} \cdot d \vec{S} - \smallint_T \vec{F} \cdot d \vec{S}$$

Now all we have to do is compute the surface integral over the two caps, & you can clearly see the normal vectors to each of the two caps point in opposite directions giving rise to the minus signs in those surface integrals, since the normal points out of the surface (as stipulated in the statement of Gauss Divergence theorem on the wiki page) - this should be clear if you graph it!

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\begin{align} \text{Stokes theorem: }\oint_C\vec{F}\cdot{d}\vec{r}&=\iint_S(\nabla\times\vec{F})\cdot{d}\vec{S}\\ \\ \\ \text{Gauss's divergence theorem:}\iint_S\vec{F}\cdot{d}\vec{S}&=\iiint_V(\nabla\cdot\vec{F})\,dV \end{align}
He's subtracting the flux through the two caps of the cylinder. The normal vectors for the caps are chosen to point outwards from the cylinder. The top lid points upwards, parallel to the z-axis: $\langle0,0,1\rangle$. The bottom lid points downard, also parallel to the z-axis but in the opposite direction: $\langle0,0,-1\rangle$.