Ytukyg

If $S$ is an oriented, smooth surface that is bounded by a simple, closed, smooth boundary curve $C$ with positive orientation, then for some vector field $vec{F}$:

$$oint_C vec{F} cdot dvec{r} = iint_S {rm curl} > vec{F} cdot dvec{S}$$

The latter integral can be written equivalently as follows for some vector $vec{n}$, given that it is normal to the surface $S$ and has the proper orientation:

$$iint_S {rm curl} > vec{F} cdot vec{n} > dS$$

Ultimately, my question is whether or not this normal vector, $vec{n}$, must be a unit normal vector or not (or if there are other constraints that must be imposed on it). The reason I ask this is because, while working on a problem involving Stokes' Theorem, I deduced that the appropriate normal vector for some surface was $hat{i} + hat{k}$, and so I normalized it, yielding $frac{1}{sqrt 2}hat{i} + frac{1}{sqrt 2}hat{k}$.

My answer ended up being off by a factor of $frac{1}{sqrt 2}$ which makes me think that how I have defined $vec{n}$ for these types of problems is incorrect.

[Edit] For those interested, the problem was to evaluate $oint_C vec{F} cdot dvec{r}$ for $vec{F}(x, y, z) = xy>hat{i} + 2z>hat{j} + 6y>hat{k}$ such that $C$ is the counterclockwise-oriented curve of intersection of the plane $x + z = 1$ and the cylinder $x^2 + y^2 = 36$.

In fact, the only constraints for the vector $bf{n}$ are

$1.$ The vector $bf{n}$ is a unit vector normal to the surface.

$2.$ It should have proper orientation depending on the orientation of the surrounding curve.

So, I think you may have made a mistake in the problem you solved and hence we may help you if you write it down in your question. :)

Verifying Stokes Theorem For Your Question

Your surface is enclosed by the intersection curve of the plane $x+z=1$ and the cylinder $x^2+y^2=36$ as the following figure shows.

The parametric equation of the intersection curve, the tangent vector, and the vector field are

$$eqalign{
& {bf{x}} = 6cos theta {bf{i}} + 6sin theta {bf{j}} + left( {1 - 6cos theta } right){bf{k}} cr
& {{d{bf{x}}} over {dtheta }} = - 6sin theta {bf{i}} + 6cos theta {bf{j}} + 6sin theta {bf{k}} cr
& F({bf{x}}) = xy{bf{i}} + 2z{bf{j}} + 6y{bf{k}} cr} $$

and hence the line integral will be

$$eqalign{
& I = intlimits_C {F({bf{x}}) cdot {{d{bf{x}}} over {dtheta }}dtheta } = int_{theta = 0}^{2pi } {left( { - 6sin theta xy + 12cos theta z + 36sin theta y} right)dtheta } cr
& ,,, = 6int_{theta = 0}^{2pi } {left( { - 36{{sin }^2}theta cos theta + 2cos theta left( {1 - 6cos theta } right) + 36{{sin }^2}theta } right)dtheta } cr
& ,,, = 6int_{theta = 0}^{2pi } {left( { - 36{{sin }^2}theta cos theta - 12{{cos }^2}theta + 36{{sin }^2}theta + 2cos theta } right)dtheta } cr
& ,,, = 6left[ { - 36int_{theta = 0}^{2pi } {{{sin }^2}theta cos theta dtheta } - 12int_{theta = 0}^{2pi } {{{cos }^2}theta dtheta } + 36int_{theta = 0}^{2pi } {{{sin }^2}theta dtheta + 2int_{theta = 0}^{2pi } {cos theta dtheta } } } right] cr
& ,,, = 6left[ { - 36left( 0 right) - 12left( pi right) + 36left( pi right) + 2left( 0 right)} right] cr
& ,,, = 144pi cr} $$

Next, compute the area element vector $dbf{S}$ and $nabla times {bf{F}}$

$$eqalign{
& {bf{x}} = x{bf{i}} + y{bf{j}} + left( {1 - x} right){bf{k}} cr
& d{bf{S}} = left( {{{partial {bf{x}}} over {partial x}} times {{partial {bf{x}}} over {partial y}}} right)dxdy = left| {matrix{
{bf{i}} & {bf{j}} & {bf{k}} cr
1 & 0 & { - 1} cr
0 & 1 & 0 cr
} } right|dxdy = left( {{bf{i}} + {bf{k}}} right)dxdy cr
& dS = left| {d{bf{S}}} right| = sqrt 2 dxdy cr
& {bf{n}} = {1 over {sqrt 2 }}left( {{bf{i}} + {bf{k}}} right) cr
& nabla times {bf{F}} = left| {matrix{
{bf{i}} & {bf{j}} & {bf{k}} cr
{{partial _x}} & {{partial _y}} & {{partial _z}} cr
{xy} & {2z} & {6y} cr
} } right| = 4{bf{i}} - x{bf{k}} cr} $$

I think you had a mistake in this part $d{bf{S}}=dS {bf{n}}$ where $sqrt2$ cancels. Finally, the surface integral will be

$$eqalign{
& I = int!!!int {nabla times {bf{F}} cdot d{bf{S}}} = int_{x = - 6}^6 {int_{y = - sqrt {36 - {x^2}} }^{sqrt {36 - {x^2}} } {left( {4 - x} right)dydx} } cr
& ,,,, = int_{x = - 6}^6 {2left( {4 - x} right)sqrt {36 - {x^2}} dx} cr
& ,,,, = int_{x = - 6}^6 {8sqrt {36 - {x^2}} dx} = 8int_{x = - 6}^6 {sqrt {36 - {x^2}} dx} cr
& ,,,, = 8left( {18pi } right) = 144pi cr} $$

In $iint_S {rm curl} > vec{F} cdot vec{n} > dS$ we have $vec{n} = frac{g_{x} times g_{y}}{leftlVert g_{x} times g_{y} rightrVert}$ and $dS = leftlVert g_{x} times g_{y} rightrVert ,dx,dy$ where $g$ parametrizes the surface. So, $vec{n},dS = (g_{x} times g_{y}),dx,dy $

That is where your mistake might be, considering the norm of the normal vector cancels out in the calculation.