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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 $\vec{F}$:

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

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 $\oint_C \vec{F} \cdot d\vec{r}$ 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 $x^2 + y^2 = 36$.

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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.

enter image description here

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

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

and hence the line integral will be

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

Next, compute the area element vector $d\bf{S}$ 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

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

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  • $\begingroup$ Haha, well I've added the problem details to my question. Even if you don't find where I've made a mistake, though, your answer will be marked correct! $\endgroup$ – nmagerko Dec 5 '15 at 19:31
  • $\begingroup$ @nmagerko: Take a look at the updated answer. I think you had a mistake in determining $dS$! :) $\endgroup$ – H. R. Dec 5 '15 at 20:37
  • $\begingroup$ Ah! I completely overlooked that. Have a green checkmark. $\endgroup$ – nmagerko Dec 6 '15 at 2:25
  • $\begingroup$ What program did you use to get that 3d graph. Nice ;) $\endgroup$ – john Sep 23 '18 at 8:58
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    $\begingroup$ @Poujh: It seems you are right. I would be happy if you make the necessary editions in my answer through the edit tap below the answer. :) $\endgroup$ – H. R. Dec 4 '18 at 21:18
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In $\iint_S {\rm curl} \> \vec{F} \cdot \vec{n} \> dS$ we have $\vec{n} = \frac{g_{x} \times g_{y}}{\left\lVert g_{x} \times g_{y} \right\rVert}$ and $dS = \left\lVert g_{x} \times g_{y} \right\rVert \,dx\,dy$ 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.

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