calculusdefinitionintuitionmultivariable-calculus
Can someone explain what Stokes' Theorem is measuring? What would taking the integral of a vector on a surface give you? When would you use it?
This is the only definition I have and I don't really understand what it's saying.
Let $S$ be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve $C$ with positive orientation. Also let $\vec{F}$ be a vector field then,
$$\int\limits_C \vec{F} \cdot d\vec{r} =
\iint\limits_S \mathrm{curl}\ \vec{F} \cdot d\vec{S}$$
Yes, that is right. The boundary of the annulus between the two concentric circles is the union of the two circles, and the natural orientation is such that the outer circle is positively oriented, and the inner circle negatively, so
$$\int_{r < \lvert (x,y)\rvert < R} \operatorname{curl} F \,dS = \int_{\lvert (x,y)\rvert = R} F\cdot d\vec{s} - \int_{\lvert (x,y)\rvert = r} F\cdot d\vec{s}.$$
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
$$\eqalign{ & {\bf{x}} = 6\cos \theta {\bf{i}} + 6\sin \theta {\bf{j}} + \left( {1 - 6\cos \theta } \right){\bf{k}} \cr & {{d{\bf{x}}} \over {d\theta }} = - 6\sin \theta {\bf{i}} + 6\cos \theta {\bf{j}} + 6\sin \theta {\bf{k}} \cr & F({\bf{x}}) = xy{\bf{i}} + 2z{\bf{j}} + 6y{\bf{k}} \cr} $$
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
$$\eqalign{ & I = \int\!\!\!\int {\nabla \times {\bf{F}} \cdot d{\bf{S}}} = \int_{x = - 6}^6 {\int_{y = - \sqrt {36 - {x^2}} }^{\sqrt {36 - {x^2}} } {\left( {4 - x} \right)dydx} } \cr & \,\,\,\, = \int_{x = - 6}^6 {2\left( {4 - x} \right)\sqrt {36 - {x^2}} dx} \cr & \,\,\,\, = \int_{x = - 6}^6 {8\sqrt {36 - {x^2}} dx} = 8\int_{x = - 6}^6 {\sqrt {36 - {x^2}} dx} \cr & \,\,\,\, = 8\left( {18\pi } \right) = 144\pi \cr} $$
Best Answer
While I can't do it justice, this video done by Khan Academy was invaluable to me. I hope you find it useful as well:
Stokes' theorem intuition