<s>[INST] Stokes theorem problem involving sketch The question is Let $S$ be that part of the surface of the paraboloid $z=x^2+y^2$ between the planes $z=1$ and $z=4$.  Using suitable diagrams, show how Stokes' theorem may be applied to this surface in order to write 
$\iint_s \nabla \times V.\hat ndS$
as the sum of two line integrals. I want to show clearly the direction of integration along the two curves assuming that the z components of $\hat n$ are positive. 
part (b) 
given $\vec{V}=x^3j+z^3k$ I want to evaluate both the surface intergal 
$\iint_s\nabla\times V.\hat n dS$ and the sum of the line intergrals $\int_{C_{1}} V.dR +\int_{C_{2}} V.dR$ where $C_1$ and $C_2$ are two curves bounding S, hence verifying stokes' theorem for this case.
My questions are:
1) How does one find the upper and lower limits of the intergrals clearly one of the integrals should be z=4 and z=1 but how can one calculate the ones for x?
2) Do we need to calculate $\hat n$ and if so how? is it just $\frac{\nabla V}{|\nabla V|}=\hat n$
3) What is the diagram supposed to look like because I'm finding it very hard to believe my diagram is correct?
4) Could someone show me the working for when evaluating the surface intergral specifically 
update
would we substitute the values of $z$ into $z=x^2+y^2$ to find the curves $C_1$ and $C_2$?.
 [/INST] We take the following parametrisation
$${\varphi ^{ - 1}}\left( {u,v} \right) = \left( {u,v,{u^2} + {v^2}} \right)$$
That is
$$\begin{gathered}
  x = u, \hfill \\
  y = v, \hfill \\
  z = {u^2} + {v^2} = {x^2} + {y^2} \hfill \\ 
\end{gathered} $$
From the given parametrisation we get a base of tangent-vectors
like
$$\begin{gathered}
  {\xi _x} = {\partial _x}{\varphi ^{ - 1}}\left( {x,y} \right) = \left( {1,0,2x} \right) \hfill \\
  {\xi _y} = {\partial _y}{\varphi ^{ - 1}}\left( {x,y} \right) = \left( {0,1,2y} \right) \hfill \\ 
\end{gathered} $$
And here is the normal and unit-normal vector
$$n = {\xi _x} \times {\xi _y} = \left( { - 2x, - 2y,1} \right)$$
with length
$$\left\| n \right\| = \left\| {{\xi _x} \times {\xi _y}} \right\| = \sqrt {4\left( {{x^2} + {y^2}} \right) + 1} $$
so
$${n_o} = \frac{1}{{\left\| n \right\|}}n = \frac{1}{{\sqrt {4\left( {{x^2} + {y^2}} \right) + 1} }}\left( { - 2x, - 2y,1} \right)$$
From given vectorfield
$$V = \left( {0,{x^3},{z^3}} \right)$$
we get the curl
$$\nabla  \times V = \left| {\begin{array}{*{20}{c}}
  {{\partial _x}}&0&i \\ 
  {{\partial _y}}&{{x^3}}&j \\ 
  {{\partial _z}}&{{z^3}}&k 
\end{array}} \right| = \left( {0,0,3{x^2}} \right)$$
Together with unit-normal vector, the dot-product is
$$\left( {\nabla  \times V} \right) \cdot {n_o} = \frac{{3{x^2}}}{{\left\| n \right\|}} = \frac{{3{x^2}}}{{\sqrt {4\left( {{x^2} + {y^2}} \right) + 1} }}$$
And now with 
$$dS = \left\| n \right\|dxdy$$
we have
$$\left( {\nabla  \times V} \right) \cdot {n_o}dS = \frac{{3{x^2}}}{{\left\| n \right\|}}\left\| n \right\|dxdy = 3{x^2}dxdy$$.
We are now going to integrate this.
Choosing polar-coordinate-system
$$x = r\cos t,y = r\sin t$$
calculating differentials
$$\begin{gathered}
  dx = \cos \left( t \right)dr - r\sin \left( t \right)dt \hfill \\
  dy = \sin \left( t \right)dr + r\cos \left( t \right)dt \hfill \\ 
\end{gathered} $$
and surface-volume form in these coordinates
$$dxdy = \left| {\begin{array}{*{20}{c}}
  {\cos \left( t \right)}&{ - r\sin \left( t \right)} \\ 
  {\sin \left( t \right)}&{r\cos \left( t \right)} 
\end{array}} \right|drdt = r \cdot drdt$$
then
$$\int_S {\left( {\nabla  \times V} \right) \cdot {n_o}dS = \int\limits_R {3{x^2}dxdy = } } \int\limits_0^{2\pi } {\int\limits_1^2 {3 \cdot {r^3}{{\cos }^2}\left( t \right)} }  \cdot drdt$$
and
$$\int\limits_0^{2\pi } {\int\limits_1^2 {3 \cdot {r^3}{{\cos }^2}\left( t \right)} }  \cdot drdt = \frac{{45}}{4}\pi $$

Next part. We are looking at
$$V \cdot \left( {dx,dy,dz} \right) = {x^3} \cdot dy + {z^3} \cdot dz$$
For the top-boundary, we take r=2 and counterclockwise orientation
$$\begin{gathered}
  x\left( t \right) = 2\cos \left( t \right) \hfill \\
  y\left( t \right) = 2\sin \left( t \right),dy\left( t \right) = 2\cos \left( t \right) \cdot dt \hfill \\
  z\left( t \right) = 4,dz\left( t \right) = 0 \cdot dt \hfill \\ 
\end{gathered} $$
This leads us to
$$\int\limits_0^{2\pi } {8{{\cos }^3}\left( t \right)}  \cdot 2\cos \left( t \right) \cdot dt = 16 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)}  \cdot dt$$
For the bottom boundary we take r=1 and clockwise orientation
$$\begin{gathered}
  x\left( t \right) = \cos \left( {2\pi  - t} \right) = \cos \left( t \right) \hfill \\
  y\left( t \right) = \sin \left( {2\pi  - t} \right) =  - \sin \left( t \right),dy\left( t \right) =  - \cos \left( t \right) \cdot dt \hfill \\
  z\left( t \right) = 1,dz\left( t \right) = 0 \cdot dt \hfill \\ 
\end{gathered} $$
Now we get
$$\int\limits_0^{2\pi } {{{\cos }^3}\left( t \right)}  \cdot \left( { - \cos \left( t \right)} \right) \cdot dt =  - \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)}  \cdot dt$$
Putting results together, we have
$$16 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)}  \cdot dt - \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)}  \cdot dt = 15 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)}  \cdot dt$$
And
$$15 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)}  \cdot dt = \frac{{45}}{4}\pi $$
that's it.

Forgot to remark the following:
Using differential-forms, we can write:
$$\omega  = {x^3} \cdot dy + {z^3} \cdot dz$$
This is a differential 1-Form.
Calculating the exterior derivative, applying $d$
this becomes
$$d\omega  = 3{x^2}dx \wedge dy + 3{z^2}dz \wedge dz = 3{x^2}dx \wedge dy$$
We have proven
$$\int\limits_S {d\omega }  = \int\limits_{\partial S} \omega  $$
Stoke's theorem.
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