# 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 $C_2$?.

• any help would be highly appreciated. – user395952 Mar 5 '17 at 2:19
• This is a really difficult topic – user395952 Mar 6 '17 at 0:48
• Thank's a lot for choosing my kind of solving equations. – Frieder Mar 14 '17 at 18:09
• Brilliant solution, not done anything to do with differential forms yet, is that something to do with Cartesian tensors? – user395952 Mar 14 '17 at 19:03
• Differential 1-Forms are Co-vectors. They belong to the dualspace of a Tangentspace. Your surface S is a two-dimensional submanifold of ${\mathbb{R}^3}$. At any point $p \in S$ we have a tangentspace ${T_p}S$. The dual for this space is $T_p^*S$ and from here linear 1-Forms comes from. – Frieder Mar 14 '17 at 19:42

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

$$\begin{gathered} x\left( t \right) = 2\cos \left( t \right) \hfill \\ y\left( t \right) = 2\sin \left( t \right),dy\left( t \right) = 2\cos \left( t \right) \cdot dt \hfill \\ z\left( t \right) = 4,dz\left( t \right) = 0 \cdot dt \hfill \\ \end{gathered}$$

$$\int\limits_0^{2\pi } {8{{\cos }^3}\left( t \right)} \cdot 2\cos \left( t \right) \cdot dt = 16 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)} \cdot dt$$

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.

• on the diagram am i write to show that n is going vertically upwards? – user395952 Mar 14 '17 at 19:24
• @Gibberish Normal vector points from the interior of the surface towards z-axis. Yes it's vertically upwards. – Frieder Mar 14 '17 at 19:29
• brilliant and is the reason why we can use stokes's theorem in this case because $S$ is bounded by both $C_1$ and $C_2$ and for the ring at the bottom of the diagram we can label the outer ring $C_2$ and the inner ring is $C_1$ with $S$ in between them, if we cut a small part of it we can see why the line integrals work? – user395952 Mar 14 '17 at 19:35
• @Gibberish That's it. You're right. Just a question of correct choosing orientation. Thank you! – Frieder Mar 14 '17 at 19:47

1) The projection of the surface down onto the $x,y$ plane is the annular region between the circles of radius 1 and 2. It would probably be easier to integrate in polar coordinates, since then the limits on the integrals would be $1$ to $2$ (for $r$) and $0$ to $2\pi$ (for $\theta$).

2) The vector field $V$ has nothing to do with the normal vector $\hat{n}$: that is just a property of the surface. The normal vector is $\left<-f_x,-f_y,1\right>$, which in your case is $\left<-2x,-2y,1\right>$.

3) For the diagram, I would just draw the indicated portion of the surface, and show that the upper boundary component (the circle of radius 2 with $z=4$) is oriented counterclockwise, and the lower boundary component (the circle of radius 1 with $z=1$) is oriented clockwise.

• okay so how can we evaluate the surface integral if we cannot calculate n? is there a way to replace dS with dxdz or something? how did you calculate that normal vector for this surface?, and for the limits of integration when when calculating line integrals for $C_1$ would it be 0 to 1 and then for $C_2$ $0-2$? would it be best to convert everything into polar coordinates ? – user395952 Mar 7 '17 at 13:55
• If you parameterize your surface as $\left< x(u,v), y(u,v), z(u,v)\right>$, then the normal vector is the cross product of the vectors $\left<x_u,y_u,z_u\right>$ and $\left<x_v,y_v,z_v\right>$. And $dS$ is replaced with the length of the normal vector times $dx dy$. This should all be in whatever textbook you are using in the section on surface integrals of scalar functions. – Nick Mar 7 '17 at 15:38
• ah this makes much more sense now thank you. – user395952 Mar 7 '17 at 16:29
• i will post my working within the next few hours if you care to see if it is correct, i was getting confused with the divergence theorem. – user395952 Mar 7 '17 at 16:37
• whenever i try to replace the n.ds i just keep getting 0.. – user395952 Mar 7 '17 at 20:03