# Stokes' Theorem 'corollary' integral in cylindrical polar coordinates

• Master1022
In summary, the conversation was about proving a formula related to Stokes' Theorem, specifically in the case of a vector \vec{a} = U\vec{b}, where \vec{b} is a constant vector. The formula derived from Stokes' Theorem was \iint_S U\vec{dS} = \iiint_V \nabla U dV. The main questions were about the type of integral and how to evaluate it. The attempt involved converting to cylindrical polar coordinates and calculating \nabla U. However, the approach of integrating component by component was incorrect and the correct method involves treating d\mathbf{S} as \mathbf{n}\,dS where \mathbf{nf

#### Master1022

Homework Statement
A scalar potential $U(x, y, z) = z + \sqrt{x^2 + y^2}$ is defined over the cylinder $x^2 + y^2 \leq a^2$, $0 \leq z \leq h$. Show that the following quantity holds true
Relevant Equations
Stokes theorem
Hi,

I was just working on a homework problem where the first part is about proving some formula related to Stokes' Theorem. If we have a vector $\vec a = U \vec b$, where $\vec b$ is a constant vector, then we can get from Stokes' theorem to the following:

$$\iint_S U \vec{dS} = \iiint_V \nabla U dV$$

My main questions:
1)
What type of integral am I looking at here?
- I think it is some sort of vector integral based on the $\vec{dS}$ and the $\nabla U$
2) How should I evaluate it?

My attempt:

The problem gives us $U(x, y, z) = z + \sqrt{x^2 + y^2}$ in cartesian coordinates which I chose to convert to cylindrical polar. This leads to:
$$U(r, \phi, z) = r + z$$
for $0 \leq r \leq a$ and $0 \leq z \leq h$. We can calculate $\nabla U$ to get:

$$\nabla U = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$$

which suggests that the volume integral becomes:
$$\iiint_V \begin{pmatrix} dV \\ 0 \\ dV \end{pmatrix} = \begin{pmatrix} V \\ 0 \\ V \end{pmatrix}$$ where $V = \pi a^2 h$

Now if we look at the surface integral: (the $\pm$ is for the top and bottom of the cylinder respectively)
$$\vec{dS} = \begin{pmatrix} a d\phi dz \\ ? \\ \pm r d\phi dr \end{pmatrix}$$

If we consider the surface integral in the $\hat z$ directions, then we get (I have used $\cdot$ just to represent normal multiplication and not anything to do with the vector dot product).

$$\int_{\phi = 0}^{2\pi} \int_{r=0}^a (r+z) \cdot r \, dr \, d\phi \hat z$$ for the top where $z = h$ and

$$- \int_{\phi = 0}^{2\pi} \int_{r=0}^a (r+z) \cdot r \, dr \, d\phi \hat z$$ for the bottom where $z = 0$

Combining them yields $\pi a^2 h \hat z$ which agrees with the volume integral.

Now if we consider the $\hat r$ direction (where $r = a$):
$$\int_{\phi = 0}^{2\pi} \int_{z=0}^h (a+z) \cdot a \, dz \, d\phi \hat r = 2\pi a \left. \left(az + \frac{z^2}{2} \right) \right|_0^h$$

which doesn't seem to lead me to the correct place. This suggests that I have approached some (if not all) of these integrals incorrectly.

Any help would be greatly appreciated.

Note: The solution states that the $\hat r$ integral should $= 0$ by symmetry. However, if that is the case, then does that not cause a mis-match between the corresponding components of the $\hat r$ of the two integrals?

Apologies, just wanted to make it clear that I did derive the quoted formula, but am now just working on the verification. Thanks

I just realized that I wrote Stokes' Theorem instead of Gauss' Theorem! Given that I cannot edit the post for some reason, can a moderator change/edit the post (relevant equations section) and the title? I really do apologize about that - I am not sure why I didn't spot that sooner.

In any case, this is where the original formula came from and perhaps providing a proof might help readers understand the situation and spot my errors.
$$\iint_S \vec a \cdot \vec{dS} = \iiint_V (\nabla \cdot \vec a) dV$$
$$\iint_S U \vec b \cdot \vec{dS} = \iiint_V (\nabla \cdot U \vec b) dV$$
Noting that:
$$\nabla \cdot U \vec b = U(\nabla \cdot \vec b) + \vec b \cdot \nabla U$$
For a constant vector, $\nabla \cdot \vec b = 0$. Thus:
$$\vec b \cdot \iint_S U\vec{dS} = \vec b \cdot \iiint_V (\nabla U) dV$$
which leads to the expression in the post

Once again, I really do apologize for this. If it is easier, I can delete this post and repost this thread with all the information corrected and put into one post? If the latter is the best way to go about it, can a moderator let me know before taking it down so I can copy all the Latex code out of the post?

Last edited:
Always treat $d\mathbf{S}$ as $\mathbf{n}\,dS$ where $\mathbf{n}$ is the (outward) unit normal:$$\int_{\partial V} U \,d\mathbf{S} \equiv \int_{\partial V} U\mathbf{n}\,dS.$$

Remember that in cylindrical polars, the basis vectors $\mathbf{e}_r = \cos\phi\,\mathbf{i} + \sin\phi\,\mathbf{j}$ and $\mathbf{e}_\phi = -\sin\phi\,\mathbf{i} + \cos\phi\,\mathbf{j}$ are functions of $\phi$. That means that you can't integrate a vector quantity component by component unless you express everything in terms of the cartesian basis vectors (which are constant).

Thus \begin{align*} \int_V \nabla U\,dV &= \int_0^{2\pi} \int_0^a \int_0^h (\mathbf{e}_r + \mathbf{k}) r\,dz\,dr\,d\phi \\ &= \int_0^{2\pi} \int_0^a \int_0^h (\cos\phi\,\mathbf{i} + \sin\phi\,\mathbf{j} + \mathbf{k})r\,dz\,dr\,d\phi \end{align*} etc.

etotheipi and Master1022
Remember that in cylindrical polars, the basis vectors $\mathbf{e}_r = \cos\phi\,\mathbf{i} + \sin\phi\,\mathbf{j}$ and $\mathbf{e}_\phi = -\sin\phi\,\mathbf{i} + \cos\phi\,\mathbf{j}$ are functions of $\phi$. That means that you can't integrate a vector quantity component by component unless you express everything in terms of the cartesian basis vectors (which are constant).

Thank you very much for replying! That makes sense I suppose - not sure why I have never heard that point expressed before in lectures. Nonetheless, thank you for clearing this up.

However, I just have a quick question about the integral in the $\hat \phi$ direction. We expect this integral to go to 0, but I am just checking to see whether I have set it up correctly? So we have:
$$\iint_S (r + z) dr dz \hat \phi = \iint_S (r + z) (-sin \phi \mathbf{i} + cos \phi \mathbf{j} ) dr dz$$

For the limits, we will have $0 \leq z \leq h$, which makes sense. Given that we are on the surface of the cylinder, should the $r$ limits be from $a$ to $a$? That is the only way that I can see the integral yielding a 0 (we aren't integrating over $\phi$, otherwise that would make it yield 0). What I have suggested feels wrong to me, but I cannot see the immediate error.

Thanks

There is no integral in the $\mathbf{e}_\phi$ direction here, because none of the surfaces have a normal component in that direction.

You have three surfaces here.
• $z = 0$ has outward normal $-\mathbf{k}$.
• $z = h$ has outward normal $\mathbf{k}$.
• $r = a$ has outward normal $\hat{\mathbf{r}}$.

You have done the first two integrals correctly, but you need to revisit the $r = a$ integral. You have mostly set it up correctly, so start from $$\int_0^h \int_0^{2\pi} (a + z)a \hat{\mathbf{r}}\,d\phi\,dz$$ and remember that as $\hat{\mathbf{r}}$ depends on $\phi$ you can't take it outside the integral.

A surface with normal $\hat{\mathbf{\phi}}$ would be a vertical half-plane. On this plane the coordinates $r$ and $z$ function as Cartesian coordinates (if $\phi$ is constant then so is $\hat{\mathbf{r}}$) and the area element would be $dS = dr\,dz$.

Master1022
There is no integral in the $\mathbf{e}_\phi$ direction here, because none of the surfaces have a normal component in that direction.

You have three surfaces here.
• $z = 0$ has outward normal $-\mathbf{k}$.
• $z = h$ has outward normal $\mathbf{k}$.
• $r = a$ has outward normal $\hat{\mathbf{r}}$.

You have done the first two integrals correctly, but you need to revisit the $r = a$ integral. You have mostly set it up correctly, so start from $$\int_0^h \int_0^{2\pi} (a + z)a \hat{\mathbf{r}}\,d\phi\,dz$$ and remember that as $\hat{\mathbf{r}}$ depends on $\phi$ you can't take it outside the integral.

A surface with normal $\hat{\mathbf{\phi}}$ would be a vertical half-plane. On this plane the coordinates $r$ and $z$ function as Cartesian coordinates (if $\phi$ is constant then so is $\hat{\mathbf{r}}$) and the area element would be $dS = dr\,dz$.

Thanks for your reply and for the clarification. I thought we might have to consider the infinitely thin surface in the hoop direction (although I suppose it is just a vertical line if we take a cross-section). I had re-done the other integral calculations using your advice already and they all yielded the correct results!