Stokes' Theorem 'corollary' integral in cylindrical polar coordinates

[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?

 

[Master1022]

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

 

[Master1022]

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.
If we start with Gauss' Theorem:
$$\iint_S \vec a \cdot \vec{dS} = \iiint_V (\nabla \cdot \vec a) dV$$
$$\iint_S U \vec b \cdot \vec{dS} = \iiint_V (\nabla \cdot U \vec b) dV$$
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?

 

[pasmith]

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.

 

[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

 

[pasmith]

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!