- #1

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- Homework Statement
- A scalar potential [itex] U(x, y, z) = z + \sqrt{x^2 + y^2} [/itex] is defined over the cylinder [itex] x^2 + y^2 \leq a^2 [/itex], [itex] 0 \leq z \leq h [/itex]. 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 [itex] \vec a = U \vec b [/itex], where [itex] \vec b [/itex] is a constant vector, then we can get from Stokes' theorem to the following:

[tex] \iint_S U \vec{dS} = \iiint_V \nabla U dV [/tex]

- I think it is some sort of vector integral based on the [itex] \vec{dS} [/itex] and the [itex] \nabla U [/itex]

The problem gives us [itex] U(x, y, z) = z + \sqrt{x^2 + y^2} [/itex] in cartesian coordinates which I chose to convert to cylindrical polar. This leads to:

[tex] U(r, \phi, z) = r + z [/tex]

for [itex] 0 \leq r \leq a [/itex] and [itex] 0 \leq z \leq h [/itex]. We can calculate [itex] \nabla U [/itex] to get:

[tex]

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

[/tex]

which suggests that the volume integral becomes:

[tex] \iiint_V \begin{pmatrix} dV \\ 0 \\ dV \end{pmatrix} = \begin{pmatrix} V \\ 0 \\ V \end{pmatrix} [/tex] where [itex] V = \pi a^2 h [/itex]

Now if we look at the surface integral: (the [itex] \pm [/itex] is for the top and bottom of the cylinder respectively)

[tex] \vec{dS} = \begin{pmatrix} a d\phi dz \\ ? \\ \pm r d\phi dr \end{pmatrix} [/tex]

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

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

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

Combining them yields [itex] \pi a^2 h \hat z [/itex] which agrees with the volume integral.

Now if we consider the [itex] \hat r [/itex] direction (where [itex] r = a [/itex]):

[tex] \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 [/tex]

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.

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 [itex] \vec a = U \vec b [/itex], where [itex] \vec b [/itex] is a constant vector, then we can get from Stokes' theorem to the following:

[tex] \iint_S U \vec{dS} = \iiint_V \nabla U dV [/tex]

**My main questions:**

1)What type of integral am I looking at here?1)

- I think it is some sort of vector integral based on the [itex] \vec{dS} [/itex] and the [itex] \nabla U [/itex]

**2)**How should I evaluate it?**My attempt:**The problem gives us [itex] U(x, y, z) = z + \sqrt{x^2 + y^2} [/itex] in cartesian coordinates which I chose to convert to cylindrical polar. This leads to:

[tex] U(r, \phi, z) = r + z [/tex]

for [itex] 0 \leq r \leq a [/itex] and [itex] 0 \leq z \leq h [/itex]. We can calculate [itex] \nabla U [/itex] to get:

[tex]

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

[/tex]

which suggests that the volume integral becomes:

[tex] \iiint_V \begin{pmatrix} dV \\ 0 \\ dV \end{pmatrix} = \begin{pmatrix} V \\ 0 \\ V \end{pmatrix} [/tex] where [itex] V = \pi a^2 h [/itex]

Now if we look at the surface integral: (the [itex] \pm [/itex] is for the top and bottom of the cylinder respectively)

[tex] \vec{dS} = \begin{pmatrix} a d\phi dz \\ ? \\ \pm r d\phi dr \end{pmatrix} [/tex]

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

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

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

Combining them yields [itex] \pi a^2 h \hat z [/itex] which agrees with the volume integral.

Now if we consider the [itex] \hat r [/itex] direction (where [itex] r = a [/itex]):

[tex] \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 [/tex]

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 [itex] \hat r [/itex] integral should [itex] = 0 [/itex] by symmetry. However, if that is the case, then does that not cause a mis-match between the corresponding components of the [itex] \hat r [/itex] of the two integrals?