Ok I am stuck yet again. Below is a synopsis of everything I have done.(adsbygoogle = window.adsbygoogle || []).push({});

D. J. Griffiths, 3^{rd}ed., Intro. to Electrodynamics, pg. 45, Problem #1.42(a) and (b):

(a) Find the divergence of the vector function:

[tex]\vec{v} = s(2 + sin^2\phi)\hat{s} + s \cdot sin\phi \cdot cos\phi \hat{\phi} + 3\cdot z\hat{z}[/tex]

So,

[tex]\nabla\circ\vec{v} = \frac{1}{s}\partial_s(s \cdot v_s) + \frac{1}{s}\partial_\phi(v_\phi) + \partial_z(v_z)[/tex]

[tex]v_s = s(2 + sin^2\phi)[/tex]

[tex]v_\phi = s \cdot sin\phi \cdot cos\phi [/tex]

[tex]v_z = 3z[/tex]

Therefore,

[tex]\nabla\circ\vec{v} = 2(2 + sin^2\phi) + (cos^2\phi - sin^2\phi) + 3[/tex]

[tex]. . . = 4 + 2 \cdot sin^2\phi + cos^2\phi - sin^2\phi + 3[/tex]

[tex]. . . = 7 + sin^2\phi + cos^2\phi [/tex]

[tex]. . . = 7 + 1 [/tex]

[tex]. . . = 8 [/tex].

Now place this in the integral over the given volume - aquarter-cylinderin the 1^{st}quadrant between: [tex]0 \le s \le 2, 0 \le \phi \le \frac{\pi}{2}, and 0 \le z \le 5 [/tex].

[tex]\int_V (\nabla \circ \vec{v}) d\tau = 8 \cdot \int_0^2 s ds \cdot \int_0^\frac{\pi}{2} d\phi \cdot \int_0^5 dz = 40 \pi[/tex]

Now for the surface integral over the closed surface. This integral is easy but more involved since there are 5 surfaces over which to chose from. Also, each surface has its own [tex]d\vec{a}[/tex] to integrate over.

For the 5 surfaces I used:

- 1: [tex] \phi = 0 [/tex]
- 2: [tex] \phi = \frac{\pi}{2} [/tex]
- 3: z = 5
- 4: z = 0
- 5: the cylindrical wall

[tex]\oint_S \vec{v} \circ d\vec{a} = \sum_{i=1}^5 \int_{S_{i}} [/tex]

SURFACE 1:

[tex]\int_{S_{1}} = [s(...) \hat{s} + (3z) \hat{z}] \circ [- ds dz \hat{\phi}] = 0[/tex].

The reason is that this surface is for [tex] \phi = 0 [/tex] and [tex]sin \phi[/tex] is zero. Thus the [tex] \hat{\phi} [/tex] term drops out of [tex] \vec{v} [/tex] and so the dot product is zero. IOW, no integral needs to be done.

SURFACE 2:

[tex]\int_{S_{2}} = [s(...) \hat{s} + (3z) \hat{z}] \circ [+dsdz \hat{\phi}] = 0[/tex].

Same reasoning here as with surface 1 but with [tex]\phi = \frac{\pi}{2}[/tex].

SURFACE 3:

[tex]\int_{S_{3}} = [s(...) \hat{s} + (s...) \hat{\phi} + (3z) \hat{z}] \circ [+sdsd\phi \hat{z}] = \int_{z=5} 3z\cdot s ds d\phi[/tex]

[tex] ... = 3 \times 5 \cdot \int_0^2sds \int_0^\frac{\pi}{2}d\phi[/tex]

[tex] ... = 15 \cdot \frac{4}{2} \frac{\pi}{2}[/tex]

[tex] ... = 15\pi[/tex]

SURFACE 4:

[tex]\int_{S_{4}} = [s(...) \hat{s} + (s...) \hat{\phi} + (3z) \hat{z}] \circ [+sdsd\phi (-\hat{z})] = 0[/tex].

The reason is that this surface is for z = 0 and since only the [tex]\hat{z}[/tex] term survives the dot product, then the integral is 0. IOW, no integral needs to be done.

SURFACE 5:

[tex]\int_{S_{5}} = [s(2 + sin^2 \phi) \hat{s} + (s...) \hat{\phi} + (3z) \hat{z}] \circ [+s d\phi dz \hat{s}] = \int_{s=2} s^2(2 + sin^2\phi) d\phi dz[/tex]

[tex] ... = 4 \cdot ( \int_0^\frac{\pi}{2} 2 d\phi + \int_0^\frac{\pi}{2} sin^2\phi d\phi ) \cdot \int_0^5dz[/tex]

[tex] ... = 4 \cdot ( \pi + \pi ) \cdot 5 [/tex]

[tex] ... = 4 \cdot 2\pi \cdot 5 [/tex]

[tex] ... = 40\pi[/tex]

Therefore, after summing all of the open surface integrals to obtain the final closed surface integral, the answer is [tex]55\pi[/tex]. This of course does not equal the [tex]40\pi[/tex] for the volume integral found above. (NOTE: I suspect the integral over the 3^{rd}open surface is wrong, but having gone over both sides of the divergence theorem I suspect the error I made can be anywhere in the entire process.

Well, that's it. I hope someone can point out yet another minor mistake which throws the whole works into disarray.

-LD

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# Test of Divergence Theorem in Cyl. Coord's.

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