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Living_Dog
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Ok I am stuck yet again. Below is a synopsis of everything I have done.
D. J. Griffiths, 3rd ed., Intro. to Electrodynamics, pg. 45, Problem #1.42(a) and (b):
(a) Find the divergence of the vector function:
\vec{v} = s(2 + sin^2\phi)\hat{s} + s \cdot sin\phi \cdot cos\phi \hat{\phi} + 3\cdot z\hat{z}
So,
\nabla\circ\vec{v} = \frac{1}{s}\partial_s(s \cdot v_s) + \frac{1}{s}\partial_\phi(v_\phi) + \partial_z(v_z)
v_s = s(2 + sin^2\phi)
v_\phi = s \cdot sin\phi \cdot cos\phi
v_z = 3z
Therefore,
\nabla\circ\vec{v} = 2(2 + sin^2\phi) + (cos^2\phi - sin^2\phi) + 3
. . . = 4 + 2 \cdot sin^2\phi + cos^2\phi - sin^2\phi + 3
. . . = 7 + sin^2\phi + cos^2\phi
. . . = 7 + 1
. . . = 8.
Now place this in the integral over the given volume - a quarter-cylinder in the 1st quadrant between: 0 \le s \le 2, 0 \le \phi \le \frac{\pi}{2}, and 0 \le z \le 5.
\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
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 d\vec{a} to integrate over.
For the 5 surfaces I used:
\oint_S \vec{v} \circ d\vec{a} = \sum_{i=1}^5 \int_{S_{i}}
SURFACE 1:
\int_{S_{1}} = [s(...) \hat{s} + (3z) \hat{z}] \circ [- ds dz \hat{\phi}] = 0.
The reason is that this surface is for \phi = 0 and sin \phi is zero. Thus the \hat{\phi} term drops out of \vec{v} and so the dot product is zero. IOW, no integral needs to be done.
SURFACE 2:
\int_{S_{2}} = [s(...) \hat{s} + (3z) \hat{z}] \circ [+dsdz \hat{\phi}] = 0.
Same reasoning here as with surface 1 but with \phi = \frac{\pi}{2}.
SURFACE 3:
\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
... = 3 \times 5 \cdot \int_0^2sds \int_0^\frac{\pi}{2}d\phi
... = 15 \cdot \frac{4}{2} \frac{\pi}{2}
... = 15\pi
SURFACE 4:
\int_{S_{4}} = [s(...) \hat{s} + (s...) \hat{\phi} + (3z) \hat{z}] \circ [+sdsd\phi (-\hat{z})] = 0.
The reason is that this surface is for z = 0 and since only the \hat{z} term survives the dot product, then the integral is 0. IOW, no integral needs to be done.
SURFACE 5:
\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
... = 4 \cdot ( \int_0^\frac{\pi}{2} 2 d\phi + \int_0^\frac{\pi}{2} sin^2\phi d\phi ) \cdot \int_0^5dz
... = 4 \cdot ( \pi + \pi ) \cdot 5
... = 4 \cdot 2\pi \cdot 5
... = 40\pi
Therefore, after summing all of the open surface integrals to obtain the final closed surface integral, the answer is 55\pi. This of course does not equal the 40\pi for the volume integral found above. (NOTE: I suspect the integral over the 3rd 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
D. J. Griffiths, 3rd ed., Intro. to Electrodynamics, pg. 45, Problem #1.42(a) and (b):
(a) Find the divergence of the vector function:
\vec{v} = s(2 + sin^2\phi)\hat{s} + s \cdot sin\phi \cdot cos\phi \hat{\phi} + 3\cdot z\hat{z}
So,
\nabla\circ\vec{v} = \frac{1}{s}\partial_s(s \cdot v_s) + \frac{1}{s}\partial_\phi(v_\phi) + \partial_z(v_z)
v_s = s(2 + sin^2\phi)
v_\phi = s \cdot sin\phi \cdot cos\phi
v_z = 3z
Therefore,
\nabla\circ\vec{v} = 2(2 + sin^2\phi) + (cos^2\phi - sin^2\phi) + 3
. . . = 4 + 2 \cdot sin^2\phi + cos^2\phi - sin^2\phi + 3
. . . = 7 + sin^2\phi + cos^2\phi
. . . = 7 + 1
. . . = 8.
Now place this in the integral over the given volume - a quarter-cylinder in the 1st quadrant between: 0 \le s \le 2, 0 \le \phi \le \frac{\pi}{2}, and 0 \le z \le 5.
\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
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 d\vec{a} to integrate over.
For the 5 surfaces I used:
- 1: \phi = 0
- 2: \phi = \frac{\pi}{2}
- 3: z = 5
- 4: z = 0
- 5: the cylindrical wall
\oint_S \vec{v} \circ d\vec{a} = \sum_{i=1}^5 \int_{S_{i}}
SURFACE 1:
\int_{S_{1}} = [s(...) \hat{s} + (3z) \hat{z}] \circ [- ds dz \hat{\phi}] = 0.
The reason is that this surface is for \phi = 0 and sin \phi is zero. Thus the \hat{\phi} term drops out of \vec{v} and so the dot product is zero. IOW, no integral needs to be done.
SURFACE 2:
\int_{S_{2}} = [s(...) \hat{s} + (3z) \hat{z}] \circ [+dsdz \hat{\phi}] = 0.
Same reasoning here as with surface 1 but with \phi = \frac{\pi}{2}.
SURFACE 3:
\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
... = 3 \times 5 \cdot \int_0^2sds \int_0^\frac{\pi}{2}d\phi
... = 15 \cdot \frac{4}{2} \frac{\pi}{2}
... = 15\pi
SURFACE 4:
\int_{S_{4}} = [s(...) \hat{s} + (s...) \hat{\phi} + (3z) \hat{z}] \circ [+sdsd\phi (-\hat{z})] = 0.
The reason is that this surface is for z = 0 and since only the \hat{z} term survives the dot product, then the integral is 0. IOW, no integral needs to be done.
SURFACE 5:
\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
... = 4 \cdot ( \int_0^\frac{\pi}{2} 2 d\phi + \int_0^\frac{\pi}{2} sin^2\phi d\phi ) \cdot \int_0^5dz
... = 4 \cdot ( \pi + \pi ) \cdot 5
... = 4 \cdot 2\pi \cdot 5
... = 40\pi
Therefore, after summing all of the open surface integrals to obtain the final closed surface integral, the answer is 55\pi. This of course does not equal the 40\pi for the volume integral found above. (NOTE: I suspect the integral over the 3rd 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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