- #1
BOAS
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Homework Statement
(3 i) Using [itex]\nabla . \mathbf{F} = \frac{\partial \mathbf{F_{\rho}}}{\partial \rho} + \frac{\mathbf{F_{\rho}}}{\rho} + \frac{1}{\rho} \frac{\partial \mathbf{F_{\phi}}}{\partial \phi} + \frac{\partial \mathbf{F_{z}}}{\partial z}[/itex] calculate the divergence of the vector field [itex]\mathbf{F} = 3\mathbf{e_{\rho}} + \cos(\phi)\mathbf{e_{\phi}} + 5\rho\mathbf{e_{z}}[/itex].
(ii) Evaluate the volume integral of [itex]\nabla . \mathbf{F}[/itex] over the cylinder defined by [itex]-H \leq z \leq H, x^{2} + y^{2} \leq R^{2}[/itex].
Homework Equations
The Attempt at a Solution
I'm not sure if I'm doing the volume integral properly.
[itex]\frac{\partial \mathbf{F_{\rho}}}{\partial \rho} = 0[/itex]
[itex]\frac{\partial \mathbf{F_{\phi}}}{\partial \phi} = -\sin(\phi)[/itex]
[itex]\frac{\partial \mathbf{F_{z}}}{\partial z} = 0[/itex]
[itex]\nabla . \mathbf{F} = \frac{3}{\rho} + \frac{1}{\rho}(-\sin(\phi) + 0 = \frac{3 - \sin(\phi)}{\rho}[/itex]
So that's part one done - Moving on to converting this cylinder to cylindrical polar coordinates.
[itex]x = \rho \cos(\phi)[/itex]
[itex]y = \rho \sin(\phi)[/itex]
[itex]\rho^{2} \cos^{2}(\phi) + \rho^2 \sin^{2}(\phi) \leq R^{2}[/itex]
[itex]\cos^{2}(\phi) + \sin^{2}(\phi) = 1[/itex]
[itex]\rho^{2} \leq R^{2}[/itex]
[itex]\rho \leq R[/itex]
[itex]\int_{V} \nabla . \mathbf{F} dV = \int^{H}_{-H} \int^{2\pi}_{0} \int^{\rho}_{0} \frac{3 - \sin(\phi)}{\rho} \rho d\rho d\phi dz[/itex]
[itex]\int_{V} \nabla . \mathbf{F} dV = \int^{H}_{-H} \int^{2\pi}_{0} \int^{\rho}_{0} 3 - \sin(\phi) d\rho d\phi dz[/itex]
[itex]\int_{V} \nabla . \mathbf{F} dV = \int^{H}_{-H} \int^{\rho}_{0} [3 - \sin(\phi)]^{2\pi}_{0} d\rho dz[/itex]
[itex]\int_{V} \nabla . \mathbf{F} dV = 6\pi \int^{H}_{-H} \int^{\rho}_{0} d\rho dz[/itex]
[itex]\int_{V} \nabla . \mathbf{F} dV = 6\pi \rho \int^{H}_{-H} dz[/itex]
[itex]\int_{V} \nabla . \mathbf{F} dV = 12 \pi \rho H[/itex]
I'm not that confident in my treatment of the limits, so it would be good to get some feedback/know if I'm doing this wrong.
Thanks!