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Problem with Spherical Surface Integral

by Noone1982
Tags: integral, spherical, surface
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Oct6-05, 10:16 AM
P: 83
[tex]A\; =\; 4\dot{r}\; +\; 3\dot{\theta }\; -\; 2\dot{\phi }[/tex]

Now the surface integral integral is:

[tex]\int_{}^{}{\left( ?\times A \right)\; \; da} [/tex]

(the ? mark is a del operator and the dot over a variable means a unit vector)

[tex]?\times A\; =\frac{\dot{r}}{r\sin \theta }\left[ \frac{\partial }{\partial \theta }\left( \sin \theta A_{\phi } \right)\; -\; \frac{\partial A_{\theta }}{\partial \phi } \right]\; +\; \frac{\dot{\theta }}{r}\left[ \frac{1}{\sin \theta }\frac{\partial A_{r}}{\partial \phi }\; -\; \frac{\partial }{\partial r}\left( rA_{\phi } \right) \right]\; +\; \frac{\dot{\phi }}{r}\left[ \frac{\partial }{\partial r}\left( rA_{\theta } \right)\; -\; \frac{\partial A_{r}}{\partial \theta } \right] [/tex]

I get:

[tex]?\times A\; =\frac{\dot{r}}{r\sin \theta }\left[ \left( -2\cos \theta \right)\; -\; 0 \right]\; +\; \frac{\dot{\theta }}{r}\left[ \frac{1}{\sin \theta }\left( 0 \right)\; +\; 2 \right]\; +\; \frac{\dot{\phi }}{r}\left[ 3\; -0 \right][/tex]

Now I dot this to da

where da is:

[tex]da\; =\; r^{2}\sin \theta \; d\theta \; d\phi \; \dot{r}\; +\; r\sin \theta \; dr\; d\phi \; \dot{\theta }\; +\; r\; dr\; d\theta \; \dot{\phi }[/tex]

I get:

[tex]\int_{}^{}{\int_{}^{}{}}-2\cos \theta r\; d\theta \; d\phi \; \; +\; \int_{}^{}{\int_{}^{}{}}2\sin \theta \; dr\; d\phi \; +\int_{0}^{ro}{\int_{\frac{\pi }{2}}^{\frac{\pi }{2}}{}}3\; dr\; d\theta \; [/tex]

which equals:

[tex]-2\sin \theta r\phi \; +\; 2\sin \theta r\phi \; +\; \frac{3}{2}\pi r_{o}\; =\; \frac{3}{2}\pi r_{o}[/tex]

The answer should be

[tex]-\pi r_{0}[/tex]
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Oct6-05, 06:06 PM
HW Helper
P: 2,566
You can't integrate over the spherical basis vectors because they change with position. You need to transform the vectors into cartesian coordinates.
Tom Mattson
Oct6-05, 06:46 PM
Sci Advisor
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Tom Mattson's Avatar
P: 5,533
He doesn't need to change basis, he just messed up his LaTeX. There are supposed to be dot products among the basis vectors in there. So while [itex]\hat{r}[/itex] does depend on position, [itex]\hat{r}\cdot\hat{r}[/itex] does not.

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