Problem with Spherical Surface Integral

[Noone1982]

A\; =\; 4\dot{r}\; +\; 3\dot{\theta }\; -\; 2\dot{\phi }

Now the surface integral integral is:

\int_{}^{}{\left( ?\times A \right)\; •\; da}

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

?\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]

I get:

?\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]

Now I dot this to da

where da is:

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 }

I get:

\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 \;

which equals:

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

The answer should be

-\pi r_{0}

 

[StatusX]

You can't integrate over the spherical basis vectors because they change with position. You need to transform the vectors into cartesian coordinates.

 

[quantumdude]

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 \hat{r} does depend on position, \hat{r}\cdot\hat{r} does not.