Problem with Spherical Surface Integral

1. Oct 6, 2005

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:

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

$$-\pi r_{0}$$

2. Oct 6, 2005

StatusX

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

3. Oct 6, 2005

Tom Mattson

Staff Emeritus
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.