1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with Spherical Surface Integral

  1. Oct 6, 2005 #1
    [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]
  2. jcsd
  3. Oct 6, 2005 #2


    User Avatar
    Homework Helper

    You can't integrate over the spherical basis vectors because they change with position. You need to transform the vectors into cartesian coordinates.
  4. Oct 6, 2005 #3

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook