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seemingly simple |
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| Sep5-05, 08:43 PM | #1 |
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seemingly simple
ok well basicly use the divergence thrm on this..
[tex] \vec{v} = rcos\theta \hat{r} + rsin\theta\hat{\theta} + rsin\theta cos \phi \hat{\phi}[/tex] so i did (remember spherical coords) i get.. [tex] 5cos \theta - sin\phi[/tex] taking that over the volume of a hemisphere resting on the xy-plane i get [tex]10R \pi[/tex] however with the surface intergral i use [tex] d\vec{a} = R^3 sin\theta d\theta d\phi \hat{r}[/tex] where R is the radius of the sphere doing that intergral gives me something obvioiusly R^3 which is not what I'm getting in the volume intergral any help or explainations (if its i need to take the surface of the bottom of the hemi sphere aka the xy-plane in a circle that will suck but let me know! |
| Sep6-05, 08:57 AM | #2 |
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[tex] d\vec{a} = R^2 sin\theta d\theta d\phi \hat{r}[/tex] not R3. |
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