seemingly simple


by Phymath
Tags: seemingly, simple
Phymath
Phymath is offline
#1
Sep5-05, 08:43 PM
P: 189
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!
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Reshma
Reshma is offline
#2
Sep6-05, 08:57 AM
P: 777
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
Your surface integral is wrong.
[tex] d\vec{a} = R^2 sin\theta d\theta d\phi \hat{r}[/tex] not R3.


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