ok well basicly use the divergence thrm on this..(adsbygoogle = window.adsbygoogle || []).push({});

[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!

**Physics Forums - The Fusion of Science and Community**

# Seemingly simple

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Seemingly simple

Loading...

**Physics Forums - The Fusion of Science and Community**