# seemingly simple

by Phymath
Tags: seemingly, simple
 P: 189 ok well basicly use the divergence thrm on this.. $$\vec{v} = rcos\theta \hat{r} + rsin\theta\hat{\theta} + rsin\theta cos \phi \hat{\phi}$$ so i did (remember spherical coords) i get.. $$5cos \theta - sin\phi$$ taking that over the volume of a hemisphere resting on the xy-plane i get $$10R \pi$$ however with the surface intergral i use $$d\vec{a} = R^3 sin\theta d\theta d\phi \hat{r}$$ 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!
P: 777
 however with the surface intergral i use $$d\vec{a} = R^3 sin\theta d\theta d\phi \hat{r}$$ where R is the radius of the sphere doing that intergral
$$d\vec{a} = R^2 sin\theta d\theta d\phi \hat{r}$$ not R3.

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