Divergence Theorem: Volume and Surface Integral Solutions

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The discussion focuses on applying the divergence theorem to a vector field in spherical coordinates, specifically analyzing the volume and surface integrals of a hemisphere. The user initially calculates the volume integral and finds a result of 10Rπ. However, they encounter a discrepancy when performing the surface integral, leading to confusion over the correct differential area element. Another participant points out that the user incorrectly used R^3 instead of the correct R^2 for the surface integral. This correction highlights the importance of accurately applying the divergence theorem in spherical coordinates.
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ok well basicly use the divergence thrm on this..
<br /> \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 explanations (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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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
Your surface integral is wrong.
d\vec{a} = R^2 sin\theta d\theta d\phi \hat{r} not R3.
 
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