Vector field flow over surface in 3D

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SUMMARY

The discussion focuses on calculating the flow of the vector field ##\vec{F}=(y^2,x^2,x^2y^2)## over the surface defined by the equation ##x^2+y^2+z^2=R^2## for ##z \geq 0## using Gauss's Law. The user correctly applies Gauss's Law, stating that since ##\nabla\vec{F}= 0##, the integral over the closed surface equals zero. The calculation of the flow through the circular surface ##O## at ##z=0## is performed using polar coordinates, leading to the result of ##-\frac{\pi R^6}{96}##. The user seeks clarification on the orientation of the normal vector, which is confirmed to be correct as it points away from the enclosed volume.

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Homework Statement


Calculate the flow of ##\vec{F}=(y^2,x^2,x^2y^2)## over surface ##S## defined as ##x^2+y^2+z^2=R^2## for ##z \geq 0## with normal pointed away from the origin.

Homework Equations


The Attempt at a Solution



The easiest was is probably with Gaussian law. I would be really happy if somebody could correct me if I am wrong and answer my question below:

Gaussian law: ##\int \int _O\vec{F}d\vec{S}+\int \int _S\vec{F}d\vec{S}=\int \int \int_{Body} \nabla\vec{F}dV## where I used notation ##O## for the circle.

Now ##\nabla\vec{F}= 0## therefore ##\int \int _O\vec{F}d\vec{S}+\int \int _S\vec{F}d\vec{S}=0## so all that remains is to calculate the floe through surface ##O##.

Using polar coordinates ##x=r \cos \varphi ## and ##y= r \sin \varphi## for ##z=0##. Than ##r_{\varphi } \times r_{r}=(0,0,-r)##

##\int \int _O\vec{F}d\vec{S}=-\int_{0}^{2\pi }\int_{0}^{R}r^{5} \cos^2 \varphi \sin^2 \varphi d\varphi dr##

That should be ##-\frac{\pi R^6}{96}##.

Question here: I am a bit confused weather I should use the other sign here ##r_{\varphi } \times r_{r}=(0,0,-r)## or is this the right one?
 
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Gauß's Law is for all the surface normal vectors pointing away from the enclosed volume. So your idea is correct.
 

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