Question; What is the flux of the vector field $F=<x,-y,z>$ out through the surface of the cylinder which is $x^{2} + y^2 = a^2 , 0<z<b$.
I can easily find the answer by divergence theorem but when I try to direct calculation of the flux through the surface , I find 0.Am I doing some algebra mistakes or do I miss anything different ?
My path;
$$\int \int <x,-y,z>.\frac{<x,y,0>}{a} dS \\ x=a*cos\theta \\ y=a*sin\theta \\ \int_0^b \int_0^{2\pi} \frac{a^2 *cos2\theta}{a}*a*d\theta dz \\ $$
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$\begingroup$The integration over divergence is correct and it is $\pi b a^2$
The integration on surface is correct except for mission the top and the bottom of the cylinder!
$$\int \vec F .d \vec S= \text{flux around} + \text{flux top} + \text{flux bottom}=\\\iint (x,-y,z).(\frac x a,\frac y a,0) dS+\iint (x,-y,b).(0,0,1)dS+\iint (x,-y,0).(0,0,-1)dS\\ =\pi b a^2$$
Using divergence:
$$\vec \nabla . \vec F=1$$
$$\iiint (\vec \nabla . \vec F) dV=V=\pi b a^2$$
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