Is This the Correct Method to Compute Flux Integral Over Cylinder Walls?

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The discussion focuses on computing the flux integral over the walls of a cylinder oriented along the z-axis. The proposed method involves using the gradient of G(x,y,z) = r² to express the flux integral as ∫∫ F·∇G dA. While some participants suggest using the Divergence Theorem for simplicity, the original poster seeks to solve it directly as a flux integral without this theorem. Clarification is requested on whether the initial approach is valid for calculating the flux over the cylinder walls. The conversation emphasizes the importance of correctly applying mathematical principles in vector calculus.
IniquiTrance
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To compute the flux integral over a cylinder's walls oriented along the z axis:

Can I do:

\int\int \vec{F}\cdot\nabla G(x,y,z) dA

G(x,y,z) = r^{2}=x^{2}+y^{2}

\nabla G = <2x, 2y, 0>

\int\int \vec{F}\cdot <2x,2y,0> dA

Is this a correct approach?
 
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IniquiTrance said:
To compute the flux integral over a cylinder's walls oriented along the z axis:

Can I do:

\int\int \vec{F}\cdot\nabla G(x,y,z) dA

G(x,y,z) = r^{2}=x^{2}+y^{2}

\nabla G = <2x, 2y, 0>

\int\int \vec{F}\cdot <2x,2y,0> dA

Is this a correct approach?

Assuming a positive orientation, the easiest way to do it is by Divergence Theorem.

(1) Find the divergence of \vec{F}

(2) Integrate this over the solid cylinder.

The other way is to split the cylinder into 3 pieces the Top, Bottom and Side and the sum the flux contributed from each piece.
 
~Death~ said:
Assuming a positive orientation, the easiest way to do it is by Divergence Theorem.

(1) Find the divergence of \vec{F}

(2) Integrate this over the solid cylinder.

The other way is to split the cylinder into 3 pieces the Top, Bottom and Side and the sum the flux contributed from each piece.

Thanks for the reply.

Yeah, I specifically want to solve it as a flux integral without the div theorem.

Also know how to split it up. Is this a proper way to compute it over the cylinder walls though?
 

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