- #1

- 190

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Can I do:

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

[tex]G(x,y,z) = r^{2}=x^{2}+y^{2}[/tex]

[tex]\nabla G = <2x, 2y, 0>[/tex]

[tex]\int\int \vec{F}\cdot <2x,2y,0> dA[/tex]

Is this a correct approach?

- Thread starter IniquiTrance
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- #1

- 190

- 0

Can I do:

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

[tex]G(x,y,z) = r^{2}=x^{2}+y^{2}[/tex]

[tex]\nabla G = <2x, 2y, 0>[/tex]

[tex]\int\int \vec{F}\cdot <2x,2y,0> dA[/tex]

Is this a correct approach?

- #2

- 45

- 0

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

Can I do:

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

[tex]G(x,y,z) = r^{2}=x^{2}+y^{2}[/tex]

[tex]\nabla G = <2x, 2y, 0>[/tex]

[tex]\int\int \vec{F}\cdot <2x,2y,0> dA[/tex]

Is this a correct approach?

(1) Find the divergence of [tex]\vec{F}[/tex]

(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.

- #3

- 190

- 0

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

(1) Find the divergence of [tex]\vec{F}[/tex]

(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.

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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