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    Surface integral

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    Double of F*dS,over S, where

    F=(x*i+y*j+z*k)/(x^2+y^2+z^2)

    and where S is the surface of the infinite cylinder x^2+y^2=1; n pointing outward

    © BrainMass Inc. brainmass.com September 28, 2022, 3:27 am ad1c9bdddf
    https://brainmass.com/physics/flux-flux-density/surface-integrals-flux-28513

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    Instead of dS I used the notation da, since ds is usually used for path integral, rather surface integral.

    We can write the surface unit as

    The normal vector is:

    And the surface element is:

    Thus:

    On the surface of the cylinder r=1 thus:

    The vector field is

    Converting to cylindrical coordinates:

    We obtain:

    Using again the fact that on the surface r=1 we get:

    Therefore:

    For the integral we have  between 0 and 2, and z between -∞ to ∞.

    The integral becomes:

    Using the substitution we obtain:

    And therefore:

    But , so:

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com September 28, 2022, 3:27 am ad1c9bdddf>
    https://brainmass.com/physics/flux-flux-density/surface-integrals-flux-28513

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