# Surface integral

Find :

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

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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:

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