Vector analysis: Find the flux of the vector field through the surface

Vector field: $A(x,y,z) = \frac{1}{x^2 + y^2 } \cdot (x,y)$

Surface $S: x^2 + y^2 = 1$ and $z = [0,1]$

The normal is pointing into the surface.

My solution: I add two surfaces, S1 and S2, to the original surface S. This will let me use the divergence theorem (Gauss theorem) on the closed surface. However, when we take the divergence of vector field A we get zero and thus the answer is that the generated flux through the surfaces S, S1 and S2 combined is zero and therefore to get the flux through S we have to subtract the flux through S1 and S2. The flux through S1 and S2 is both zero and thus we get that the answer must be zero. The real solution to this problem is apparently: $-2 \pi$

The two surfaces S1 and S2 I added:

$S1: x^2+y^2 <= 1 , z=0$

$S2: x^2+y^2 <= 1 , z=1$

• Are you sure the divergence is zero? – Paul Aljabar Oct 8 '17 at 17:45
• I get: Div A = $\frac{x^2+y^2 - x \cdot 2x}{(x^2+y^2)^2} + \frac{x^2+y^2 - y \cdot 2y}{(x^2+y^2)^2} = 0$ – SwedeGustaf Oct 8 '17 at 17:48
• Apologies, I misread the function as having $\sqrt{x^2 + y^2}$ in the denominator. – Paul Aljabar Oct 8 '17 at 18:11
• No problem! I am glad you answered. Have a great day. – SwedeGustaf Oct 8 '17 at 18:14

The vector field ${\bf A}(x, y, z) = \frac{1}{x^2 + y^2} (x, y, 0)$ is not defined at the points $(0, 0, 0)$ and $(0, 0, 1)$, which are respectively contained in the surfaces $S_1$, $S_2$. Therefore the Divergence Theorem does not apply.
Notice that at every point on $S$, $\bf A$ is nonzero and parallel to but pointing in the direction opposite of the normal vector $\bf n$. Thus, the flux $\iint_S {\bf A} \cdot {\bf n} \,dS$ is given by the integral over a surface of the negative function ${\bf A} \cdot {\bf n}$ and hence must itself be negative.