try it 5

FindRoot[r xskin[s,t] == 0 && r yskin[s,t] == 1 && r zskin[s,t] == 0, {s, 0, Pi}, {t, tlow, thigh}, {r, 0, 1}]

12.5664/(2Pi)

0.131892 {xskin[0,1.46608],yskin[0,1.46608],zskin[0,1.46608]}

Yes. As shown above, the singularity {0, 1, 0} occurs inside the domains of s and t, although r just needs to be reduced (indicating that the singularity is inside of the surface rather than on the surface).



This is a poor idea since calculating the divField ignores the presence of singularities. Since it has been shown that there is a singularity inside of the surface, you would need to check for the flow along at the singularity in order to conclusvely say that the net flow of the field across the surface is 0.



divField1[x_, y_, z_] = Together[D[Field1[x,y,z][[1]], x] + D[Field1[x,y,z][[2]], y] + D[Field1[x,y,z][[3]], z]]

divField2[x_, y_, z_] = Together[D[Field2[x,y,z][[1]], x] + D[Field2[x,y,z][[2]], y] + D[Field2[x,y,z][[3]], z]]

divField[x_, y_, z_] = Together[D[Field[x,y,z][[1]], x] + D[Field[x,y,z][[2]], y] + D[Field[x,y,z][[3]], z]]

Since divField1[x,y,z] does not have a singularity, I can assume that the flow across the region will be 0. However, I still have to calculate divField2[x,y,z], as it has a singularity that falls inside the surface.

singularity = {0, 1, 0};

Clear[x, y, z, s, t, normal];
{x[s_, t_], y[s_, t_], z[s_, t_]} = singularity + {Sin[s] Cos[t], Sin[s] Sin[t], Cos[s]};

{{slow, shigh}, {tlow, thigh}} = {{0, Pi}, {0, 2 Pi}};

normal[s_, t_] = Simplify[TrigExpand[Cross[D[{x[s, t], y[s, t], z[s, t]}, s], D[{x[s, t], y[s, t], z[s, t]}, t]]]]

NIntegrate[Field2[x[s, t], y[s, t], z[s, t]] . normal[s, t], {s, slow, shigh}, {t, tlow, thigh}]

To find the flow across the singularity, I had to center a small sphere around it when calculating the double integral. By doing so, I discovered that the flow of Field[x,y,z] across the plotted region is inside to outside.



No, as the singularity is {12, 0, 0}, which is way outside of the region. A simple glance at the plotted region clearly reveals this.



divField is incorrect when a singularity is present inside of a surface. In this instance, the singularity is outside of the region so I can safely assume that a divField[x,y,z] of 0 also indicates a net flow across of 0.