Net flux calculation using symmetry

Question

Asked to find net flux of $ \underline f = x^2 \underline i + y^2 \underline j + z^2 \underline k$ through a surface $S$ defined by $x^2 + y^2 +z^2 = a^2$ by symmetry. I recognise that $x^2$ etc. are all symmetric functions, but surely for each octant there will therefore be positive outward flux - not cancellation of fluxes and a net outward flux of zero? I can't seem to visualise why these flux contributions cancel.

Answer 1

If you consider the direction of the flux, it is always positive (i.e. pointing along the non-negative sense of the i-j-k unit vectors) since each of the variables are squared. Consider a point on the x-y plane: (1,0,0). The flux is then (1,0,0). Now consider an opposite point (-1,0,0) which will also have the flux (1,0,0). Both these flux vectors point in the same direction, so you can imagine them cancelling if evaluated crossing the boundary of a circle of radius 1 in the x-y plane. I believe you can generalize the argument along these lines for your case. I hope this helps.

Answer 2

Note that the unit normal at $(x,y,z)$ is $\dfrac1a(x,y,z)$, so flux integral becomes $$\frac1a\iint_S (x^3+y^3+z^3)\,d\sigma.$$ Now you can invoke symmetry. Since $x^3$, for example, is an odd function, and since the sphere is symmetric about the plane $x=0$, the integral over that portion of $S$ with $x<0$ will be the negative of the integral over that portion of $S$ with $x>0$.