(Not) just your average problem

The average value of a consistant function f defined on a curve K can be determined by:

$\frac{\int_{K} fds}{\int_{K} ds}$

First we note that the plane cuts out a large circle from the sphere which position is defined by the normal vector (1, 1, 1). This induces the idea of a linear transformation of the cartesian coordinates into coordinates regarding the cut circle. We find 3 unit vectors of an orthonormal system: $(0, \sqrt{2}/2, -\sqrt{2}/2) / (-2/\sqrt{6}, 1/\sqrt{6}, 1/\sqrt{6}) / (1/\sqrt{3}, 1/\sqrt{3},1/\sqrt{3})$

The matrix having these basis vectors in its columns transforms coordinates regarding the new basis (above) into cartesian coordinates whereat the coordinate regarding the third basis vector above has to be equal 0.

Implementation into the integrand and transitition to polar coordinates delivers for the integrand: $1/6cos^{4}\phi-1/9sin^{4}\phi-1/6cos^{6}\phi+7/54sin^{6}\phi$

Integration fom 0 to $2\pi$ and dividing through $2\pi$ (=length of circle with radius 1) after doing the reciprocal finally gives us 108.

As we have seen here , there exists an arclength parameterization of $C$ as $x=a\cos(t)-b\sin(t),y=-a\cos(t)-b\sin(t),z=2b\sin(t)$ , where $a=\frac{1}{\sqrt{2}}$ and $b=\frac{1}{\sqrt{6}}$ . This gives us $x^2y^2z^2=\frac{1}{54}\sin^2(3t)$ , for $0\leq t\leq 2\pi$ . Thus the average value is $\frac{1}{2\pi}\int_{0}^{2\pi}x^2y^2z^2dt=\frac{1}{108\pi}\int_{0}^{2\pi}\sin^2(3t)dt=\frac{1}{108}$ .

The answer is $\boxed{108}$ .