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The square wave alternates between a low interval of length five and a high interval of length five; its integral average is three. In the real Fourier series conventionintegration over one period givesandThus only odd harmonics occur:The Dirichlet-Jordan convergence theorem makes the sum equal to the function at its continuity points and to the mean of its one-sided limits at a jump. Consequently the sum is 3 at each of $x=-5,0,5$, regardless of any separately assigned values of the original function at those endpoints.

For $h=x^2/a^2+y^2/b^2+z^2-1$, the gradient $\nabla h=2(x/a^2,y/b^2,z)$ is a normal vector to the ellipsoid. The outward unit normal vector is $(x/a^2,y/b^2,z)/\sqrt{x^2/a^4+y^2/b^4+z^2}$. The conditions $a,b>\sqrt{2}$ ensure that $z=\sqrt{1-x^2/a^2-y^2/b^2}>0$ everywhere on the rectangular projection.

The dot product $F\cdot \nabla h=2xy(1/b^2-1/a^2)$ vanishes pointwise when $a=b$. In that case $F$ has no normal component, so the surface flux is zero. For unequal axes, the given vector surface element givesPut $u=x^2/a^2$, $v=y^2/b^2$. Then $xydxdy=a^2{b}^{2}dudv/4$, andTwo elementary integrations now giveThis is the rotational flux through an ellipsoidal graph patch with unit vertical semiaxis and unit projection lengths. It also tends continuously to zero when $a=b$, agreeing with the pointwise argument.

Use Gaussian elimination. The last equation gives $y=c-z$, and the second then gives $x=b-4c-2z$. Substitute these into the first equation to obtain $3b-10c-z=a$. Back-substitution yieldsThe linear system has matrix formThe coefficients of the arbitrary right-hand side in the solved expressions are the rows of the matrix inverse. ThereforeMultiplying the original matrix by this candidate gives the identity matrix; its determinant is $-1$, so the solution is unique for every $(a,b,c)$.