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Use spherical coordinates $x=s\sin \theta \cos \varphi$, $y=s\sin \theta \sin \varphi$, $z=s\cos \theta$. The density becomes ${\rho }_{0}s\sin \theta$, and the Jacobian determinant supplies $dV=s^2\sin \theta dsd\theta d\varphi$. Therefore the mass is the volume integralUsing ${\sin }^2\theta =(1-\cos 2\theta )/2$, the three integrals are $2\pi$, $\pi /2$ and $1/4$. Hence

The complementary function solves the associated homogeneous linear differential equation. Substituting $y=e^{mx}$ gives the characteristic equationIts distinct polynomial roots giveThe forcing affects the particular solution, which is not needed for the complementary function.

The product-to-sum formula immediately gives the Fourier series:In the convention $f(x)=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$, the only nonzero Fourier coefficients are $a_2=a_8=1/2$. This is a finite trigonometric polynomial, and the identity holds at both endpoints as well.

No. Every one of the three vectors has equal first and second Cartesian coordinates, so their linear span is contained in the plane $x=y$. That linear subspace has dimension two and cannot equal ${\mathbb{R}}^3$.

More explicitly, call the three given vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ in their listed order. ThenThis nontrivial linear relation proves linear dependence, whereas a basis must be linearly independent. Being unit vectors alone does not ensure that they form a basis.