<s>[INST] Contradictions when do integral $\unicode{x222F} ~f(x,y,z)~dS $, given $f(tx,ty,tz)=t^4 f(x,y,z)$ and $f_{xx}+f_{yy}+f_{zz}=x^2+y^2+z^2$ If $f(tx,ty,tz)=t^4 f(x,y,z)$ and $f_{xx}+f_{yy}+f_{zz}=x^2+y^2+z^2$, then compute the surface integral on the unit sphere $x^2+y^2+z^2=1$:
$$\unicode{x222F} ~f(x,y,z)~dS $$
One particular solution is $f_p=\frac{1}{12}(x^4+y^4+z^4)$,
Another particular solution is $f_p=\frac{1}{4}(x^2y^2+y^2z^2+z^2x^2)$.
Any solutions of the form $f=f_p+h$ where $h$ is a harmonic function and homogenous with degree $4$ (for example, $h=x^4-6x^2y^2+y^4$) will satisfy the given conditions.
Take solution $f=f_p+h$ by Mean Value Theorem, the integral on the harmonic function $h$ equals the value $4\pi h(0,0,0)$. Due to $h$ is homogeneous of degree $4$, we have $h(0,0,0)=0$, so the integral equals
$$I=\unicode{x222F} ~f_p~dS=\int_0^{2\pi} \int_0^\pi f_p(\theta, \phi) \sin(\phi)d\phi d\theta$$
But the problem is this integral seems depends on the choice of $f_p$
If let $f_p=\frac{1}{12}(x^4+y^4+z^4),~~I=\frac{7\pi^2}{64}$
If let $f_p=\frac{1}{4}(x^2y^2+y^2z^2+z^2x^2),~~I=\frac{11\pi^2}{128}$
Do I make some mistakes?
 [/INST] Both integrals should have a value of $\dfrac\pi5$.

Using the first choice of $f_p$:
$$\iint\limits_{x^2+y^2+z^2=1} \frac{x^4+y^4+z^4}{12} \, dS \\ = \int_0^{2\pi} \int_0^\pi \frac{\cos^4(\theta)\sin^4(\phi) + \sin^4(\theta)\sin^4(\phi) + \cos^4(\phi)}{12} \sin(\phi) \, d\phi \, d\theta$$
Now,
$$\begin{align*}
I &= \int_0^{2\pi} \int_0^\pi \cos^4(\theta) \sin^5(\phi) \, d\phi \, d\theta \\[1ex]
&= \left(\int_0^{2\pi} \cos^4(\theta)\,d\theta\right) \left(\int_0^\pi \sin^5(\phi) \, d\phi\right) \\[1ex]
&= \left(\int_0^{2\pi} \frac{3 + 4\cos(2\theta) + \cos(4\theta)}8 \, d\theta\right) \left(\int_0^\pi \sin(\phi) (1 - \cos^2(\phi))^2 \, d\phi\right) \\[1ex]
&= \frac{3\pi}4 \times \frac{16}{15} = \frac{4\pi}5
\end{align*}$$
The integrals of $\sin^4(\theta)\sin^4(\phi)$ and $\cos^4(\phi)$ have the same value, so the surface integral is $\dfrac{I+I+I}{12} = \dfrac\pi5$.

Using the second choice of $f_p$:
$$\iint\limits_{x^2+y^2+z^2=1} \frac{x^2y^2 + x^2z^2 + y^2z^2}4 \, dS \\ = \int_0^{2\pi} \int_0^\pi \frac{\cos^2(\phi)\sin^2(\phi) + \cos^2(\theta)\sin^2(\theta)\sin^2(\phi)}4 \sin(\phi) \, d\phi \, d\theta$$
Now,
$$\begin{align*}
I_1 &= \int_0^{2\pi} \int_0^\pi \cos^2(\phi)\sin^3(\phi) \, d\theta \, d\phi \\[1ex]
&= 2\pi \int_0^\pi \cos^2(\phi) (1 - \cos^2(\phi)) \sin(\phi) \, d\phi \\[1ex]
&= 2\pi\times\frac4{15} = \frac{8\pi}{15} \\[2ex]
I_2 &= \int_0^{2\pi} \int_0^\pi \cos^2(\theta) \sin^2(\theta) \sin^5(\phi) \, d\phi \, d\theta \\[1ex]
&= \left(\int_0^{2\pi} \left(\frac{\sin(2\theta)}2\right)^2 \, d\theta\right) \left(\int_0^\pi(1-\cos^2(\phi))^2\sin(\phi)\,d\phi\right) \\[1ex]
&= \frac14 \left(\int_0^{2\pi} \frac{1-\cos(4\theta)}2 \, d\theta\right) \left(\int_0^\pi(1-\cos^2(\phi))^2\sin(\phi)\,d\phi\right) \\[1ex]
&= \frac14\times\pi\times\frac{16}{15} = \frac{4\pi}{15}
\end{align*}$$
Then the surface integral is again $\dfrac{I_1+I_2}4 = \dfrac\pi5$.
</s>