Evaluate $$\iint_S \left (x^4+y^4+z^4 \right )\ dS,$$ where $S=\left \{(x,y,z)\ :\ x^2+y^2+z^2 = a^2 \right \},\ a > 0.$
My attempt $:$ I have tried to find the integral using Gauss' divergence theorem which states that
Suppose $V$ is a subset of $\Bbb R^n$ (in this case $n=3,$ which represents the volume of the sphere of radius $a$ centered at the origin) which is compact and has a piecewise smooth boundary $S.$ If $\textbf {F}$ is a continuously differentiable vector field defined on a neighbourhood of $V,$ then
$$\begin{align*} \iiint_V\left(\mathbf{\nabla}\cdot\mathbf{F}\right)\,dV & = \iint_{S}(\mathbf{F}\cdot\mathbf{n})\,dS . \end{align*} $$
In this case $\textbf {F} = x^3 \hat {i} + y^3 \hat {j} + z^3 \hat {k},$ which is a continuously differentiable vector field on a neighbourhood of $V$ and $\textbf {n} (x,y,z) = \frac {x \hat i + y \hat j + z \hat k} {a}.$ Hence we have \begin{align*} \iint_S \left (x^4 + y^4 + z^4 \right )\ dS & = a \iiint_V \left (\mathbf {\nabla} \cdot \mathbf {F} \right)\ dV\\ & = a \iiint_V 3 \left (x^2+y^2+z^2 \right )\ dV\\ & = 3a^3 \iiint_V dV \\ & = 3a^3 \times \frac 4 3 \pi a^3 \\ & = 4 \pi a^6. \end{align*}
But the answer given as $\frac {12 \pi} {5} a^6 \neq 4 \pi a^6.$ Where did I do mistake? Any help in this regard will be highly appreciated.
Thank you very much.
EDIT $:$ Here is another way to find the integral. Let $\varphi : [0, \pi] \times [0, 2 \pi] \longrightarrow \Bbb R^3$ be the continuously differentiable parameterization of the given sphere into spherical coordinates i.e. $$\varphi (u,v) = \left (a \sin u \cos v , a \sin u \sin v , a \cos u \right ),\ (u,v) \in [0, \pi] \times [0, 2 \pi].$$ Then the above surface integral can be evaluated as follows $:$
Let $F(x,y,z) = x^4 + y^4 + z^4 ,\ (x,y,z) \in \Bbb R^3.$ Then we have
$$\begin{align*} \iint_S F(x,y,z)\ dS & = \int_{0}^{2 \pi} \int_{0}^{\pi} F \left (\varphi (u,v) \right ) \left \| \varphi_u \times \varphi_v \right \|\ du\ dv \\ & = a^6 \int_{0}^{2 \pi} \int_{0}^{\pi} \left (\sin^5 u \cos^4 v + \sin^5 u \sin^4 v + \sin u \cos^4 u \right )\ du\ dv. \end{align*}$$
The computation of the last integral is not so easy though.
Best Answer
Recognize the symmetry to simplify the integral first,
$$I=\int_S \left (x^4+y^4+z^4 \right )\ dS,=3\int_S z^4 \ dS$$
Then, integrate with spherical coordinates $z=a\cos\theta$, $dS = a^2\sin\theta d\theta d\phi$, $$I=3 \int_0^{2\pi}\int_0^{\pi}(a\cos\theta)^4 a^2\sin\theta d\theta d\phi =-6\pi a^6 \int_0^{\pi}\cos^4\theta d(\cos\theta) =\frac{12\pi}5a^6$$
Edit:
$$I= 3a \int_V \left (x^2+y^2+z^2 \right )dV =9a\int_V z^2 dV$$ $$=9a\int_0^{2\pi}\int_0^{\pi}\int_0^a (r\cos\theta)^2 r^2\sin\theta drd\theta d\phi=\frac{12\pi}5a^6$$