Compute the integral $$\oint (ax+by+cz)^{2018} \,d\operatorname{vol} (S)$$ where $S = \{x^2 +y^2 +z^2 = 1\}$ (Unit sphere on $\mathbb{R^3}$) and $a,b,c\in \mathbb{R}$.
I thought to use Divergence theorem with the field $F(x,y,z) = (a,b,c)$, because the normal on $S$ is $N(x,y,z) = (x,y,z)$.
and we get inside the bracket $\langle F(x,y,z),N(x,y,z)\rangle$
The problem is that I don't know how to deal with the power 2018.
Any help will be appreciated.
Thanks.
If we let $$\hat n=\frac{\langle a,b,c\rangle}{\sqrt{a^2+b^2+c^2}}$$ Then $$\begin{align}\int_S(ax+by+cz)^{2018}d^3V&=(a^2+b^2+c^2)^{1009}\int_S(\hat n\cdot\vec r)^{2018}d^3V\\ &=(a^2+b^2+c^2)^{1009}\int_Sz^{2018}d^3V\end{align}$$ If we change to a spherical coordinate system where the $z$-axis is aligned with $\hat n$. Then $$\begin{align}\int_S(ax+by+cz)^{2018}d^3V&=(a^2+b^2+c^2)^{1009}\int_0^{\pi}\sin\theta\int_0^{2\pi}\int_0^1r^2r^{2018}\cos^{2018}\theta dr\,d\phi\,d\theta\\ &=(a^2+b^2+c^2)^{1009}\left(\frac2{2019}\right)(2\pi)\left(\frac1{2021}\right)\\ &=\frac{4\pi(a^2+b^2+c^2)^{1009}}{4080399}\end{align}$$ Now, it's not clear whether you wanted the surface integral or the volume integral, so I just did the volume integral. If you need the surface integral, just omit the integration over $r$.