Suggestion for Computing an Integral Let $$A=\left\{(x,y,z)\in \mathbb R^3:\dfrac{x^2}{2}+\dfrac{y^4}{4}+\dfrac{z^6}{6}\leq1\right\}.$$
Then I want to compute the following integral:
$$\frac{1}{\operatorname{vol}(A)}\displaystyle\int_{\partial A}^{}\!\frac{1}{\sqrt{x^2+y^6+z^{10}}}\, d(x,y,z)$$
Should I use spherical coordinates or something like that? 
I am a beginner to this topic, so any help would be nice.
Edit: I put the factor $\frac{1}{\operatorname{vol}(A)}$ before the integral (that's exactly my task now), but there should be no different concerning our problem...
$$I := \frac{1}{\lambda_3(A)} \int\limits_{\partial A} \frac{1}{\sqrt{x^2 + y^6 + z^{10}}} \, dS_{\partial A}$$
 A: Let us express the surface element in terms of $y,z$:
$$dS=\sqrt{1+\left(\frac{\partial x}{\partial y} \right)^2+\left(\frac{\partial x}{\partial z} \right)^2} \,dy dz,\tag{1}$$
where 
$$\frac{x^2}{2}+\frac{y^4}{4}+\frac{z^6}{6}=1. \tag{2}$$
Now from (2) it follows that
$$\frac{\partial x}{\partial y}=-\frac{y^3}{x},\qquad \frac{\partial x}{\partial z}=-\frac{z^5}{x},$$
and therefore the formula (1) transforms into
$$dS=\frac{\sqrt{x^2+y^6+z^{10}}}{|x|}dydz.$$
The surface integral we want to compute (without the inverse volume prefactor) then becomes
$$I_S=4\sqrt{2}\iint_D\frac{dydz}{\sqrt{\displaystyle1-\frac{y^4}{4}-\frac{z^6}{6}}},$$
where $\displaystyle D=\left\{(y,z)\in\mathbb{R}^2:\frac{y^4}{4}+\frac{z^6}{6}\leq1,y\geq0,z\geq0\right\}$. Next define
$$a=\frac{y^2}{2},\qquad b=\frac{z^3}{\sqrt{6}}\qquad
\Longleftrightarrow \qquad y=\sqrt{2a},\qquad z=6^{\frac16}b^{\frac13}$$
and rewrite the integral as
$$I_S=8\cdot 6^{-5/6}\iint_{D'}\frac{a^{-1/2}b^{-2/3}dadb}{\sqrt{1-a^2-b^2}},$$
with $$D'=\{(a,b)\in\mathbb{R}^2:a^2+b^2\leq 1,a\geq0,b\geq0\}.$$
Now it becomes helpful to introduce polar coordinates $a=r\cos\theta,b=r\sin\theta$ and rewrite the integral as
$$I_S=8\cdot 6^{-5/6}\underbrace{\int_0^1\frac{r^{-1/6}dr}{\sqrt{1-r^2}}}_{I_1}\;\underbrace{\int_0^{\pi/2}(\cos\theta)^{-1/2}(\sin\theta)^{-2/3}d\theta}_{I_2}.$$
It is not difficult to express $I_{1,2}$ in terms of gamma functions:
$$I_1=-6\sqrt{\pi}\frac{\Gamma(\frac{5}{12})}{\Gamma(-\frac{1}{12})},\qquad
I_2=\frac12\frac{\Gamma(\frac14)\Gamma(\frac16)}{\Gamma(\frac5{12})},$$
and therefore
$$\boxed{\displaystyle I_S=-24\sqrt{\pi}\cdot 6^{-5/6}\cdot\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{6})}{\Gamma(-\frac{1}{12})}}\tag{3}$$

The volume of $A$ can be computed very similarly. Indeed, 
\begin{align}\operatorname{vol}A&=\iiint_A dx dy dz=\\&=8\sqrt2\iint_D \sqrt{\displaystyle1-\frac{y^4}{4}-\frac{z^6}{6}} \,dy dz=\\&=
16\cdot 6^{-5/6}\iint_{D'}a^{-1/2}b^{-2/3}\sqrt{1-a^2-b^2}\,dadb=\\
&=16\cdot 6^{-5/6}\cdot I_2\cdot \int_0^1 r^{-1/6}\sqrt{1-r^2}dr=\\
&=16\cdot 6^{-5/6}\cdot I_2\cdot\frac{\sqrt\pi}{4}\frac{\Gamma(\frac5{12})}{\Gamma(\frac{23}{12})},
\end{align}
so that
$$\boxed{\displaystyle I=\frac{I_S}{\operatorname{vol}A}=\frac{11}{12}}$$
A: Notice that the normal vector of $A's$ boundary is $\hat{n}=\frac{(x, y^3, z^5)}{\sqrt{x^2+y^6+z^10}}$. Then we choose $F(x,y,z)=(\frac{x}{2},\frac{y}{4},\frac{z}{6})$, on the boundary, we have $$F\cdot \hat{n}=\frac{\frac{x^2}{2}+\frac{y^4}{4}+\frac{z^6}{6}}{\sqrt{x^2+y^6+z^{10}}}=\frac{1}{\sqrt{x^2+y^6+z^{10}}}$$ So the integral
$$\frac{1}{vol(A)}\int_{\partial A}\frac{1}{\sqrt{x^2+y^6+z^{10}}}ds=\frac{1}{vol(A)}\int_{\partial A} F\cdot \hat{n} ds=\frac{1}{vol(A)}\int_{ A}\nabla\cdot F dxdydz  \quad= \frac{1}{vol(A)}\int_{ A} \frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}$$
A: Hint: I suspect that you should use the Divergence theorem. First, note that the normal vector to the surface under consideration ($\partial A$) is$$\begin{array}{c}\hat n = \frac{{\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)}}{{\sqrt {{{\left( {\frac{{\partial f}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial f}}{{\partial y}}} \right)}^2} + {{\left( {\frac{{\partial f}}{{\partial z}}} \right)}^2}} }}\\ = \frac{{\left( {x,{y^3},{z^5}} \right)}}{{\sqrt {{x^2} + {y^6} + {z^{10}}} }}\end{array}$$Note that the denominator is already present in your question.
