<s>[INST] 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...
$$I := \frac{1}{\lambda_3(A)} \int\limits_{\partial A} \frac{1}{\sqrt{x^2 + y^6 + z^{10}}} \, dS_{\partial A}$$
 [/INST] 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
$$\frac{1}{vol(A)}\int_{\partial A}\frac{1}{\sqrt{x^2+y^6+z^{10}}}ds=\frac{1}{vol(A)}\int_{\partial A} F\cdot \hat{n} ds=\frac{1}{vol(A)}\int_{ A}\nabla\cdot F dxdydz  \quad= \frac{1}{vol(A)}\int_{ A} \frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}$$
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