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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...

$$I := \frac{1}{\lambda_3(A)} \int\limits_{\partial A} \frac{1}{\sqrt{x^2 + y^6 + z^{10}}} \, dS_{\partial A}$$

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Do you really mean to have $dx$ at the end of the integral? – JimmyK4542 Jul 19 '14 at 19:01
No, I just correct it, thanks – Marc Jul 19 '14 at 19:04
Cool looking surface! – TylerHG Jul 19 '14 at 19:13
Another route might be to look for a substitution like $(x,y,z)\to(x^r,y^p,z^q)$. Though I don't know whether it's smarter to make the integrand nice or the bounds nice. – Semiclassical Jul 19 '14 at 19:31
Where did this problem come from @Xtk ? – TylerHG Jul 19 '14 at 20:04
up vote 5 down vote accepted

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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That's the exact kind of answer I was hoping someone would provide rather than some complicated special-functions route. (+1) – Semiclassical Jul 21 '14 at 11:36
Very nice answer (+1) I feel a bit stupid :) – Start wearing purple Jul 21 '14 at 12:01
thanks for the answer. thats what i wanted! Could you explain me how you get that normal vector? – Marc Jul 21 '14 at 12:33
Haha, you are welcome. I saw @Seyed Mohsen Ayyoubzadeh, the third answerer, have already shown the process of how to find the normal vector, so I just skipped it. We know if $F(x,y,z)=0$ is boundary curve, then $(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z})$ is in the same direction as the normal vector, so just normalize it by dividing by its length. – Shine Jul 21 '14 at 14:16

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 $$\boxed{\displaystyle I=\frac{I_S}{\operatorname{vol}A}=\frac{11}{12}}$$

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I'm not sure about the $4\sqrt{2}$ constant in front of the integral. Should it be $1/\sqrt{2}$? – TylerHG Jul 19 '14 at 21:46
@TylerHG Additional symmetry factor $8$ comes from the condition $x,y,z\geq 0$. – Start wearing purple Jul 19 '14 at 22:17
Oh I see! I didn't account for $x$ @O.L. – TylerHG Jul 19 '14 at 22:38
Could just leave it as a beta function @O.L. – TylerHG Jul 20 '14 at 2:18

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.

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I strongly suspect that this is the correct way to go: the beta function method may be correct, but it's probably far more circuitous than finding a direct Divergence Theorem approach. – Semiclassical Jul 20 '14 at 14:18

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