Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

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

share|improve this question
    
Do you really mean to have $dx$ at the end of the integral? –  JimmyK4542 Jul 19 at 19:01
    
No, I just correct it, thanks –  Benjamin Jul 19 at 19:04
    
Cool looking surface! –  TylerHG Jul 19 at 19:13
1  
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 at 19:31
2  
Where did this problem come from @Xtk ? –  TylerHG Jul 19 at 20:04

3 Answers 3

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

$$\frac{1}{vol(A)}\int_{\partial A}\frac{1}{\sqrt{x^2+y^6+z^{10}}}ds=\frac{1}{vol(A)}\int_{\partial A} F\cdot \hat{n} ds=\frac{1}{vol(A)}\int_{ A}\nabla\cdot F dxdydz \quad= \frac{1}{vol(A)}\int_{ A} \frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}$$

share|improve this answer
1  
That's the exact kind of answer I was hoping someone would provide rather than some complicated special-functions route. (+1) –  Semiclassical Jul 21 at 11:36
1  
Very nice answer (+1) I feel a bit stupid :) –  O.L. Jul 21 at 12:01
    
thanks for the answer. thats what i wanted! Could you explain me how you get that normal vector? –  Benjamin Jul 21 at 12:33
1  
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 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 $$\boxed{\displaystyle I=\frac{I_S}{\operatorname{vol}A}=\frac{11}{12}}$$

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

share|improve this answer
1  
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 at 14:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.