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I will describe the problem then show what I tried to solve it.

I need to find the area of the cone defined as follows:

$$z^2=a^2(x^2+y^2)$$ $$0\leq z\leq bx+c$$

where $a,b,c>0$ and $b<a$.

For this I considered the parametrization $x=r\rm{cos}\theta$, $y=r\rm{sin}\theta$, $z=ar$, with $0\leq\theta\leq2\pi$. To find the interval which $r$ varies, I used the fact that $0\leq z\leq bx+c$, then $0\leq ar\Rightarrow 0\leq r$, and $ar\leq br\rm{cos}\theta+c\Rightarrow r\leq c/(a-b\rm{cos}\theta)$, therefore $0\leq r\leq c/(a-b\rm{cos}\theta)$.

The area of this surface is $$\int_0^{2\pi}\int_0^{\frac{c}{a-b\rm{cos}\theta}}\bigg\lvert(\rm{cos}\theta, \rm{sin}\theta, a)\times(-r\rm{sin}\theta,r\rm{cos}\theta, 0)\bigg\lvert dr \ d\theta=$$

$$=\int_0^{2\pi}\int_0^{\frac{c}{a-b\rm{cos}\theta}}r\sqrt{a^2+1} dr \ d\theta=\sqrt{a^2+1}\int_0^{2\pi}\bigg[\frac{r^2}{2}\bigg]_0^{\frac{c}{a-b\rm{cos}\theta}} \ d\theta=$$

$$=\sqrt{a^2+1}\int_0^{2\pi}\frac{c^2}{2(a-b\rm{cos}\theta)^2} \ d\theta=\frac{c^2\sqrt{a^2+1}}{2}\int_0^{2\pi}\frac{1}{(a-b\rm{cos}\theta)^2} \ d\theta .$$

Well, this last integral is quite difficult, in fact i can't integrate this, and wolframaplha showed me a terrible solution. Now I think I've made some mistake but don't know where. Any help is very welcome. Thanks.

PS: I tried to find something like a^2(x^2+y^2) or a^2\cdot(x^2+y^2) in the search but that was useless. Later I did some tests and verified that this search is not that good for TeX search, is there some better way to better searching when using TeX language? Again, thanks.

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Sorry if I changed the meaning of your question when I edited. I don't think it's right with $\theta$ ranging over $[0,0\textbf{]}$ though. –  Git Gud Apr 6 '13 at 22:23
    
no problem, I was doing some editing myself...not sure if I covered your edit. –  Integral Apr 6 '13 at 22:25
    
Check the edit history to see my edit. –  Git Gud Apr 6 '13 at 22:27
    
I already did.. –  Integral Apr 6 '13 at 22:46
    
"is there some better way to better searching when using TeX language?" Try Google writing mathematics stack exchange a^2(x^2+y^2) –  Américo Tavares Apr 8 '13 at 0:55

2 Answers 2

up vote 4 down vote accepted

I have not looked at the complete problem in detail yet, but you did mention difficulty with the integral

$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2}$$

Evaluation of this integral is not too bad using the Residue theorem. Let $z=e^{i \theta}$, $d\theta = dz/(i z)$, $\cos{\theta} = (z + z^{-1})/2$, and the integral becomes

$$\oint_{|z|=1} \frac{dz}{i z} \frac{1}{(a-b(z+z^{-1})/2)^2} = -i \frac{4}{b^2} \oint_{|z|=1} dz \: \frac{z}{(z^2 - 2 (a/b) z + 1)^2}$$

By the residue theorem, this integral is $i 2 \pi$ times the sum of the residues of the poles inside the unit circle. The poles of the integrand are at

$$z_{\pm} = \frac{a}{b} \pm \sqrt{\left ( \frac{a}{b} \right )^2 - 1}$$

Note that only $z_-$ is inside the unit circle, so we need only evaluate the residue at that pole to get the value of the integral. Note also that this is a double pole, so the residue calculation looks like

$$\begin{align}\text{Res}_{z=z_-} \frac{z}{(z^2 - 2 (a/b) z + 1)^2} &= \lim_{z \rightarrow z_-} \frac{d}{dz} \left [(z-z_-)^2 \frac{z}{(z^2 - 2 (a/b) z + 1)^2} \right ] \\ &= \left[\frac{d}{dz} \frac{z}{(z-z_+)^2}\right]_{z=z_-} \\ &= \frac{1}{(z_- -z_+)^2} - \frac{2 z_-}{(z_--z_+)^3}\\ &= \frac{z_++z_-}{(z_+-z_-)^3}\\ &=\frac{2 (a/b)}{(2 \sqrt{(a/b)^2-1})^3}\\ &= \frac{1}{4} \frac{(a/b)}{[(a/b)^2-1]^{3/2}}\end{align}$$

We then multiply this residue by $(i 2 \pi)(-i 4/b^2)$ and the result is

$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2} = 2 \pi \, a \, (a^2-b^2)^{-3/2}$$

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I did all the calculations following what you did. It was just fine until you left the algebra to the reader XD I found the residue $b^2/(2(a^2-b^2)) - ab^2$ which is real. When I multiply by $i2\pi$ I get the complex number $i\pi b^2/(a^2-b^2) - i2\pi ab^2$. –  Integral Apr 8 '13 at 0:29
    
See my edits above; I hope they explain how the result comes about more clearly. –  Ron Gordon Apr 8 '13 at 0:39
    
thank you R. Gordon, sorry for my lack of skills. –  Integral Apr 8 '13 at 0:41
2  
@Integral: no apology necessary - I try to gauge the OP's skills and leave the detail accordingly. You need to learn how to do these things somehow. –  Ron Gordon Apr 8 '13 at 0:42

One way to do it is to write $z = f(x,y) = a\sqrt{x^2+y^2}$, then integrate $\sqrt{1 +\|\nabla f(x,y)\|^2}$ over a region in the $xy$-plane. The integrand will be constant because $\|\nabla f(x,y)\|$ will be constant. It looks like you will have to integrate over an ellipse in the $xy$-plane. You will need to multiply the integrand by the area of the ellipse. Finding the area of that ellipse is the only step that may not be easy if you do it this way.

To find the area of the ellipse, substitute $z=bx+c$ into the cone equation, simplify, and complete the square in $x$. Your ellipse is a disc that has been stretched in the $x$ and $y$ directions. Try to figure out what the "stretching factor" is in each direction.

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This way was my first trial, but there is two things I dont know. Why do I have to double? I'm asking this because by hypothesis $z\geq 0$, then there is no way to consider $z=-a\sqrt{x^2+y^2}$. And second, what are the limits of the integrals? I know for $z$, but for $x$ I have no ideia, in fact this is what led me to consider another parameterization. –  Integral Apr 7 '13 at 15:00
    
@Integral : Sorry, I made a mistake. I didn't notice $z \geq 0$, so you don't have to double anything. To integrate a constant over a region in the $xy$-plane, you just multiply the constant by the area of the region. You don't need any limits of integration. Here your region will be a disc, and I assume you know how to find the area of a disc. –  Stefan Smith Apr 7 '13 at 15:33
    
@Integral : Sorry, I noticed another mistake I made. The upper limit on $z$ is not a constant, so you will need to multiply the integrand by the area of an ellipse, not a disc. This area might be easiest to compute using polar coordinates. Although it was nice of you to include it, I didn't read your solution attempt too closely, because it leads to an integral that is quite difficult. –  Stefan Smith Apr 7 '13 at 15:53
    
@Integral : MSE is giving me some problems and will not let me improve my previous comment. To find the area of the ellipse, you can substitute $z = bx + c$ into $z^2 = a^2(x^2+y^2)$, simplify, complete the square in $x$, and use the fact that the ellipse is a disc that has been stretched horizontally and vertically. –  Stefan Smith Apr 7 '13 at 16:01
    
I will think about it. –  Integral Apr 7 '13 at 16:02

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