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 September 18th, 2013, 05:28 AM #1 Senior Member   Joined: Sep 2011 Posts: 140 Thanks: 0 Length of intersection between two surfaces How do you find the total length of intersection between two planes. For example, in my textbook there is a question escribe the intersection of the sphere x^2+y^2+z^2=1 and the elliptic cylinder x^2+2z^2=1. Find the total length of this intersection curve. Can anyone please tell me how to do this? thank you
 September 18th, 2013, 09:43 AM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Length of intersection between two surfaces $x^2= 1 - 2z^2$. Substituting this into the sphere's equation, we observe that $y^2= z^2$ so that $y= \pm z$. We can parametrize the elliptic cylinder more easily: $x= cos \theta \ ; \ y = t \ ; \ z = \dfrac{sin \theta}{\sqrt{2}}$ where $\theta \in [0 , 2\pi]$. Using our observation above, our intersection curve must be split into $C_1 , C_2$ as follows; $C_1 : [0, 2\pi] \rightarrow \mathbb{R}^3 \ ; \ \theta \mapsto \left(cos \theta , \dfrac{sin \theta}{\sqrt{2}}, \dfrac{sin \theta}{\sqrt{2}} \right)$ $C_2 : [0, 2\pi] \rightarrow \mathbb{R}^3 \ ; \ \theta \mapsto \left(cos \theta , \dfrac{-sin \theta}{\sqrt{2}}, \dfrac{sin \theta}{\sqrt{2}} \right)$ Using the formula $L= \int _{[0, 2\pi]}|c #39;(\theta)| d\theta$ for $c= C_1$ and also for $c= C_2$, we get both arclengths and then sum up. This should be straightforward for you.
September 18th, 2013, 05:22 PM   #3
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Re: Length of intersection between two surfaces

Quote:
 Originally Posted by gaussrelatz How do you find the total length of intersection between two planes.
The intersection of two planes is an infinitely long line!

Quote:
 For example, in my textbook there is a question escribe the intersection of the sphere x^2+y^2+z^2=1 and the elliptic cylinder x^2+2z^2=1. Find the total length of this intersection curve. Can anyone please tell me how to do this? thank you
First find the intersection! (These are NOT planes, of course.) Since $x^2+ 2z^2= 1$, $x^2+ z^2= 1- z^2$ so the equation of the sphere becomes $y^2+ 1- z^2= 1$ or $y^2- z^2= (y- z)(y+ z)= 0$. Either y= z or y= -z. Of course, $x^2+ 2z^2= 1$ is an ellipse and we can write it as parametric equations $x= cos(\theta)$, $z= (1/2)sin(\theta)$ and then $y= (1/2)sin(\theta)$ or [latex]y= (-1/2)sin(\theta).

Now the length is $\int_0^{2\pi}\sqrt{x^2+ y^2+ z^2}d\theta= 2\int_0^{2\pi}\sqrt{cos^2(\theta)+ (1/2)sin^2(\theta)}d\theta$ (the "2" is to get the lengths of both "y= z" and "y= -z" parts).

 September 18th, 2013, 08:05 PM #4 Senior Member   Joined: Sep 2011 Posts: 140 Thanks: 0 Re: Length of intersection between two surfaces I am sorry i just started to learn multivariable calculus and this part is seemingly more difficult. I have no idea on how the intercepts are found, i cant get a clear picture yet. Thank you for the help, but im still confused are there some external references on this part "finding the total length of intersection of two surfaces"? Also i have a question on parametrizing the curves. How do you know what to use as a parameter? I mean i know you have to use the cylindrical and spherical coordinates based on the form they resemble, however for example questions like: parametrize: z=x^2+y^2 and z-2x-4y+4=0, how do you know what to put in for z or x or y as the parameter? Thank you
 September 19th, 2013, 09:32 AM #5 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Length of intersection between two surfaces HallsofIvy, your final integral will not give the correct answer, because you made an error in the parametric equation of the intersection curve. Verify that it should be $\left( cos\theta , \pm \dfrac{sin \theta}{\sqrt 2},\dfrac{sin \theta}{\sqrt 2} \right)$ for $\theta \in [0 , 2\pi]$ to get one full cover of the intersection curve. Gaussrelatz, what you need is just practice. You must know how to parametrize the most common curves which include the circle and ellipse; and also the most common surfaces which include the plane, cylinder and sphere. In general, a curve will be parametrized by a single parameter while a surface will be parametrized by two parameters. Since multivariable calculus and coordinate geometry are not so easy, keep posting problems on this forum when you get stuck. That's what it's here for anyway.

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