Finding the Line of Intersection for Two Cylinders

[OpheliaM]

Homework Statement

I am looking for the line of intersection of two cylinders (x^2+y^=1 and x^2+z^2=1) to find the curve for a line integral. However, I am getting z=y, which is a patently false answer. Mathematically, it seems to work out though.

Homework Equations

x^2+y^=1
x^2+z^2=1

The Attempt at a Solution

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[Charles Link]

I have not previously worked such a problem, but suggestion is let $x=t$. Then $y=\sqrt{1-t^2}$ and $z=\sqrt{1-t^2}$, (omitting $\pm 's$) will parameterize the curve.

 

[SammyS]

Homework Statement

I am looking for the line of intersection of two cylinders (x^2+y^=1 and x^2+z^2=1) to find the curve for a line integral. However, I am getting z=y, which is a patently false answer. Mathematically, it seems to work out though.

Homework Equations

x^2+y^=1
x^2+z^2=1

The Attempt at a Solution

[ ATTACH=full]216873[/ATTACH] [/B]

Hello @OpheliaM . Welcome to Physics Forums !

Actually you get y² = z², which is somewhat different than y = z .

 

[OpheliaM]

@SammyS Could you explain further?

 

[Charles Link]

Hello @OpheliaM . Welcome to Physics Forums !

Actually you get y² = z², which is somewhat different than y = z .

There is a constraint on what $x$ is doing that you are disregarding in simply saying that $y^2=z^2$.

 

[SammyS]

@SammyS Could you explain further?

Graph the following function, or tabulate some values.

$f(x)=\sqrt{x^2\,}$

 

[OpheliaM]

Specifically, I am looking for the “boundary of the region common to the cylinders x^2+y^2<1 and x^2+z^2<1 in the first octant”. Can you explain the restriction further?

 

[SammyS]

Specifically, I am looking for the “boundary of the region common to the cylinders x^2+y^2<1 and x^2+z^2<1 in the first octant”. Can you explain the restriction further?

Please use the "Reply" link, so we know who you and what you addressing.

 

[Charles Link]

The boundary region is a curved line in space. Something like $y^2=z^2$ gives you basically a couple of planes in three dimensions.

 

[Charles Link]

Generally, curved lines in 3 dimensions are described by $x=f(t)$, $y=g(t)$, and $z=h(t)$. They are parameterized with the dummy variable $t$.

 

[OpheliaM]

The boundary region is a curved line in space. Something like $y^2=z^2$ gives you basically a couple of planes in three dimensions.

That makes sense. What is the specific restrixrion I ignored with y=z?

 

[Charles Link]

That makes sense. What is the specific restrixrion I ignored with y=z?

You are ignoring the original equations that contain $x$. See my post 2 again.

 

[OpheliaM]

The original equations that contain $x$. See my post 2 again.

Thank you for your help. However, I am having trouble understanding specificiallg why am I unable to equate those two equationsin your post. Why can i not disregard x? Does it have to do with the +- signs? If so, specifically where were they disregarded?

 

[Charles Link]

Thank you for your help. However, I am having trouble understanding specificiallg why am I unable to equate those two equationsin your post. Why can i not disregard x? Does it have to do with the +- signs? If so, specifically where were they disregarded?

It is correct that $y= \pm z$ along the curve. In getting this result though, you can't throw away both of the original equations. This is kind of a peculiar system of equations, and I don't have a really good answer for why what you did doesn't work, but it just doesn't...

 

[Charles Link]

To look at something that is somewhat related to the type of result you need for this problem, examine what a straight line looks like in 3 dimensions: You will typically see something like $\frac{x-5}{3}=\frac{y-2}{5}=\frac{z-6}{4}=t$ and this line passes through $(5,2,6 )$ and it is in the direction of the vector $\vec{v}=3 \hat{i}+5 \hat{j}+ 4 \hat{k}$. You can even normalize this vector $\hat{v}$ to a unit vector in that direction. $\\$ Basically the equation of this line is the parameterization $x=3t+5$, $y=5t+2$, and $z=4t+6$. It might appear clumsy, but we can't do a simple line in 3 dimensions like we do a line in two dimensions. e.g. the equation $3x+5y +2z+4=0$ represents a plane, rather than a line.

 

[OpheliaM]

To look at something that is somewhat related to the type of result you need for this problem, examine what a straight line looks like in 3 dimensions: You will typically see something like $\frac{x-5}{3}=\frac{y-2}{5}=\frac{z-6}{4}=t$ and this line passes through $(5,2,6 )$ and it is in the direction of the vector $\vec{v}=3 \hat{i}+5 \hat{j}+ 4 \hat{k}$. You can even normalize this vector $\hat{v}$ to a unit vector in that direction. $\\$ Basically the equation of this line is the parameterization $x=3t+5$, $y=5t+2$, and $z=4t+6$. It might appear clumsy, but we can't do a simple line in 3 dimensions like we do a line in two dimensions. e.g. the equation $3x+5y +2z+4=0$ represents a plane, rather than a line.

Thanks!

 

[LCKurtz]

@OpheliaM: Your original result that $z=y$ (in the first octant) is correct and tells you that the required curve lies in that plane. Here are a couple of hints:
1. Do you know how to parameterize the $xy$ cylinder in terms of $\theta$ and $z$, where $\theta$ is the polar angle? If so, then do that and
2. Use the fact that $z = y$.
That should give you a parameterization in the form $\vec R(\theta) = \langle x(\theta),y(\theta),z(\theta) \rangle$, which is the parametric form of a curve.