# Intersection of Two Cylinders

1. Dec 15, 2017

### OpheliaM

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
x^2+y^=1
x^2+z^2=1

3. The attempt at a solution

2. Dec 15, 2017

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.

Last edited: Dec 15, 2017
3. Dec 15, 2017

### SammyS

Staff Emeritus
Hello @OpheliaM . Welcome to Physics Forums !!!

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

4. Dec 15, 2017

### OpheliaM

@SammyS Could you explain further?

5. Dec 15, 2017

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

6. Dec 15, 2017

### SammyS

Staff Emeritus
Graph the following function, or tabulate some values.

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

7. Dec 15, 2017

### 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?

8. Dec 15, 2017

### SammyS

Staff Emeritus

9. Dec 15, 2017

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.

10. Dec 15, 2017

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 $t$.

11. Dec 15, 2017

### OpheliaM

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

12. Dec 15, 2017

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

13. Dec 15, 2017

### OpheliaM

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?

14. Dec 15, 2017

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...

15. Dec 15, 2017

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.

16. Dec 15, 2017

### OpheliaM

Thanks!

17. Dec 15, 2017

### 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.