# Intersecting cylinders!

1. Apr 19, 2004

### Xishan

Can anybody help me finding the surface area of two intersecting cylinders with different radii and NOT perpendicular to each other (the axes of the two are in the same plane)?

Xishan

2. Apr 19, 2004

### Chen

Welcome to Physics Forums.

Do you have the angle between the axes of the cylinders? I assume it is safe to assume that the cylinders are both high enough so the intersection is "complete".

3. Apr 19, 2004

### Xishan

Yes, Chen! the cylinders are COMPLETELY intersecting each other, and the angle can be assumed to be any value other than 90 degrees. I've solved the problem with the perpendicular case and verified the results with my CAD software.

4. Apr 20, 2004

### Theelectricchild

5. Apr 20, 2004

### HallsofIvy

Staff Emeritus
No, Xishan was asking about an angle other than 90 degrees. The problem you give is the 90 degree case.

6. Apr 20, 2004

### Xishan

The problem still remains unsolved! the intersection area and the surface area ...

7. Apr 20, 2004

### arildno

Just a suggestion:
We let the cylinder with largest radius be identified by uppercase letters,
while the cylinder with the least radius be identified with lowercase letters.
Going into the plane defined by the cylinder axes, we have the relation between the vertical variables: Z=z (X and x go along the respective axes).
In general, the region is the intersection between the regions:
Y^(2)+Z^(2)<=R^(2), y^(2)+z^(2)<=r^(2).

We let the axis of the cylinder with largest radius be rotated an angle w to the other axis, the intersection of the axes is the origin.
We have therefore:
X=x*cos(w)+y*sin(w), Y=-x*sin(w)+y*cos(w),
or we have the inequalities in x,y,z:
y^(2)+z^(2)<=r^(2)
z^(2)+(-x*sin(w)+y*cos(w))^(2)<=R^(2)

The last inequality gives the limits on x as functions of z and y

Last edited: Apr 20, 2004
8. Apr 21, 2004

### Xishan

Thanks!

That is what I was doing but with some mistake in the limits. I initially thought the maxima (in x) of the intersecting curve in the xy plane will be at y=0; and that was wrong. Due to the inclined nature of the surface this point moves a little on the other side. Having corrected that error in the limit I've managed to solive the problem and verified it with my CAD software.

Thanks for paying this much attention to my question. I never thought this many people would view this thread.

Last edited: Apr 21, 2004
9. Apr 21, 2004

### arildno

Xishan, just a question:
Did you solve it numerically, or did you find some exact solution that has escaped me?

10. Apr 22, 2004

### Xishan

Well, I did try my best to solve it numerically and eventually got the solution in terms of elliptical integrals but since I had to program that solution in my VB routine I employed Simpson's rule for its numerical solution. Following are the two integrals that i had at the end:

intg(sqrt((r^2-x^2)/(R^2-x^2))) 'for the intersection area

and

intg(sqrt((R^2-x^2)/(r^2-x^2))) 'for surface area

where R > r and limits on x are: x=0 to x=r in both cases

the second integral poses an added problem of being improper since it becomes infinite at the upper limit. I had to transform it in to a managable form by substituting 'x = r cos(a)' (from the geometry of the problem) and ended up with the following integral

intg(sqrt((R/r)^2-cos(a)^2))

with limits of a: 0 to pi/2

I will appreciate any one giving me an analytical solution or at least tell me a better (yet easy to program) and accurate numerical technique than Simpson's rule.