| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 528.1 | TOO easy? | CACHE::MARSHALL | beware the fractal dragon | Fri Jul 11 1986 10:34 | 7 |
| Is this problem so simple that none of you wizards deem it worthy
of your attention? Probably.
Well let's make it harder: Can you do it without calculus?
sm
|
| 528.2 | A geometric hint | MODEL::YARBROUGH | | Mon Jul 14 1986 09:46 | 7 |
| Here's a hint: Consider the cross-section made in the intersection
by a plane parallel to the axes of the cylinders. It is a square
whose sides vary from 0 to the diameter of the cylinders, the edges
being also parallel to the axes. The area of the square that is
a distance h from the axis is determined by the Pythagorean theorem
as a difference of two other squares. What other simpler solid figure
has the same cross-sectional area? Its volume is easy to compute.
|
| 528.3 | it is ba-loooon | CACHE::MARSHALL | beware the fractal dragon | Mon Jul 14 1986 14:36 | 16 |
| re .2:
"what other simpler solid figure has the same cross-sectional area?"
well...are you thinking of an octahedron? it has the same cross-section
but the function of side-length to h is not the same.
this object (the shape of the volume removed) is like an inflated
octahedron.
BTW, I have calculated the volume, using calculus. I won't post
it here, it was pretty trivial.
BUT can it be done without calculus?
sm(ile)
|
| 528.4 | Well, it's related to an octahedron | MODEL::YARBROUGH | | Mon Jul 14 1986 15:29 | 9 |
| The simpler figure, it turns out, is a cube with pyramids hollowed
out from opposite sides and meeting with their common apex at the
center. This figure and the intersection of two circular cylinders
have exactly the same cross-sectional area at the same distance
from the center, therefore the same volume. I think it works out
to 2/3 d^3, or 16/3 r^3.
If you cut the cube in half and flip the halves over, you get a
cube with an enclosed octahedral cavity.
|
| 528.5 | | CLT::GILBERT | $ no /nono vaxnotes | Mon Jul 14 1986 20:37 | 24 |
| Let the radius of the rod be R and the radius of the hole be r, with R >= r.
| ROD |
| |
----+-------+----
| |hole
----+-------+----
| |
| |
Now as we take cross-sections, each is a rectangle (above is the
cross-section at its widest). If the distance from the center is x,
the sides of the rectangle are 2 sqrt(r�-x) and 2 sqrt(R�-x�). The
volume of material removed is given by the integral:
_ +r
4 / sqrt(r�-x) sqrt(R�-x�) dx
_/ -r
However, this seems too dificult, so we consider only the case R=r.
_ +r ] +r
4 / (r�-x�) dx = 4(r�x-x�/3) ] = 16 r� / 3
_/ -r ] -r
|