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Hi, i have a little problem for you. I have a solid tube and have drilled a circular hole straight through the side of it. I want to be able to find the volume of the hole. I have attached a picture to help describe the problem. the top image is looking down on to the hole and the bottom image is a cross-sectional view. Thank you in advance for any help. :)
http://i62.photobucket.com/albums/h1...f88/Puzzle.jpg
Welcome to the forums. Seems to me the hole is a cylinder with spherical caps on both ends. This should help with the volume of the spherical caps:
http://mathforum.org/dr.math/faq/for...aq.sphere.html
are you sure they are spherical caps?
Duh, of course they aren't! The intersection of 2 perpendicular cylinders is a complex contour. Years ago, these types of problems were handled with Bezier curves/surfaces or Non-Uniform Rational B-Splines (NURBS). I don't know if there is a simple solution.
.
There is an interesting solution to this; it applies if the cylinders are the same size but may with a bit of thought be adaptable.
Imagine a sphere inscribed in both the cylinders, i.e. in the intersection volume and then take a horizontal section through the common volume, i.e. parallel to the axes of the cylinders. It will look like a circle circumscribed by a square, where the circle is the cross-section of the sphere and the square is the cross-section of the intersection solid.
If the radius of the cylinders is taken as 1 unit, then the sum of all the circles must be 4/3 Pi (x 1^3), i.e. the volume of the sphere.
The sum of all the squares is the volume of the common solid - i.e. the area of the hole in this case.
But at each stage, the area of the square (4r^2 = 4) and the area of the circle (Pi r^2 = Pi) are in the same ratio, namely 4 : Pi, so the same applies to the volume of the solids.
Hence if the volume of the sphere is 4/3 Pi, then the volume of the common solid is 16/3.
Interestingly, this doesn't involve Pi!
If one cylinder is smaller than the other, then the sphere will naturally be smaller and you'll have rectangles rather than squares.
I haven't yet worked out if this means that the ratios stay the same throughout (I think not), or if you can cobble something using a square and a rectangle.
But I thought I'd post it anyway because it is a neat solution to a specific example of this problem, and I like it.
zaza