Hole in the Sphere
A cylindrical hole six inches long has been drilled straight through the center of a solid sphere. What is the volume of the solid material remaining in the sphere?
Labels: funphysics
posted by Rajesh Lal at
9:28 PM
13 Comments:
No change, it's still 36*Pi
Cam
The volume of the holed sphere is = 36π cubic inches
Detailed explanation of remaining volume of sphere with cylindrical hole
The volume of a sphere with a cylindrical hole can be determine as follows:
-quick analysis for a cylindrical hole with 0 diameter
-integrating the areas of rotation over the y axis
-hole with 0 diameter:
-we know that the volume of a sphere =4/3*Pi*R^3
-if a cylindrical hole with 0 diameter is cut the cylindrical hole has a volume of 0
-If the cylindrical hole has a 0 diameter. The height of the hole would extend tip to tip through the centre of the sphere, thus the height would be = to the diameter of the sphere (2R). In this case 2R=6”. Thus R=3”
-V=4/3*Pi*R^3= 4/3*Pi*27= 36*Pi
-not sure how one can infer that this applies to all diameter cyllinders, but it does provides one possible result
-integration technique
We integrate infinitesmally thin segments over y, with a volume of A*dy
-In the case of a circle with a hole in the centre, A=Pi*Xo^2-Pi-Xo^2 where Xo = outer radius and Xi = inner radius
-by symmetery we only need to consider half the sphere and multiply by 2
-V=2*integral from a to b (Pi*Xo^2-Pi*Xi^2) dy
Now to solve for Xo in terms of R and y, we imagine the cross section of the sphere with the x axis horizontal and the y axis vertical. The sphere’s crosss section before the hole is cut is circular.
To define the outer edge of the cross section of the sphere we define the centre of the sphere at x=0, y=0.
From the formula of a circle Xo^2+y^2= R^2, or Xo=sqrt(R^2-y^2).
For Xi we define the point where the vertical line at Xi from the centre intersect the circle (the intersection of the cyllinderical hole and the sphere for the cross section).
The point is Xi from the centre and y is H at this point (H is defined as 3” by the problem). We then draw a line from the centre of the circle to this point, which must be R away. Thus Xi^2+y^2=R^2, or Xi=sqrt(R^2-H^2)
Plugging into formula for V we have
V=2*integral from a to b (Pi*Xo^2-Pi*Xi^2) dy
V=2*integral from a to b (Pi*(R^2-y^2)-Pi*(R^2-H^2))
V=2*integral from a to b (Pi*(H^2-y^2))
For terms of a and b to integrate from/to. We set a to H and b to 0 for the volume remaining .
V=2* (from H to 0) (Pi*H^2*y-1/3*Pi*y^3)
V=2*(Pi*H^2*(H)-1/3*Pi*(H^3)) =2*(2/3*Pi*H^3)
V=4/3*Pi*H^3
Which is an interesting result as it is independent of the sphere’s radius R.
The problem has defined H as 3” thus
V=4/3*Pi*3^3=36*Pi
Answer = 36*Pi
Cam
The solution(copied from "beady eyed"): Read for a bit then make yourself a 2D sketch (I'll use 3D terminology though). Imagine a planet similar to the Earth. It has two identical circular icecaps. Drill a hole from North to South, exactly cutting out the icecaps. The length of this hole is measured from the outer edge of the ex-Northern icecap to the outer edge of the ex-Southern icecap and its length is L. Let h= L/2. Draw the axis from N to S. Draw the equatorial plane. Let the radius of the hole be a, and the radius of the sphere be r. As the distance between the centre of the sphere/circle and an edge of either ex-icecap is r we have a^2+h^2=r^2 (Pythagoras).
Now draw a plane parallel to the equatorial plane, but about half-way to the North pole. Let the distance between this plane and the equatorial plane be y, and the radius of the new plane be x. Note that x^2+y^2=r^2, and the area of the annulus (the washer shape) is pi(x^2-a^2) = pi((r^2-y^2)-(r^2-h^2)) = pi(h^2-y^2)
Now consider a similar situation, but with a whole sphere of radius h. The radius of the non-equatorial plane is x^2+y^2 = h^2 and the area of the non-equatorial disc is pi*x^2, where the plane is at the same distance y from the equatorial plane as before. So area of disc = pi(h^2-y^2).
Comparing the two equations of area immediately shows that the areas of the corresponding annulus and disc are identical. So the volumes of the holed and whole sphere are the same, as one can be constructed from the other by deformation. Note it was important to convert the equations into functions of y for the comparison.
NB in this case L = 6" => h = 3"
What is the diameter of the hole?
If it is zero, like everyone has assumed, then it is not a hole.
Hi Ragknot. The diameter of the hole is whatever it needs to be. No-one has assumed 0.
However. If the original sphere was 6" diameter (which corresponds to the smallest possible original sphere), then the diameter of the hole is 0 - this is the limiting case. For very large spheres, the diameter of the hole is a little shy of the diameter of the original sphere. You will be left with a ring with a "D" cross-section, 6" wide on the flat side.
If you want to be pedantic about it, choose an initial sphere infinitesimally larger than 6", then the hole is infinitesimally wide.
Ragknot,
This is off topic for this puzzle, but could you please post the solution for "Peer to Peer", or provide a fairly detailed explanation of how you see the P2P network working ?
You indicated that the way I described it was incorrect. I'm hoping that with a more detailed description, of your view of it, that I could tackle that problem.
Thanks,
Cam
Here's a nice approach:
Let R be the radius of the sphere. The radius of the cylindrical hole will then be the square root of
R^2-9, an the altitude ot the spherical caps at each end of the cylinder will be R-3.To determine the residue after the cylinder and caps have been removed, we add the volume of the cylinder,6╥(R^2-9), to twice the volume of the spherical cap, and subtract the total from the volume of the sphere
4╥R^3/3. the volume of the cap is obtained by the following formula, in which A stands for its altitude and r for its radius:
╥A(3r^2+A^2)/6 . When this computation is made, all terms cancel out except 36╥.
Sorry if some of the symbols don't look right ... I rarely type formulas.
This post has been removed by the author.
Hi Zaux. Your formula for a cylindrical cap is hardly a standard or trivial expression (I don't recall ever having seen it before). I expect that it's derivation is of similar complexity to directly solving the problem.
Chris ... here's a nice explanation for the formula of a spherical cap:
http://mathworld.wolfram.com/SphericalCap.html
Hi Zaux. I'd already seen it.
... but I've taken another look, I'll give you that the final equations are quite sweet, but I doubt that I'd ever remember either.
Note that in that article, they assume other equations which, no doubt, are also tricky to derive.
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