think.ms.tink
New member
- Joined
- Apr 3, 2009
- Messages
- 18
A cylindrical drill with radius 2 is used to bore a hole throught the center of a sphere of radius 5. Find the volume of the ring shaped solid that remains. :?:
A cylindrical drill with radius 2 is used to bore a hole
- Thread starter think.ms.tink
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,582
As a "type", this is apparently refered to as "the napkin ring problem". You can review archived discussions online:
. . . . .Wikipedia: Napkin ring problem
. . . . .Everything2: old chestnut: hole in sphere
. . . . .Ask Dr. Math
Have fun!
. . . . .Wikipedia: Napkin ring problem
. . . . .Everything2: old chestnut: hole in sphere
. . . . .Ask Dr. Math
Have fun!
Hello, think.ms.tink!
This can be solved with Cylindrical Shells.
\(\displaystyle \text{Formula: }\;V \;=\;2\pi\int^b_a\text{(radius)(height)}\,dx\)
\(\displaystyle \text{The radius is }x.\)
\(\displaystyle \text{The height is }2y,\:\text{ where }y \:=\:\sqrt{25-x^2}\)
\(\displaystyle \text{The volume is: }V \;=\; 2\pi\int^5_2 x\cdot 2\sqrt{25-x^2}\,dx \;=\;4\pi\int^5_2 x\left(25-x^2\right)^{\frac{1}{2}}\,dx\)
Go for it!
This can be solved with Cylindrical Shells.
Code:
|
* * *
* - - + - - *
* | | |:*
* | | y|::*
| | |:::
* | | x |:::*
- - * - + - - + - - + - * - -
* | | 2|:::*5
| | |:::
* | | |::*
* | | |:*
* - - + - - *
* * *
|\(\displaystyle \text{Formula: }\;V \;=\;2\pi\int^b_a\text{(radius)(height)}\,dx\)
\(\displaystyle \text{The radius is }x.\)
\(\displaystyle \text{The height is }2y,\:\text{ where }y \:=\:\sqrt{25-x^2}\)
\(\displaystyle \text{The volume is: }V \;=\; 2\pi\int^5_2 x\cdot 2\sqrt{25-x^2}\,dx \;=\;4\pi\int^5_2 x\left(25-x^2\right)^{\frac{1}{2}}\,dx\)
Go for it!
A cylindrical drill with radius 2 is used to bore a hole throught the center of a sphere of radius 5. Find the volume of the ring shaped solid that remains.
A golden oldie that usually takes the following form..
A hole is drilled clear through the center of a solid sphere. The length of the hole, as measured between the points where the hole pierces the surface of the sphere, is 6 inches. What is the volume of material remaining? After deriving your answer (and nothing has been left out), try giving a physical explanation and/or interpretation of the result.
Geometry or calculus will lead you to the correct answer. Geometrically, let the radius of the sphere be R. The height of the remaining volume is given as 6. The radius of the cylindrical hole created is then sqrt(R^2 - 9). The height of the spherical end caps at the ends of the cylindrical hole is (R - 3). The remaining volume is then the total volume of the sphere minus the volume of the cylindrical hole minus the volume of the two end caps. The volume of the cylindrical hole is Vh = 6Pi(R^2 - 9). The volume of the end cap is Vc = Pih(3R^2 + h^2)/6. Substituting, expanding, and simplifying, the remaining volume becomes 36Pi, a quantity totally independent of the radius of the sphere. Working it out in general terms with a hole length L, the remaining volume becomes Vr = PiL^3/6.
John W. Cambell Jr., editor of Astounding Science Fiction, probably offered the best explanation of the hole in the sphere problem for the untrained mathematical eye. The problem would never have been created unless it had a unique solution. If it has a unique solution, the volume must be a constant which would hold even when the hole is reduced to zero radius. Therefore the residue must equal the volume of a sphere with a diameter of six inches, namely 36Pi. In other words, the residue is constant regardless of the hole's diameter or the size of the sphere.
While your problem statement did not specify the drilled height, it is easily determined from the given information.
A golden oldie that usually takes the following form..
A hole is drilled clear through the center of a solid sphere. The length of the hole, as measured between the points where the hole pierces the surface of the sphere, is 6 inches. What is the volume of material remaining? After deriving your answer (and nothing has been left out), try giving a physical explanation and/or interpretation of the result.
Geometry or calculus will lead you to the correct answer. Geometrically, let the radius of the sphere be R. The height of the remaining volume is given as 6. The radius of the cylindrical hole created is then sqrt(R^2 - 9). The height of the spherical end caps at the ends of the cylindrical hole is (R - 3). The remaining volume is then the total volume of the sphere minus the volume of the cylindrical hole minus the volume of the two end caps. The volume of the cylindrical hole is Vh = 6Pi(R^2 - 9). The volume of the end cap is Vc = Pih(3R^2 + h^2)/6. Substituting, expanding, and simplifying, the remaining volume becomes 36Pi, a quantity totally independent of the radius of the sphere. Working it out in general terms with a hole length L, the remaining volume becomes Vr = PiL^3/6.
John W. Cambell Jr., editor of Astounding Science Fiction, probably offered the best explanation of the hole in the sphere problem for the untrained mathematical eye. The problem would never have been created unless it had a unique solution. If it has a unique solution, the volume must be a constant which would hold even when the hole is reduced to zero radius. Therefore the residue must equal the volume of a sphere with a diameter of six inches, namely 36Pi. In other words, the residue is constant regardless of the hole's diameter or the size of the sphere.
While your problem statement did not specify the drilled height, it is easily determined from the given information.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
Here is the disk method (actually washer), soroban showed the shell method.
\(\displaystyle V = \pi\int_{a}^{b}({[R(x)]}^2-{[r(x)]^2})dx\)
\(\displaystyle x^{2}+y^{2} = 25, \ y=2 \ -> \ x = + or- \ \sqrt(21), \ limit \ of \ integration\)
\(\displaystyle Hence \ V = \pi\int_{\sqrt-21}^{\sqrt21}[({\sqrt{25-x^{2}}})^2-{2}^2}]dx \ =2\pi\int_{0}^{\sqrt21}(21-x^{2})dx\)
\(\displaystyle Go \ for \ it.\)
\(\displaystyle V = \pi\int_{a}^{b}({[R(x)]}^2-{[r(x)]^2})dx\)
\(\displaystyle x^{2}+y^{2} = 25, \ y=2 \ -> \ x = + or- \ \sqrt(21), \ limit \ of \ integration\)
\(\displaystyle Hence \ V = \pi\int_{\sqrt-21}^{\sqrt21}[({\sqrt{25-x^{2}}})^2-{2}^2}]dx \ =2\pi\int_{0}^{\sqrt21}(21-x^{2})dx\)
\(\displaystyle Go \ for \ it.\)
think.ms.tink
New member
- Joined
- Apr 3, 2009
- Messages
- 18
Thank you.