Volume of sphere, after a hole has been drilled.

Bandicoot

New member
Joined
May 15, 2018
Messages
7
I have gotten this question I'm not sure how to go about.

"A cyllinder shaped hole, 10 inches long, is drilled through a sphere.
Explain the remainder of the volume of the sphere."

Cyllinder width is not defined
Sphere is not defined
Hole placement is not defined


I'm completely lost on this one. The question dosent take into account if I drill the hole on the lower half of the sphere. It dosent state if the 10 inch is to be measured from the center of the cyllinder.

So I'm thinking the question wants me to explain what considerations needs to be taken in order to make a simulation.

What do you think?
 
Last edited:
I have gotten this question I'm not sure how to go about.

"A cyllinder shaped hole, 10 inches long, is drilled through a sphere.
Explain the remainder of the volume of the sphere."

Cyllinder width is not defined
Sphere is not defined
What are your thoughts regarding the assignment?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for
 
I'm completely lost on this one. The question dosent take into account if I drill the hole on the lower half of the sphere. It dosent state if the 10 inch is to be measured from the center of the cyllinder.

So I'm thinking the question wants me to explain what considerations needs to be taken in order to make a simulation.

What do you think?
 
Assume the sphere has radius R. Then we can write it as \(\displaystyle x^2+ y^2+ z^2= R^2\). We can orient the coordinate system so that the axis of the cylindrical hole is the z-axis. Taking the radius of the cylindrical hole to be r, it has equation \(\displaystyle x^2+ y^2= r^2\).

We could, then, by drawing a line from the center of the sphere to the edge of the opening of the cylindrical hole, use the Pythagorean theorem to calculate that \(\displaystyle r^2+ l^2= R^2\) where "l" is half the length of the hole. We are told that the length of the hole is 10 cm so l= 5 cm. We have \(\displaystyle r^2+ 25= R^2\). We could then calculate the volume of the sphere in terms of R, the volume of the cylindrical hole in terms of r, and the two "endcaps", the parts of the sphere that would have to be cut away at each end of the cylinder when we drill the hole. The volume remaining is, of course, the volume of the sphere minus the volume of the cylinder and the volumes of the two endcaps.

But a much easier way to do this problem is to assume that, since we are not told the radius of the sphere or the radius of the cylindrical hole, they are unnecessary! So imagine a sphere that is exactly 10 cm in radius. The "cylindrical hole" through it is a line with radius 0. That does not remove any volume from the sphere and the answer is just the volume of the sphere, \(\displaystyle \frac{4}{3}\pi (10^3)= \frac{4000}{3}\pi\). If you do the problem the "hard way", as above (I have), you will find that, in fact, the "R" and "r" terms cancel giving that answer.
 
Not sure what you mean by "lower half". There is no lower half in a sphere.
It's important that the resulting hole is a cylinder. Draw it in 2D - there are many ways to drill a hole through a sphere, but the cylinder shape requirement eliminates a whole (sorry) lot of them. What remains?
 
I have gotten this question I'm not sure how to go about.
"A cyllinder shaped hole, 10 inches long, is drilled through a sphere.
Explain the remainder of the volume of the sphere."

Cyllinder width is not defined
Sphere is not defined
Hole placement is not defined

I'm completely lost on this one. The question dosent take into account if I drill the hole on the lower half of the sphere. It dosent state if the 10 inch is to be measured from the center of the cyllinder.

So I'm thinking the question wants me to explain what considerations needs to be taken in order to make a simulation.

What do you think?

This is a standard mind-bender, where there doesn't seem to be enough information, so you just have to try something and see what happens. When you aren't told some dimensions, you just assign variables.

But this version is missing something extra, as you identify: hole placement is not defined. Most versions explicitly say that the hole goes through the center of the sphere, and often include a picture to clarify what the length of the hole means.

In this case, though, if the hole were not centered, what would "the length of the hole" even mean? Where would you measure it, since it would be longer on one side than the other? So I think we have to assume they do mean a centered hole, in order to make sense of the question.

And in that case, HallsofIvy has told you all about the problem. (But as this problem is stated, you can't really just assume it has a constant answer; you are told to explain, and there is no reason the answer couldn't be a function of R, and maybe other variables.)
 
It does state that the hole is a cylinder.

True; but I didn't see it that way because to me a cylinder is really an infinite surface, not including circular ends, so a hole can still be called cylindrical as long as it was made by an ordinary drill.

But certainly your interpretation is what the author intended; that is probably even why they said "cylinder-shaped" rather than "cylindrical".
 
Considering the author, yes, it is a mind-bender, made to see how I would try and approach the problem.


The way I image it, is a 3D simulation, with a sphere of varible size, and cylinder shaped hole going all the way through. The "hole" can be moved around and is not locked to the center. The size of the sphere, the placement of the hole, and the diameter of the hole, is all varibles, and the only constant being that the "hole" is of 10 inch or cm in length. Moving, or changing size, would therefore alter other paremeters realtime.

What I was hoping; to cook up a formular that subtracts the volume of the cylinder from the sphere, with these considerations..

But one of the problem is, like you said, the "length" of the hole is also unspecified. Is it the center of the hole? Can we assume that?
 
Let the radius of the cylinder be r

, the volume of the cylinder is 10pi*r^2. The radius of the sphere is 0.5((10^2+(2r)^2)^.5=(25+r^2)^.5, denote it as R. So the sphere with a hole has volume 4/3*Pi*R^3-10*pi*r^2. The center of the cylinder is the same point as the center of the sphere since a cylinder is symmetry to its center.
 
@Yma16 you assume the hole has been drilled through the spheres center. Also wouldn't you be missing the endcaps?

The volume remaining is, of course, the volume of the sphere minus the volume of the cylinder and the volumes of the two endcaps.


How would you go about to calculate the volume of the two endcaps?
 
Last edited:
Let the radius of the sphere be R so that the equation of the sphere is \(\displaystyle x^2+ y^2+ z^2= R^2\). Let the radius of the hole be r and take its axis to be the z-axis so that its equation is \(\displaystyle x^2+ y^2= r^2\). The hole cuts the surface of the sphere where \(\displaystyle r^2+ z^2= R^2\) so \(\displaystyle z= \sqrt{R^2- r^2}\). The circular base of the end cap is given by \(\displaystyle x^2+ y^2= r^2\), \(\displaystyle z= \sqrt{R^2- r^2}\). To find the volume of the end cap integrate \(\displaystyle \sqrt{R^2- x^2- y^2}- \sqrt{R^2- r^2}\) over that disk. Using polar coordinates, \(\displaystyle x= \rho cos(\theta)\), \(\displaystyle y= \rho sin(\theta)\) that integral becomes \(\displaystyle \int_{\rho= 0}^r\int_{\theta= 0}^{2\pi} \sqrt{R^2- \rho^2}- \sqrt{R^2- r^2} d\theta d\rho\).
 
A cylinder has two flat bases

@Yma16 you assume the hole has been drilled through the spheres center. Also wouldn't you be missing the endcaps?



How would you go about to calculate the volume of the two endcaps?

Therefore after the drill, the two end caps are gone. The end caps are not part of the remainder of the sphere. After the drill, the sphere should be able to stay on a flat surface and will not be able to roll when the radius of the cylinder is reasonably large. If you drill a hole off the center from the sphere, the missing piece is not a cylinder. I explained it using the symmetry reasoning. Drilling through the center of the sphere is a given condition, not a assumesion. If you drill a hole off the center, the hole is not a cylinder because one side of the hole is longer than the other. You cannot use base x height to compute the hole's volume.
 
Last edited:
Top