Drilling a hole through the center of a solid sphere

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Homework Statement:

What volume of material is removed from a solid sphere of radius 2r by drilling a hole of radius r through the center

Relevant Equations:

integration formula
The volume of the sphere = [itex] \frac{32{\pi}r^{3}}{3}[/itex]

The answer given at the back of the book is [itex] (\frac {32}{3} - 4\sqrt{3}){\pi}r^3[/itex]

To drill a hole completely through the sphere, the hole would have to have a length of 4r.

To get the answer in the back of the book, it requires setting the length of the hole to [itex] r = \sqrt{4r^2-x^2} [/itex] and solving for r. So the hole has a radius of r, and a length of [itex]2\sqrt3r[/itex]

But this doesn't account for the material of the sphere that is greater than [itex]x=\sqrt3r[/itex] and less than [itex]x=-\sqrt3r[/itex]

Not only that, but the answer seems to be the volume of the sphere - the volume of a hole that doesn't completely go through the sphere. The question says "what volume of material is removed," The subtraction mentioned previously give you the volume of the sphere that has the given hole, not the volume of the material that was removed.

http://www.stumblingrobot.com/2015/08/11/volume-of-cylindrical-hole-removed-from-a-sphere/

This website has the same answer as the back of the book. The math is there for your convenience. Thank you!
 
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  • #2
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Just to clarify my doubt, how does the bounds that the author used ([itex]-/+\sqrt3r[/itex]) account for the circled part in the attachment
 

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  • #3
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It seems to me like the author is giving, as an answer, the volume of the area shaded in red in my attachment, revolved around the x axis. Either I'm really stupid or his answer doesn't match the question being asked.
 

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  • #4
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So the question is " What volume of material is removed from a solid sphere of radius 2r by drilling a hole of radius r through the center," that's implying we are starting with a sphere, that is solid, and we are drilling a hole through it. Thus the part circled in red, in post #2, is being drilled through. But the author is using the area of the part shaded in red in post #3, which does not include the part in post #2.
 
  • #5
phyzguy
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I think the book's answer is correct. I think you are making wrong assumptions about how the answer was arrived at. Think about the part that is removed. It consists of a cylinder plus two "caps" on each end. Each cap has a flat bottom and a top which is a part of the sphere. Try calculating the volumes of these three pieces and adding them up and see if the answer matches what the book gives. Post your work here and we'll see if we can help.
 
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  • #6
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Okay, will do. On a quick glance, it does seem like what I wrote in post #3 is NOT true. Please give me a moment to work through the problem again.
 
  • #7
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Huh, I guess I was wrong. The answer is indeed what the book wrote. And I arrived at the answer differently than the website that I linked to as well.

Note to self: just because someone arrived at the answer in a different way than you did, does not mean you're wrong. Also note to self: don't rush through, and take your time, when looking at someones answer. I guess I didn't properly understand what the person who's website I linked, was actually doing. I just kinda assumed that he was wrong and that's that. Not a good way to approach things.

Anyways, I will write my answer for the benefit and interest of anyone who may read this thread.

Like used phyzguy mentioned, you can break this problem down into 2 parts.

1) The box of height r, which has a length of [itex]2\sqrt3r[/itex]. This length was obtained from setting [itex] r = \sqrt{4r^2-x^2} [/itex]. This is the volume of the material removed not including the material remove of the two "caps." The integral is: [itex]2\int_0^{\sqrt3r} {\pi}r^2 dx[/itex]

2) Next, you have two caps. The integral is [itex] 2\int_{\sqrt3r}^{2r} {\pi}(4r^2-x^2)dx[/itex]

If you add both integrals, you will get the answer [itex]({\pi}r^3)(\frac{32}{3} - 4{\sqrt3})[/itex]

So yeah, to whoever is reading this, check you work, take your time, don't assume your way of doing something is the only way.
 
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  • #9
haruspex
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It is a curious fact that in order to find the volume remaining after drilling a hole centrally through a sphere you only need to know the length of the hole, measured between the apertures at each end. The answer is the same regardless of the radius of the sphere.
Armed with this knowledge, you can find the answer to the question in this thread very quickly: volume of sphere radius 2r minus volume of sphere of diameter equal to the length of the hole (r√3).
 

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