drilled sphere volume

[frederick]

Ah, I used to be able to do this easily but I really have forgot by now...

 

Say you have a sphere with radius 10, you are drilling out a hole along the diameter with a radius of 3 inches... Find the volume of what's left of the sphere...

 

So the way I thought to do it was just y = sqrt(100-x^2) in range y>3 rotated around x axis, but I really did forget how to write this...

 

It was something like [math]pi \int^{10}_{3}(100-x^2)[/math]?

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Actually, sorry for wasting your time, that seems to be about right.

[hermanntrude]

there's a quicker way to answer this without calculus, too:

 

mathworld says that the volume of a spherical ring is equal to

 

[math]V = \frac{4}{3}(r^2-R^2)^\frac{3}{2}[/math]

 

where r is the radius of the drilled hole and R is the radius of the sphere

 

Interestingly you can also get the volume of the spherical ring knowing only the length of the hole:

 

[math] V = \frac{1}{6}\pi L^3[/math]

 

where L is the length of the hole (measured after drilling... in other words not including the depth of the spherical caps.)

Edited May 2, 2009 by hermanntrude

[Bignose]

Interestingly you can also get the volume of the spherical ring knowing only the length of the hole:

 

[math] V = \frac{1}{6}\pi L^3[/math]

 

where L is the length of the hole (measured after drilling... in other words not including the depth of the spherical caps.)

 

Just a quick note here that be sure to go to that Mathworld link and look at the specific definition of L. (Not that I am blaming you herman, because you got it from the webpage, but what I thought was L from "length of the hole" was NOT what L actually was!) The diagram on the Mathworld site cleared it up quite a bit for me.