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.

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, 200916 yr by hermanntrude

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.