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April 18, 2015

April 18, 2015

Posted by **Jane** on Thursday, December 8, 2011 at 1:34pm.

- calculus -
**Steve**, Thursday, December 8, 2011 at 4:50pmA little investigation will lead you to the formula that if a hole of radius r is drilled through a sphere of radius R, the remaining volume is just 4pi/3 (R^2 - r^2)^(3/2)

In this case, that would be 4pi/3 * 3√3 = 4√3 pi

__________________________

You might also look up the napkin ring problem, where it is shown that if a hole of length h is drilled through a sphere, the remaining volume is independent of the radius of the sphere!

In this case, h/2 = R^2 - r^2 = √3, so h = 2√3

The remaining volume is pi/6 h^3 = pi/6 * 24√3 = 4√3 pi

- calculus -
**Jane**, Friday, December 9, 2011 at 3:07pmthanks!

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