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November 27, 2014

November 27, 2014

Posted by **K** on Tuesday, January 29, 2013 at 12:14am.

- calculus -
**Steve**, Tuesday, January 29, 2013 at 10:59amSince the radius of the surface of the water is r^2 = x^2+y^2, it looks to me like the volumes are

paraboloid: v = ∫[0,4]πz dz = 8π

cone: v = ∫[0,4]πz^2 dz = 64/3 π

hyperboloid: v = ∫[1,5]π(z^2-1) dz = 112/3 π

Hmm. Not what I expected, the cone having greater volume than a round-bottomed paraboloid. However, it makes sense when you see that the cone's radius at the top of the water is √8, while the paraboloid's is only 2.

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