calculus
posted by K on .
three bowls are filled to a depth of four units. a paraboloid, z=x^2 +y^2 for 0<z<4, a cone z=(x^2 + y^2)^(1/2) for 0<z<4, and a hyperboloid z=(1+x^2+y^2)^(1/2) for 1<z<5. which bowl holds more water?

Since 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^21) dz = 112/3 π
Hmm. Not what I expected, the cone having greater volume than a roundbottomed 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.