What is the Volume Between Two Paraboloids?

[damoj]

We have two paraboloids
z = 18 + x^2 + y^2
and z = 3x^2 + 3y^2 + 10

i know that the intersection of the two is where
18 + x^2 + y^2 = 3x^2 + 3y^2 + 10

which gives us 4= x^2 + y^2 which is the radius of the paraboloids at that intersection.
we find that the intersection is at z = 22 by substituting 4 for the x^2 + y^2 in both equations

It seems to me that the volume between the two paraboloids show be the volume of the paraboloid z = 18 + x^2 + y^2 from z=18 (there x^2 + y^2 = 0) to z= 22(where x^2 + y^2 = 4)

which would give us 0.5pi r^2 h
which would give us 0.5 pi 4^2 4 = 32pi
but the answer is 16pi

i know that to solve it normally you would use a double integral with polar coordinates
but i can't figure out why the volume isn't the volume of the first paraboloid from z= 18 to z= 22


can someone explain why?

 

[lanedance]

won't it be the volume of z = 3x^2 + 3y^2 + 10, from z=10 to 22 minus the volume you described for z = 18 + x^2 + y^2

also try drawing a cross section to help visualise what is going on

 

[damoj]

yeah you're right
i realized that after i posted.

interesting that a calc 3 problem on a past exam that can be solved with basic algebra and geometry, ha:)