Volume of region bounded by cone and parabloid

  • Thread starter adi_butler
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  • #1
I dont know if anyone will be able to help me, im really stuck on this question!

"Show that the volume of the region bounded by the cone
z=sqrt((x*x)+(y*y)) and the parabloid z=(x*x)+(y*y) is
PI/6"

The bits in the brackets (ie x*x and y*y) are x squared and y squared respectively and sqrt is square root.

Any help would be very much appreciated!

Cheers,

Adi
 

Answers and Replies

  • #2
[tex]
\int \sqrt{(x^2 + y^2)} - (x^2 + y^2) dx
=~30\deg
[/tex]
 
  • #3
Originally posted by PrudensOptimus
[tex]
\int \sqrt{(x^2 + y^2)} - (x^2 + y^2) dx
=~30\deg
[/tex]
30 degrees?
 
  • #4
334
1
Pi/6 is not 30 degrees. Pi/6 radians is 30 degrees. And you set up the integral wrong. There's more than one variable.
 
  • #5
1,036
1
Those two guys intersect at z=1, directly above the circle (on the x-y plane) x2 + y2 = 1, and at the origin.
Within that region, (i.e. inside the cylinder x2 + y2 = 1) the surface of the cone lies above the surface of the paraboloid, so you want the volume bounded by the cone, the cylinder, and the plane z=0
MINUS the volume bounded by the cylinder, the paraboloid, and the plane z=0.

Put your two equations into polar coordinate form & you'll have two very simple double integrals that will give you the result you're looking for.
 
  • #6
HallsofIvy
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In cylindrical coordinates (in other words, the polar coordinates gnome recommended), the bounding surfaces are z= √(r2)= r (since r is positive in polar coordinates) and z= r2. You "z-difference" between the two surfaces is r- r2. The differential in polar coordinates is r dr dθ. Since there is no θ in the integrand and we have circular symmetry, the integral with respect to theta is 2π. This is the integral 2π(r2- r3)dr. The anti derivative of r2- r3 is (1/3)r3- (1/4)4.

Since z= r and z= r2 intersect at r= 1, the integration is from 0 to 1. (1/3)r3- (1/4)4 evaluated between 0 and 1 is (1/3)- (1/4)= 1/12.

The volume is 2π/12= π/6.
 
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