Volume of region bounded by cone and parabloid (1 Viewer)

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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
 
[tex]
\int \sqrt{(x^2 + y^2)} - (x^2 + y^2) dx
=~30\deg
[/tex]
 
Originally posted by PrudensOptimus
[tex]
\int \sqrt{(x^2 + y^2)} - (x^2 + y^2) dx
=~30\deg
[/tex]
30 degrees?
 
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.
 
1,032
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.
 

HallsofIvy

Science Advisor
41,626
821
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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