Volume of region bounded by cone and parabloid

[adi_butler]

I don't know if anyone will be able to help me, I am 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

 

[PrudensOptimus]

<br /> \int \sqrt{(x^2 + y^2)} - (x^2 + y^2) dx<br /> =~30\deg<br />

 

[Guybrush Threepwood]

Originally posted by PrudensOptimus
<br /> \int \sqrt{(x^2 + y^2)} - (x^2 + y^2) dx<br /> =~30\deg<br />

30 degrees?

 

[Lonewolf]

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.

 

[gnome]

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]

In cylindrical coordinates (in other words, the polar coordinates gnome recommended), the bounding surfaces are z= &radic;(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 d&theta;. Since there is no &theta; in the integrand and we have circular symmetry, the integral with respect to theta is 2&pi;. This is the integral 2&pi;(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 2&pi;/12= &pi;/6.