# Volume of region bounded by cone and parabloid

1. Nov 26, 2003

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,

2. Nov 27, 2003

### PrudensOptimus

$$\int \sqrt{(x^2 + y^2)} - (x^2 + y^2) dx =~30\deg$$

3. Nov 27, 2003

### Guybrush Threepwood

30 degrees?

4. Nov 27, 2003

### 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.

5. Nov 27, 2003

### 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.

6. Nov 27, 2003

### HallsofIvy

Staff Emeritus
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.

Last edited: Nov 28, 2003