# Voulme of an ice cream cone bound by a sphere

1. Dec 4, 2007

### zimbob

1. The problem statement, all variables and given/known data
Find the volume of an ice cream cone bounded by the sphere x^2+y^2+z^2=1 and the cone z=sqrt(x^2+y^2-1)

2. Relevant equations
The two simultaneous equations yield x^2+y^2=1

3. The attempt at a solution

Attached

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Last edited: Dec 4, 2007
2. Dec 5, 2007

### HallsofIvy

Staff Emeritus
z=sqrt(x^2+y^2-1) is NOT the equation of a cone- it is a hyperboloid.

z= sqrt{x^2+ y^2) would be (the upper nappe of) a cone with vertex at the origin with sides making angle $\pi/4$ with the xy-plane.

3. Dec 5, 2007

### zimbob

Thanks for your response, so is it logical to re-arrange the integral limits such that it becomes:
Volume of cone =integral(limits theta= 0 to pi/4)integral(limits r=0 to 1/sqrt2)[sqrt((1-r^2)-r)dr d theta.

4. Dec 5, 2007

### HallsofIvy

Staff Emeritus
First you are going to have to define the cone part! If it is z= sqrt{x^2+ y^2}, then yes, you take, in polar coordinates, $\theta= 0$ to $\pi/4$. However, r goes from 0 to 1, not $1/\sqrt{2}$ because you are going up to the spherical cap.

5. Dec 5, 2007

### zimbob

The "cone" part is given as z= sqrt{x^2+ y^2-1} which I agree is not an equation for a cone but a hyperboloid as you mentioned above. What is troubling me is how to deal with the (-1) inside the sqrt.

6. Dec 6, 2007

### zimbob

Any ideas please, I am stuck.

7. Apr 21, 2008

### obperryo

Did you ever get this figured out?
I am working on the same problem with the exact same issue .. the -1.