Voulme of an ice cream cone bound by a sphere

[zimbob]

Homework Statement

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)

Homework Equations

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

The Attempt at a Solution

Attached

 

[HallsofIvy]

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.

 

[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.

 

[HallsofIvy]

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.

 

[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.

 

[zimbob]

Any ideas please, I am stuck.

 

[obperryo]

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