Voulme of an ice cream cone bound by a sphere

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SUMMARY

The discussion focuses on calculating the volume of an ice cream cone bounded by the sphere defined by the equation x²+y²+z²=1 and the hyperboloid z=sqrt(x²+y²-1). Participants clarify that the equation z=sqrt(x²+y²-1) does not represent a cone but a hyperboloid. They suggest using polar coordinates for integration, with limits adjusted to account for the spherical cap, specifically integrating from θ=0 to π/4 and r from 0 to 1. The challenge lies in addressing the negative one within the square root in the hyperboloid equation.

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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



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

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