Question

Find the volume of the "ice cream cone" bounded by the sphere x^2+y^2+z^2=1 and the cone \(z=\sqrt(x^2+y^2-1)\)
Please, help

Answer 1

@mathmate @Sam

Answer 2

Can you check if the cone is actually \(\sqrt{x^2+y^2}-1\) and not \(\sqrt{x^2+y^2-1}\)

Answer 3

all under radical.

Answer 4

Then the cone doesn't close at the bottom where x^2+y^2<1. :(

Answer 5

Wolfram gives me this image. https://www.wolframalpha.com/input/?i=z%3Dsqrt(x%5E2%2By%5E2-1)

Answer 6

What I don't get is what part we need to find the volume?
is it the volume of the sphere - volume of the cone ?

Answer 7

We need first find the intersection of the cone and sphere, and take it from there. The tangent plane/surface is not vertical.

Answer 8

That's a weird ice cream cone! lol
Where does it end at the top? :(

Answer 9

the cone is defined by \(z=\sqrt {x^2+y^2-1}\)
so that the term under radical must be \(\geq 0\) that is the \(x^2+y^2\geq1\)

Answer 10

This is a truncated cone, that fits above the sphere.
Then we need a cover, i.e. the upper limit of the cone.
I expect it to look like this.
The volume would be V1+V2. V1 is volume of the sphere above the tangent plane A-A, and V2 is the cone below A-A.