Find the volume above the cone `z=sqrt(x^2+y^2)` and below the sphere `x^2+y^2+z^2=1`
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The volume between two surfaces can be calculated by
`V=int int_D (z_(t o p)-z_(b o t)) dA`
Where `D` is the area contained by the boundary of the volume projected onto the xy-plane.
To find `D` we need to solve when they two surfaces meet. Solve for `z` on both equations and set them equal to each other.
We have `z=sqrt(x^2+y^2)`
and
`z^2+x^2+y^2=1`
`z=+-sqrt(1-(x^2+y^2))`
Notice that the bottom half of the sphere `z=-sqrt(1-(x^2+y^2))` is irrelevant here because it does not intersect with the cone. The following condition is true to find the curve of intersection.
`z=z`
`sqrt(x^2+y^2)=sqrt(1-(x^2+y^2))`
Switch to polar coordinates to simplify the problem.
`r=sqrt(1-r^2)`
(The entire section contains 351 words.)
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