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)`

`2r^2=1`

`r=1/sqrt(2)`

The boundary of the volume is a circle of radius `1/sqrt(2)` . Then `D` is

`D={(r,theta)|0lt= r =lt1/sqrt(2), 0lt= theta lt= 2pi}`

Now preform the double integral.

`V=int int_D [(sqrt(1-(x^2+y^2)))-(sqrt(x^2+y^2))] dxdy`

`V=int_0^(2pi) int_0^(1/sqrt(2)) ((sqrt(1-r^2))-r)r dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2))((1-r^2)^(3/2)-r^2)dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2))(r(1-r^2)^(1/2)-r^2)dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2))(r(1-r^2)^(1/2))dr d(theta)-int_0^(2pi) int_0^(1/sqrt(2))r^2dr d(theta)`

On the first integral, let `u=1-r^2` `->` `du=-2r dr`

`V=int_0^(2pi) int_0^(1/sqrt(2)) r(u)^(1/2)((du)/(-2r)) d(theta)-int_0^(2pi) int_0^(1/sqrt(2))r^2dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2)) (-1/2)(u)^(1/2) du d(theta)-int_0^(2pi) int_0^(1/sqrt(2))r^2dr d(theta)`

Integrate `u` and substitute back in for `r` .

`V=int_0^(2pi) [(-1/3)(1-r^2)^(3/2)-(1/3)r^3]|_0^(1/sqrt(2)) d(theta)`

`V=int_0^(2pi) [(-1/3)(1-(1/sqrt(2))^2)^(3/2)-(1/3)(1/sqrt(2))^3+(1/3)]d(theta)`

`V=int_0^(2pi) [(-1/3)(1/(2sqrt(2)))-(1/(6sqrt(2)))+(1/3)]d(theta)`

`V=int_0^(2pi) [-1/(6sqrt(2))-1/(6sqrt(2))+(2sqrt(2))/(6sqrt(2))] d(theta)`

`V=int_0^(2pi) (-2+2sqrt(2))/(6sqrt(2))d(theta)`

`V=(2pi)((-2+2sqrt(2))/(6sqrt(2)))`

`V=(2pi)/6*2((-1+sqrt(2))/sqrt(2))`

`V=(2pi)/3*(1-1/sqrt(2))`

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