Find the volume above the cone z=sqrt(x^2+y^2) and below the sphere x^2+y^2+z^2=1

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))