Find the volume of the solid that lies within the sphere x^2+y^2+z^2=1

Question

above the xy plane, and outside the cone z=6*sqrt(x^2+y^2)?

Answer 1

The Sphere is x² + y² + z² = 1

And cone z = 6√(x² + y²)

The volume of the Sphere is

Where x,y and z are in Cartesian coordinates and Ѳ, φ and ρ are in Spherical co-ordinate system.

Where,

x = ρ sin(φ)cos(Ѳ)

y = ρ sin(φ)sin(Ѳ)

z = ρ cos(φ)

then consider

z = 6√(x² + y²)

φ = 9.46°.

Since it is outside the cone, φ varies fom 9.46° to π/2

And

x² + y² + z² = ρ²

Then ρ² = 1

ρ = 1

Therefore ρ varies from 0 to 1.

And Ѳ varies from 0 to 2π.

Volume of the Solid

Volume of the Solid = 2.065

Therefore Volume of the Solid is 2.065.