Find the volume of the solid bounded below by the circular cone z=1.5\sqrt(x^(2)+y^(2)) and above by the sphere x^(2)+y^(2)+z^(2)=9.75z.

Question

Find the volume of the solid bounded below by the circular cone z=1.5\sqrt(x^(2)+y^(2)) and above by the

sphere x^(2)+y^(2)+z^(2)=9.75z.

Answer 1

There are two ways to do this question: Using cylindrical coordinates or spherical coordinates. However, the fact that the question deals with spheres and cones, the latter is better to use. In fact, this question is the "prototype" for which spherical coordinates are built for. So, let's begin. To for which spherical coordinates we need to know what ρ, θ, and φ are. The first two are always found via the sphere, whereas the last is found via the cone. With some work you can see that the equation of the sphere given is one that has a radius of , and do get the complete sphere we need 0<θ<2π. Again, we use the cone to figure out φ as follows: z=3/2sqrt(x^2+y^2) becomes ρcosφ=3/2sqrt(ρ^2sin^2φ)=3/2ρsinφ by using the transformation of Cartesian to sphere coordinates. This equation becomes 2/3=tanφ. Which gives 0<φ<tan^{-1}(2/3). Now the volume is given by ∫∫∫dV= ∫_{0}^{2π} ∫_{0}^{tan^{-1}(2/3)} ∫_{2/3ρsinφ}^{sphre} ρ^2sinφdρdφdθ. Which I suspect you can do!