4x^2-3x+4y^2+4z^2=0 here we shall proceed as follows: x=ρcosθsinφ y=ρsinθsinφ z=ρcosφ thus 4x^2-3x+4y^2+4z^2= 4(ρcosθsinφ)^2-3(ρcosθsinφ)+4(ρsinθsinφ)^2+4(ρcosφ) but ρ=1/4cosθsinφ
hence we shall have: 4x^2-3x+4y^2+4z^2 =1/4cosθsinθ(cosθ(4-3sinφ))+4sin^2(φ)
