1. The problem statement, all variables and given/known data Find the volume of the solid region that lies inside the cone φ= pi/6 and inside the sphere ρ=4. Use rectangular coordinates. 2. Relevant equations x=ρ sinφ cos θ y=ρsinφ sin θ z=ρ cos φ ρ^2=x^2+y^2+z^2 x= r cos θ y= r sin θ r^2=x^2+y^2 3. The attempt at a solution at first, I tired to use rectangular coordinates, but I dont even know how to express this cone in rectangular coordinates. so I i divided them into 2 parts and used polar coordinates, where the lower part is just a straightforward volume of a cone, the upper part is the small dome where the sphere and the cone intersects each other. since ρ=4, z=4cosφ = 4cos (π/6) = 2√3 which is the height when the sphere and the cone intersects at. 16=x^2+y^2+z^2 z=√(16-x^2-y^2) set them equal to each other : √(16-x^2-y^2)=2√3 , by simplify, i get 4=x^2+y^2 which means radius =2 for the R where the sphere and the cone meet. the volume of the cone in the lower part: 1/3*π*(2)^2*(2√3)= 14.51 [using volume formula for a cone] the upper dome volume= ∫∫ [(√(16-r^2)- 2√3 ] r dr dθ where r goes from 0 to 2, and θ goes from 0 to 2π. I get V=3.449, adding it with the volume of the cone, i get 17.958. I am confused on how to do this problem in x.y,z coordinates( without dividing it into 2 parts). more precisely, I dont know how to get the expression of the cone in x,y,z coordinates from the given φ=pi/6 .