- #1
qq545282501
- 31
- 1
Homework Statement
Find the volume of the solid region that lies inside the cone φ= pi/6 and inside the sphere ρ=4. Use rectangular coordinates.
Homework 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
The Attempt at a Solution
at first, I tired to use rectangular coordinates, but I don't 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 don't know how to get the expression of the cone in x,y,z coordinates from the given φ=pi/6 .