Turn spherical coordinates into rectangular coordinates

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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.

Snapshot.jpg

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 .
 
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Hint: the equation of a cone, in rectangular coords, pointing along the z axis, is the equation of a circle whose radius depends linearly on z.
 
Simon Bridge said:
Hint: the equation of a cone, in rectangular coords, pointing along the z axis, is the equation of a circle whose radius depends linearly on z.

thank you for the hint

i found the equation of a general cone from wolfram:
NumberedEquation2.gif


where

NumberedEquation3.gif

in my case, the equation of the cone would be z= sqrt ( (x^2)/3 + (y^2)/3 )
since, r=2, h = 2√3 , c^2= 1/3 , correct?
 
Do you not understand the equation wolfram gives you?

The equation of a circle whose radius depends on z is: ##x^2 + y^2 = f^2(z)##
It varies linearly with z so f has the form of a straight line. ##f(z) = az + b##
Now its a matter of finding a and b using the information you have.
Is this consistent with what you got by rote from wolfram?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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