Turn spherical coordinates into rectangular coordinates

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Homework Help Overview

The problem involves finding the volume of a solid region defined by a cone and a sphere, specifically converting spherical coordinates to rectangular coordinates. The original poster seeks to express the cone defined by φ= π/6 in rectangular coordinates.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to express the cone in rectangular coordinates but struggles with the conversion from spherical coordinates. They initially consider using polar coordinates and divide the problem into two parts.
  • Some participants provide hints regarding the equation of a cone in rectangular coordinates and suggest a linear relationship between the radius and height.
  • There is a discussion about the general form of the cone's equation and how it relates to the specific case presented by the original poster.

Discussion Status

Participants are exploring the relationship between the cone's geometry and its representation in rectangular coordinates. Hints have been offered regarding the form of the cone's equation, and the original poster has begun to derive an expression based on these hints. However, there is no explicit consensus on the correctness of the derived equation.

Contextual Notes

The original poster expresses confusion about converting the cone's equation from spherical to rectangular coordinates and is seeking clarification on the assumptions involved in this conversion.

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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?
 

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