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Spherical Coordinates

  • Thread starter MozAngeles
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  • #1
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Homework Statement



Set up the triple integral for the volume of the given solid using spherical coordinates:

The solid bounded below by the sphere ρ=6cosθ and above by the cone z=sqrt(x2+y2)

Homework Equations





The Attempt at a Solution



I thought i had this set up right where ρ goes from 6cosθ to sqrt(3)/2 ( how i got sqrt(3)/2 was by z=sqrt(x2+y2 is th esame as ρcos∅=ρsin which means ∅=pi/4 plugging that into ρ=6cosθ, gives me sqrt3/2) i don't know if this is right but for ∅ it goes from 0 to pi/4 and for θ it goes from 0 to 2pi

thanks in advance
 

Answers and Replies

  • #2
ehild
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What is ρ, and what is θ?

How can be ρ=6cosθ the equation of a sphere?


ehild
 
  • #3
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ρ is the distance from P to the origin

∅ is the angle that the line makes with the positive x axis

θ is the angle from the cylindrical coordinates (0<θ<2pi)

i know that the equation for a sphere is when ρ= constant, in this case she told us it was cos∅
 
  • #4
ehild
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Last edited:
  • #5
ehild
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is that the volume you nee to find?

http://img811.imageshack.us/img811/4920/conesphere.jpg [Broken]

ehild
 
Last edited by a moderator:
  • #6
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i couldnt get a picture but that is how the sphere looks with a radius of six and the cone is just the top half as looking as any cone would look
 
  • #7
ehild
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Is the volume as shown in the picture in my previous post? If so, you have to integrate the volume element dV=r^2sin(∅)dr d∅ dθ from 0 to 2pi for θ,
from 0 to pi/4 for ∅ (it is the angle of the cone it makes with the positive y axis) and for r from the top of the cone to the radius of the sphere r=ρ=6.

ehild
 

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