CLYINDRICAL coordinates of volume bound by z=r and z^2+y^2+x^2=4

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The discussion focuses on finding the volume bounded by the cone z=r and the sphere z^2+y^2+x^2=4 using cylindrical coordinates. The initial attempt provided limits for r, theta, and z, resulting in a volume calculation of 8 pi, which did not match the spherical coordinate solution. Participants emphasize the need for correct limits, particularly for r, which should depend on z. Clarifications suggest that the limits for r should be linear rather than curvilinear. The conversation highlights the importance of accurately determining these limits for proper volume calculation.
Unemployed
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Homework Statement



Find the smaller volume bound by cone z=r and sphere z^2+y^2+x^2=4 using cylindrcal coordinates

Homework Equations



dV=r-dr d-theta dz

The Attempt at a Solution



Limits on r: z to sqrt (4-z^2)
limits on theta: 2pi to 0
limits on z: 2-0

Did this and got 8 pi, not the same answer with spherical.
Need guidance on limits.

 
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Hi Unemployed! :smile:

(have a theta: θ and a pi: π and a square-root: √ and try using the X2 icon just above the Reply box :wink:)
Unemployed said:
Limits on r: z to sqrt (4-z^2)

Not for the cone bit :wink:
 
tiny-tim said:
Hi Unemployed! :smile:

(have a theta: θ and a pi: π and a square-root: √ and try using the X2 icon just above the Reply box :wink:)


Not for the cone bit :wink:


For r = 0 to √2 ?
 
No, the limit for r will still depend on z, but linearly instead of "curvily". :wink:
 

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