Setting up triple integral in cylindrical coords (looking to check my answer)

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SUMMARY

The discussion focuses on setting up a triple integral in cylindrical coordinates to compute the volume of the solid bounded by the sphere defined by the equation x² + y² + z² = 12 and the cone defined by 3z² = x² + y², with the condition z ≥ 0. The integral is expressed as I[0,2π], I[0,3] I[r/sqrt[3],sqrt[12-r²] r dz dr dθ]. The limits for θ are from 0 to 2π, and for r from 0 to 3, confirming the integration of the function '1' to find the volume.

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Homework Statement



set up an integral in cylindrical coords to compute the volume of the solid S bounded by the sphere x^2+y^2+z^2=12 and the cone 3z^2=x^2+y^2 where z>=0



The Attempt at a Solution



i will post my answer here. please let 'I' stand for integral:

i get,

I[0,2pi] , I[0,3] I[r/sqrt[3],sqrt[12-r^2] r dz dr dtheta.



so theta goes from 0 to 2pi
r goes from 0 to 3 etc

thank you
 
Last edited:
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It says to find volume. So the function you are integrating is '1'. That's a little hard to read, but it looks ok to me.
 

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