Volume Generated by Hyperboloid & Plane

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The discussion focuses on calculating the volume generated by the solid enclosed by the hyperboloid defined by the equation -x² - y² + z² = 1 and the plane z = 2. The integral used for volume calculation is ∫∫ (1 + r²) dr dθ, with r integrated from 0 to √3 and θ from 0 to 2π, yielding a volume of 15π/2. However, a participant pointed out an error in the integration approach, suggesting that the correct integral should include an additional r factor, leading to the correct formulation of ∫∫ z * r dr dθ instead of ∫∫ z² dr dθ.

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Volume generated[Solved]

Use polar coordinates to find the volume of the given solid enclosed by the hyperboloid -x2-y2+z2=1 and the plane z=2.

I wrote out the integral \int\int 1+r^2 dr d\vartheta integrated from 0 to 2\pi and r integrated from 0 to sqrt3 and volume enclosed gotten to be 15\pi/2. So this is the volume enclosed by the hyperboloid and the xy-plane correct? So I wanted to use the volume of the cyclinder to subtract off the volume found previously. But the volume is 6\pi so if I subtract i would get negative volume?
 
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semc said:
Use polar coordinates to find the volume of the given solid enclosed by the hyperboloid -x2-y2+z2=1 and the plane z=2.

I wrote out the integral \int\int 1+r^2 dr d\vartheta integrated from 0 to 2\pi and r integrated from 0 to sqrt3 and volume enclosed gotten to be 15\pi/2. So this is the volume enclosed by the hyperboloid and the xy-plane correct? So I wanted to use the volume of the cyclinder to subtract off the volume found previously. But the volume is 6\pi so if I subtract i would get negative volume?

It looks like you are integrating z^2 instead of z. Also shouldn't there be an extra r in the integral?

ie. \int\int z*r dr d\vartheta whereas it looks like you are doing \int\int z^2 dr d\vartheta
 

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