Find the volume enclosed by the hyperboloid -x^2-y^2+z^2=1 and plane z=2. I correctly figure the bounds to be D={(r,theta)|0 leq theta leq 2pi, 0 leq r leq sqrt(3)}

Nayeli Osborne 2022-10-25 Answered
Integrating over or under this object
There is clearly a kink in my understanding of double integrals with polar coordinates.
Our problem is to find the volume enclosed by the hyperboloid x 2 y 2 + z 2 = 1 and plane z = 2. I correctly figure the bounds to be D = { ( r , θ ) | 0 θ 2 π , 0 r 3 } .
But I have trouble finding if we are integrating between the section of the hyperboloid and the disk r = 1 at z = 0 or z = 2.
While r 1 r 2 d r d θ would make some sense as an integrand, I essentially dont know if I am integrating (ie, finding volume) under or over the surface of the hyperboloid.
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Answers (1)

blogpolisft
Answered 2022-10-26 Author has 10 answers
Step 1
Since you've said in the comments that you wish to find the volume, this can be done using a triple integral. The region in the xy plane can be written in polar coordinates as the set D you have said. As for the z limits, we'll have 1 + x 2 + y 2 z 2, therefore
V = d V = 0 3 0 2 π 1 + r 2 2 r d z d θ d r .
Step 2
This ensures that you are computing the volume between the hyperboloid and the plane z = 2.
In general when you compute volumes you compute from the xy plane up to whatever height you wish. When we set 1 + x 2 + y 2 z 2 we are essentially doing two computations: the volume of the cylinder of height 2 and subtracting the volume up to the hyperboloid, giving the volume desired.
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