- #1

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## Homework Statement

Find the volume of the solid that lies between

z=x

^{2}+y

^{2}and

x

^{2}+y

^{2}+z

^{2}=2

## Homework Equations

z=r

^{2}

z=√(2-r

^{2})

## The Attempt at a Solution

So changing this into cylindrical coordinates, I get

z goes from r

^{2}to √(2-r

^{2})

r goes from 0 to √2

theta goes from 0 to 2π

so we get [itex] \int_0^{2π} \int_0^{\sqrt{2}} \int_{r^2}^{\sqrt{2-r^2}} r dzdrdt[/itex]

I dunno... something feels off about this. I think this way it would take into account the negative area on the outside of the curve... How else could I set up the integral?