# Using cylindric coordinates, what is the volume of the solid bounded above by the sphere r^2+z^2=2 and under by the paraboloid z=r^2?

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### 1 Answer

Determine the boundaries:

Let's find r where the surfaces intersect.

` `` `

in this case, `z=r^2` and `r^2+z^2=2`

Hence `r^2+r^4-2=0`

`(r^2-1)(r^2+2)=0`

Only one positive real solution

`r=1. `

r belongs to `[0, 1]`

For a given r, the area is under the sphere therefore `z^2<=2-r^2`

the area is above the paraboloid therefore `z>=r^2`

No condition on `theta` therefore `theta in[0,2pi]`

The area is determined by

`A=int_0^1int_(r^2)^sqrt(2-r^2)int_0^(2pi) 1 rd theta dzdr`

integrate `d theta`

`A=``int_0^1int_(r^2)^sqrt(2-r^2)r2pidzdr`

integrate `dz`

` <br> `

`A=int_0^(1)2pi r(sqrt(2-r^2)-r^2)dr`

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` `

`A=int_0^1 2 pi rsqrt(2-r^2)-2pi r^3 d r` ` `

`A=[-2/3 pi (2-r^2)^(3/2)-2pi r^4/4]_0 ^1`

`A=-2/3 pi +2/3 pi sqrt(2)^3-pi /2= 4sqrt(2)pi/3-7pi/6`

**The area is **`A=-7pi/6+4sqrt(2) pi/3 ~~2.26`