Question

Find the volume of the solid.

The solid in the first octant is above the plane z=2 and below the paraboloid z=4-x^2-y^2

Answer 1

In Cartesian coordinates, the region (call it
`R`) is the set

`R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}`

In the plane
`z=2`, we have

`2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2`

which is a circle with radius
`\sqrt2`. Then we can better describe the solid by

`R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le √(2 - x^2) \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}`

so that the volume is

`\displaystyle \iiint_R dV = \int_0^(\sqrt2) \int_0^(√(2-x^2)) \int_2^(4-x^2-y^2) dz \, dy \, dx`

While doable, it's easier to compute the volume in cylindrical coordinates.

`\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}`

Then we can describe
`R` in cylindrical coordinates by

`R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\frac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}`

so that the volume is

`\displaystyle \iiint_R dV = \int_0^(\pi/2) \int_0^(\sqrt2) \int_2^(4-r^2) r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^(\sqrt2) \int_2^(4-r^2) r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^(\sqrt2) r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^(\sqrt2) (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}`