Find the volume of a solid bounded by the paraboloids \displaystyle{z=4x^2+4y^2} and \displaystyle{z=5-6x^2-y^2}

Question

Find the volume of a solid bounded by the paraboloids
`\displaystyle{z=4x^2+4y^2}` and
`\displaystyle{z=5-6x^2-y^2}`

Answer 1

The two surfaces meet in an ellipse in some plane with
`0\le z\le5`, since

`4x^2 + 4y^2 = 5 - 6x^2 - y^2 \implies 10x^2 + 5y^2 = 5 \implies 2x^2 + y^2 = 1`

so the solid is bounded by the elliptical cylinder with this equation. In cylindrical coordinates, with

`\begin{cases}x = \frac1{\sqrt2} r \cos(\theta) \\ y = r \sin(\theta) \\ z = \zeta \\ dV = dx\,dy\,dz = \frac1{\sqrt2} r \, dr \, d\theta \, d\zeta\end{cases}`

the solid is the set

`\begin{array}{c} R = \left\{(r,\theta,\zeta) ~:~ 0\le r\le1 \text{ and } 0 \le\theta\le2\pi \text{ and } \right. \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \left.3r^2 - r^2 \cos(2\theta) \le \zeta \le 5 - 2r^2 - r^2 \cos(2t)\right\} \end{array}`

and the volume is given by the integral

`\displaystyle \iiint_R dV = \frac1{\sqrt2} \int_0^(2\pi) \int_0^1 \int_(3r^2 - r^2 \cos(2\theta))^(5 - 2r^2 - r^2 \cos(2\theta)) r \, d\zeta \, dr \, d\theta \\\\ ~~~~ = \frac1{\sqrt2} \int_0^(2\pi) \int_0^1 r(5 - 5r^2) \, dr \, d\theta \\\\ ~~~~ = (10\pi)/(\sqrt2) \int_0^1 (r - r^3) \, dr \\\\ ~~~~ = 5\sqrt2\,\pi \left(\frac12 - \frac14\right) = \boxed{(5\pi)/(2\sqrt2)}`