Question

Calculate the volume of the solid, bounded by the surfaces: z=4x2 +4y2; z=x2+y2; z=4.

Answer 1

Split up the integral over the bounded region
`D` as


`\displaystyle\iiint_D\mathrm dV=\left\{\iiint_{\text{region between }z=4\text{ and }z=x^2+y^2}+\iiint_{\text{region between }z=4x^2+4y^2\text{ and }z=x^2+y^2}\right\}\mathrm dV`

Converting to cylindrical coordinates, you then have a volume of


`\displaystyle\left\{\int_(\theta=0)^(\theta=2\pi)\int_(r=1)^(r=2)\int_(z=r^2)^(z=4)+\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=1)\int_(z=r^2)^(z=4r^2)\right\}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta`

`=\displaystyle2\pi\left(\int_(r=1)^(r=2)rz\bigg|_(z=r^2)^(z=4)\,\mathrm dr+\int_(r=0)^(r=1)rz\bigg|_(z=r^2)^(z=4r^2)\,\mathrm dr\right)`

`=\displaystyle2\pi\left(\int_(r=1)^(r=2)(4r-r^3)\,\mathrm dr+\int_(r=0)^(r=1)(4r^3-r^3)\,\mathrm dr\right)`

`=6\pi`