Question

THIS QUESTION IS KILLING ME
Calculate the volume of the object by using the triple integral.

Answer 1

The volume of the solid (call it S) in Cartesian coordinates is

`\displaystyle\iiint_S\mathrm dV=\int_(-1)^1\int_(-√(1-x^2))^(√(1-x^2))\int_((x^2+y^2)^2-1)^(4-4(x^2+y^2))\mathrm dz\,\mathrm dy\,\mathrm dx`

but I suspect converting to cylindrical coordinates would make the integral much more tractable.

Take

`\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=z\end{cases}\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz`

Then

`4-4(x^2+y^2)=4-4r^2=4(1-r^2)`

`(x^2+y^2)^2-1=(r^2)^2-1=r^4-1`

and the integral becomes

`\displaystyle\iiint_S\mathrm dV=\int_0^(2\pi)\int_0^1\int_(r^4-1)^(4(1-r^2))r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta`

`=\displaystyle2\pi\int_0^1r(4(1-r^2)-(r^4-1))\,\mathrm dr`

`=\displaystyle2\pi\int_0^1r(5-4r^2-r^4)\,\mathrm dr`

`=\displaystyle2\pi\int_0^15r-4r^3-r^5\,\mathrm dr`

`=2\pi\left(\frac52-1-\frac16\right)=\boxed{\frac{8\pi}3}`