1 Answer

Insict · Answer 1 · 2021-01-23T13:40:13+0000

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

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

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

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

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

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

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