# Compite the integral &#x222B;<!-- ∫ --> V </msub> 4 &#x2212;<!--

Compite the integral
${\int }_{V}\frac{4-{z}^{2}}{\left({x}^{2}+{y}^{2}{\right)}^{3}}\mathrm{d}x\mathrm{d}y\mathrm{d}z,$
where V is the solid enclosed by the paraboloids ${x}^{2}+{y}^{2}=z,{x}^{2}+{y}^{2}=2z$ and cones ${x}^{2}+{y}^{2}=\left(z-2{\right)}^{2},{x}^{2}+{y}^{2}=4\left(z-2{\right)}^{2}.$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Antonio Dickerson
Let ${D}_{-}=\left\{\left(r,z\right):r\ge 0,z\le 2,z\le {r}^{2}\le 2z,\left(z-2{\right)}^{2}\le {r}^{2}\le 4\left(z-2{\right)}^{2}\right\}$, and consider the following change of variables: $u=z/{r}^{2}$ and $v=\left(z-2{\right)}^{2}/{r}^{2}$, then
$|det\left(\begin{array}{cc}\frac{\mathrm{\partial }u}{\mathrm{\partial }r}& \frac{\mathrm{\partial }u}{\mathrm{\partial }z}\\ \frac{\mathrm{\partial }v}{\mathrm{\partial }r}& \frac{\mathrm{\partial }v}{\mathrm{\partial }z}\end{array}\right)|=2\frac{4-{z}^{2}}{{r}^{5}}.$
Then
${\iiint }_{V\cap \left\{z\le 2\right\}}\frac{4-{z}^{2}}{\left({x}^{2}+{y}^{2}{\right)}^{3}}dxdydz=2\pi {\iint }_{{D}_{-}}\frac{4-{z}^{2}}{{r}^{6}}rdrdz=2\pi {\int }_{u=1/2}^{1}{\int }_{v=1/4}^{1}\frac{1}{2}dudv=\frac{3\pi }{8}.$