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How do I calculate the volume V of D :

D = {$(x,y,z)\in R^3$ : $x^2+y^2<=z^2+9,|z|\le 3$}

I know that i have to use the triple integral:

$\int\int\int_V div(F)dV$

$\int\int\int_Vdxdydz= Vol(V)$

how do I procede from here ? what do I need to put here '*'

$\int_{*}\int_{*}\int_{*}***d*d*d* = Vol(D)$

2 Answers 2

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\iiint_{\mathbb{R}^{3}}\bracks{x^{2} + y^{2} \leq z^{2} + 9} \bracks{\verts{z} \leq 3}\dd x\,\dd y\,\dd z\ =\ \overbrace{\iiint_{\mathbb{R}^{3}}\bracks{r^{2} \leq z^{2} + 9} \bracks{\verts{z} \leq 3}r\,\dd r\,\dd \theta\,\dd z} ^{\ds{\mbox{Cylindrical Coordinates}}} \\[5mm] = &\ \int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty} \bracks{r^{2} \leq z^{2} + 9} \bracks{\verts{z} \leq 3}r\,\dd r\,\dd \theta\,\dd z \\[5mm] = &\ 2\pi\int_{-\infty}^{\infty}\int_{0}^{\infty} \bracks{r \leq z^{2} + 9} \bracks{\verts{z} \leq 3}\,{1 \over 2}\,\dd r\,\dd z\label{1}\tag{1} \\[5mm] = &\ 2\pi\int_{0}^{\infty}\bracks{z \leq 3}\int_{0}^{\infty} \bracks{r \leq z^{2} + 9}\,\dd r\,\dd z\label{2}\tag{2} \\[5mm] & = 2\pi\int_{0}^{3}\pars{z^{2} + 9}\,\dd z = \bbx{\ds{72\pi}} \end{align}

  • 0
    why the outer integral is between -inf and +inf ? should't be -3 --- +3 ? – 2017-02-20
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    @NPLS You can see in line $\color{#f00}{1}$ that the integrand is an $even$ function of $z$. So, multiply by $2$ and integrate over $\left(0,\infty\right)$. It yields a prefactor $\color{#f00}{2} \times 2\pi \times 1/2 = \color{#f00}{2\pi}$ in line $\color{#f00}{2}$. – 2017-02-20
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You may simply integrate along sections. For any $z_0\in(-3,3)$ the section is a circle with squared radius $z_0^2+9$, hence area $\pi(z_0^2+9)$. By Cavalieri's principle the volume is given by $$ \int_{-3}^{3}\pi(z^2+9)\,dz = \color{red}{72\,\pi}.$$