# Volume using Triple Integrals

Find the volume of solid enclosed by surfaces $x^2+y^2=9$ and $x^2+z^2=9$

I understand that these are two cylinders in XY and XZ planes respectively, that will cut each other above the XY plane. I get the following limits for triple integrals

$$\int_{\theta=0}^{2\pi}\int_{r=0}^3\int_{z=0}^{\sqrt{ 9-r^2\cos^2\theta}}rdzdrd\theta$$

Is it correct ?? If yes, the integral itself looks so complicates. some hints on solving it please !!

#### Solutions Collecting From Web of "Volume using Triple Integrals"

I think it’s better not to transform to cylindrical coordinates. Instead, first note that your region is $$\left\{(x,y,z)\in\mathbb R^3|-3\leq x\leq 3, -\sqrt{9-x^2}\leq y\leq\sqrt{9-x^2},-\sqrt{9-x^2}\leq z\leq\sqrt{9-x^2}\right\},$$ so your integral is equal to $$\int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}dzdydx=8\int_0^3\int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2}}dzdydx=8\int_0^3(9-x^2)\,dx,$$ which is equal to $144$.

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\expo}[1]{{\rm e}^{#1}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$

Hereafter, $\ds{\Theta:\mathbb{R}\setminus\braces{0} \to \mathbb{R}}$ is the Heaviside Step Function. Namely,
$\ds{\Theta\pars{x} \equiv \left\{\substack{\ds{0}\quad\mbox{if}\quad\ds{x < 0} \\[3mm] \ds{1}\quad\mbox{if}\quad\ds{x > 0}}\right.}$

\begin{align}
V
&\equiv
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\Theta\pars{9 – x^{2} – y^{2}}\Theta\pars{9 – x^{2} – z^{2}}\,\dd x\,\dd y\,\dd z
\\[3mm]&=
\int_{-\infty}^{\infty}\dd x
\int_{-\infty}^{\infty}\dd y\,\Theta\pars{9 – x^{2} – y^{2}}\,
\int_{-\infty}^{\infty}\Theta\pars{9 – x^{2} – z^{2}}\,\dd z
\tag{1}
\end{align}

\begin{align}
\int_{-\infty}^{\infty}\Theta\pars{9 – x^{2} – z^{2}}\,\dd z
&=
2\Theta\pars{9 – x^{2}}\sqrt{\vphantom{\Large A}9 – x^{2}\,}
\end{align}

By replacing this result in $\pars{1}$, we get
\begin{align}
V
&=
2\int_{-3}^{3}\dd x\,\sqrt{\vphantom{\Large A}9 – x^{2}\,}
\int_{-\infty}^{\infty}\dd y\,\Theta\pars{9 – x^{2} – y^{2}}\,
=
2\int_{-3}^{3}\dd x\,\sqrt{\vphantom{\Large A}9 – x^{2}\,}
\times 2\,\sqrt{\vphantom{\Large A}9 – x^{2}\,}
\\[3mm]&=
8\int_{0}^{3}\dd x\,\pars{9 – x^{2}}
=
8\pars{9\times 3 – {3^{3} \over 3}}
=
8\pars{27 – 9} =\ \bbox[10px,border:1px solid #000]{\displaystyle 144}
\end{align}

$\int^{2π}_{θ=0}\int^3_{r=0}\int^\sqrt{{9−r^2cos^2θ}}_{z=0}rdzdrdθ=\int^{2π}_{θ=0}\int^3_{r=0}r\sqrt{{9−r^2cos^2θ}}drd\theta=\int^{2π}_{θ=0}[-\frac{1}{3}sec^2(\theta)(9-r^2cos^2(\theta))^{3/2}]^3_0d\theta=144$

The last part was done by Mathematica since it is quite complicated to do so by hand. Thus, cylindrical coordinate is necessary to calculate the above integral. Yet, your formula is correct, whatever coordinate is used.