# 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 !!

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 $144$.

• Yeah, makes sense, Just for practice sake, is my cylindrical cordinate also correct ? – Aman Mittal Oct 13 '13 at 7:24
• Yes, it is correct, though $z$ should go from $-\sqrt{9-r^2\cos^2\theta}$ to $\sqrt{9-r^2\cos^2\theta}$. – detnvvp Oct 13 '13 at 8:57
• Why ? The complete figure is above the XY plane. ? – Aman Mittal Oct 13 '13 at 16:51
• But, from what the problem says, $z$ could take negative values. For example, the point $(1,2,-2)$ is inside both of those cylinders. – detnvvp Oct 13 '13 at 20:45

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

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}

• What does $\Theta\pars{9 - x^{2} - y^{2}}$ mean, and why did you change to infinite limits. – john May 24 '17 at 1:38
• @john I add some text at the very beginning. We can set the limits to the whole $\mathbb{R}$ because the $\Theta$ is 'in charge' of setting the boundaries. It's convenient. For example, we don't have to rely in help from some graphic, etc$\ldots$ – Felix Marin May 24 '17 at 2:16

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

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.