How can we find the limits of this integration $\iiint_{G}f(x,y,z)\,dx\,dy\,dz$ with cylindrical coordinates?

Question

Let $$G=\left\{(x,y,z)\in\mathbb{R}^3:\ x^2+y^2+z^2\leq a^2,\ (x^2+y^2)^2\geq a^2(x^2-y^2),\ z\geq 0\right\}$$

If we apply cylindral coordinates on $G$ we have that $0\leq z \leq \sqrt{4a^{2}-r^{2}}$ but if $0\leq \theta \leq \frac{\pi}{4}$ we have that $a\sqrt{\cos{2\theta}}\leq r \leq 2a$ and if $-\frac{\pi}{4}\leq \theta \leq 0$ we have that $0\leq r \leq 2a$.

How can find the limis of integration $\iiint\limits_{G}f(x,y,z)\,dx\,dy\,dz$ with the use of cylindral coordinates ?

Answer 1

The first condition gives you points inside a sphere of radius a. Take square root of second condition , you see that gives you the points outside a cylinder, radius a^{1/4}.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \bbox[5px,#ffd]{\mbox{Lets}\ \vec{r} \equiv x\,\hat{x} + y\,\hat{y} + z\,\hat{z}}\,\,\, \mbox{and} \\ & \bbox[5px,#ffd]{G\! \equiv\! \braces{\!\vec{r} \in \mathbb{R}^{3}:\! r^{2}\! \leq a^{2}, \pars{x^{2} + y^{2}}^{2}\! \geq\! a^{2}\pars{x^{2} - y^{2}}, z \geq 0\!}} \end{align}

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Hereafter, I'll be using Cylindrical Coordinates. Namely, \begin{align} &\bbox[5px,#ffd]{\iiint_{\large\mathbb{R}^{3}} \bracks{x^{2} + y^{2} +z^{2} \leq a^{2}} \bracks{\pars{x^{2} + y^{2}}^{2} \geq a^{2}\pars{x^{2} - y^{2}}}} \\ &\ \bbox[5px,#ffd]{% \phantom{\iiint_{\large\mathbb{R}^{3}}\,}\bracks{z \geq 0}\dd^{3}\vec{r}} \\[5mm] = &\ \int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty} \bracks{r^{2} + z^{2} \leq a^{2}} \bracks{r^{4} \geq a^{2}r^{2}\cos\pars{2\phi}} \\ &\ \phantom{\int_{0}^{\infty}\int_{0}^{2\pi} \int_{0}^{\infty}\,\,\,} r\,\dd r\,\dd\phi\,\dd z \\[5mm] \stackrel{r^{2}\ \mapsto\ r}{=} &\ {1 \over 2} \int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty} \bracks{a^{2}\cos\pars{\phi} \leq r \leq a^{2} - z^{2}} \dd r\,\dd\phi\,\dd z \\[5mm] = &\ {1 \over 2}\verts{a}^{3} \int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\infty} \bracks{\cos\pars{\phi} \leq r \leq 1 - z^{2}} \dd r\,\dd\phi\,\dd z \end{align}

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Note that $\ds{\int_{0}^{2\pi}\on{f}\pars{\cos\pars{\phi}}\,\dd\phi = 2\int_{0}^{\pi/2}\bracks{% \on{f}\pars{\sin\pars{\phi}} + \on{f}\pars{-\sin\pars{\phi}}}\dd\phi}$.

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Then, \begin{align} &\bbox[5px,#ffd]{\iiint_{\large\mathbb{R}^{3}} \bracks{x^{2} + y^{2} +z^{2} \leq a^{2}} \bracks{\pars{x^{2} + y^{2}}^{2} \geq a^{2}\pars{x^{2} - y^{2}}}} \\ &\ \bbox[5px,#ffd]{% \phantom{\iiint_{\large\mathbb{R}^{3}}\,}\bracks{z \geq 0}\dd^{3}\vec{r}} \\[5mm] = &\ \verts{a}^{3}\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\infty} \bracks{\sin\pars{\phi} \leq r \leq 1 - z^{2}}\dd r\,\dd\phi\,\dd z \\[2mm] + & \verts{a}^{3} \int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\infty} \bracks{-\sin\pars{\phi} \leq r \leq 1 - z^{2}}\dd r\,\dd\phi\,\dd z \\[5mm] = &\ \verts{a}^{3}\int_{0}^{1}\int_{0}^{\pi/2} \bracks{\sin\pars{\phi} \leq 1 - z^{2}}\bracks{% 1 - z^{2} - \sin\pars{\phi}}\dd\phi\,\dd z \\[2mm] + & \verts{a}^{3}\ \underbrace{\int_{0}^{1}\int_{0}^{\pi/2} \pars{1 - z^{2}}\dd\phi\,\dd z} _{\ds{\pi \over 3}} \\[5mm] = &\ \verts{a}^{3}\int_{0}^{\pi/2}\int_{0}^{\root{1 - \sin\pars{\phi}}} \bracks{1 - z^{2} - \sin\pars{\phi}}\dd z\,\dd\phi \\[2mm] + & {1 \over 3}\,\pi\verts{a}^{3} \\[5mm] = &\ {2 \over 3}\verts{a}^{3}\int_{0}^{\pi/2} \bracks{1 - \sin\pars{\phi}}^{3/2}\,\dd\phi + {1 \over 3}\,\pi\verts{a}^{3} \\[5mm] = &\ {2 \over 3}\verts{a}^{3}\int_{0}^{\pi/2} {1 - \sin^{2}\pars{\phi} \over \bracks{1 + \sin\pars{\phi}}^{3/2}}\,\cos\pars{\phi}\,\dd\phi + {1 \over 3}\,\pi\verts{a}^{3} \\[5mm] = &\ {2 \over 3}\verts{a}^{3}\int_{0}^{1} {1 - t^{2} \over \pars{1 + t}^{3/2}}\dd t + {1 \over 3}\,\pi\verts{a}^{3} \\[5mm] = &\ {2 \over 3}\verts{a}^{3}\int_{1}^{2} \pars{2t^{-1/2} - t^{1/2}}\dd t + {1 \over 3}\,\pi\verts{a}^{3} \\[5mm] = &\ \bbx{{3\pi + 16\root{2} - 20 \over 9}\,\verts{a}^{3}} \approx 1.3391\,\verts{a}^{3} \\ & \end{align}