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 Post subject: Triple integral and ellipsoidPosted: Sat Jun 24, 2017 9:58 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Let ${\rm E}$ be the solid ellipsoid

$${\rm E} = \left\{(x,y,z)\in\mathbb{R}^3 \; \bigg|\; \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1 \right \}$$

where $a > 0,\: b > 0,\: c > 0$

Evaluate $\displaystyle \iiint xyz \, {\rm d}(x, y, z)$ over:

(a) the whole ellipsoid

(b) that part of it in the first quadrant, $x \ge 0,\: y \ge 0,\: z \ge 0$

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 Post subject: Re: Triple integral and ellipsoidPosted: Sun Jun 25, 2017 5:18 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Considering the change of cooordinates \begin{align*}
\left\{\begin{array}{l}
x=a\,r\sin\vartheta\cos\varphi\\
y=b\,r\sin\vartheta\sin\varphi\\
z=c\,r\cos\vartheta
\end{align*} with Jacobian
\begin{align*}
\biggl|{\frac{\partial(x,y,z)}{\partial(r,\vartheta,\varphi)}}\biggr|&=\left|{\begin{array}{ccc}
a\sin\vartheta\cos\varphi & a\,r\cos\vartheta\cos\varphi & -a\,r\sin\vartheta\sin\varphi\\
b\sin\vartheta\sin\varphi & b\,r\cos\vartheta\sin\varphi & b\,r\sin\vartheta\cos\varphi\\
c\cos\vartheta & -c\,r\sin\vartheta & 0
\end{array}
}\right|\\
&=abc\,r^2\sin\vartheta\,,
\end{align*}
1. the solid ellipsoid ${\rm{E}}=\Big\{{(x,y,z)\in{\mathbb{R}}^3\;\big|\;\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\leqslant1\,}\Big\}$ can be represented as ${\rm{E}}=\Big\{{\big(a\,r\sin\vartheta\cos\varphi,\, b\,r\sin\vartheta\sin\varphi,\,\,c\,r\cos\vartheta\big)\in{\mathbb{R}}^3\;\big|\;r\in[0,1]\,,\; \vartheta\in[0,\pi], \; \varphi\in[0,2\pi]\,}\Big\}.$
So \begin{align*}
\mathop{\iiint}\limits_{\rm{E}}{xyz\,d(x,y,z)}&=\mathop{\iiint}\limits_{\rm{E}}{abc\,r^3\sin^2\vartheta\cos\vartheta\cos\varphi\sin\varphi\,\big|abc\,r^2\sin\vartheta\big|\,d(r,\vartheta,\varphi)}\\
&=(abc)^2\int_{0}^{1}\int_{0}^{\pi}\int_{0}^{2\pi}{r^5\sin^3\vartheta\cos\vartheta\cos\varphi\sin\varphi\,d\varphi\,d\vartheta\,dr}\\
&=\frac{(abc)^2}{2}\int_{0}^{1}r^5\int_{0}^{\pi}\sin^3\vartheta\cos\vartheta\cancelto{0}{\bigg(\int_{0}^{2\pi}{\sin(2\varphi)\,d\varphi}\bigg)}\,d\vartheta\,dr\\
&=0\,.\end{align*}
2. Similarly the part ${\rm{E}}_{+}=\Big\{{(x,y,z)\in{\mathbb{R}}^3\;\big|\;\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\leqslant1\,,\; x\geqslant0\,,\; y\geqslant0\,,\; z\geqslant0\,}\Big\}$ of ${\rm{E}}$ can be represented as ${\rm{E}}_{+}=\Big\{{\big(a\,r\sin\vartheta\cos\varphi,\, b\,r\sin\vartheta\sin\varphi,\,\,c\,r\cos\vartheta\big)\in{\mathbb{R}}^3\;\big|\;r\in[0,1]\,,\; \vartheta\in\big[0,\tfrac{\pi}{2}\big], \; \varphi\in\big[0,\tfrac{\pi}{2}\big]\,}\Big\}.$ So
\begin{align*}
\mathop{\iiint}\limits_{{\rm{E}}_{+}}{xyz\,d(x,y,z)}&=\mathop{\iiint}\limits_{{\rm{E}}_+}{abc\,r^3\sin^2\vartheta\cos\vartheta\cos\varphi\sin\varphi\,\big|abc\,r^2\sin\vartheta\big|\,d(r,\vartheta,\varphi)}\\
&=(abc)^2\int_{0}^{1}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}{r^5\sin^3\vartheta\cos\vartheta\cos\varphi\sin\varphi\,d\varphi\,d\vartheta\,dr}\\
&=\frac{(abc)^2}{2}\int_{0}^{1}r^5\int_{0}^{\frac{\pi}{2}}\sin^3\vartheta\cos\vartheta\cancelto{1}{\bigg(\int_{0}^{\frac{\pi}{2}}{\sin(2\varphi)\,d\varphi}\bigg)}\,d\vartheta\,dr\\
&=\frac{(abc)^2}{2}\int_{0}^{1}r^5\cancelto{\frac{1}{4}}{\bigg(\int_{0}^{\frac{\pi}{2}}\sin^3\vartheta\cos\vartheta\,d\vartheta\bigg)}\,dr\\
&=\frac{(abc)^2}{8}\int_{0}^{1}r^5\,dr\\
&=\frac{(abc)^2}{8}\,\frac{1}{6}\\
&=\frac{(abc)^2}{48}\,.\end{align*}

Additional question: For $n\in\mathbb{N}$ evaluate
$\displaystyle\mathop{\iiint}\limits_{\rm{E}}\big(a^2b^2-b^2x^2-a^2y^2\big)^{n-\frac{1}{2}}\,d(x,y,z)\,,$ where ${\rm{E}}$ is the above solid ellipsoid.

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