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We are given the elipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1$ with density function $p(x,y,z)=x^2+y^2+z^2$.

Find the mass of the elipsoid.

what I did:

I used the transformation $x=ar\sin\theta \cos\phi$, $y=br\sin \theta \sin \phi$ and $z=cr\cos\theta$. Spherical coordinates but I multiplied $x,y,z$ by $a,b,c$ respectively.

I calculate the jacobian of this transformation, it is $abcr^2\sin\theta$.

So the mass is $$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta drd\theta d\phi$$

But I'm having trouble calculating this integral. I thought maybe we can use that $\sin^2x+\cos^2x=1$ but because the coefficients are different it's difficult to use that here. How would I calculate this integral?

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  • $\begingroup$ It is normal that the expression does not simplify because the density has spherical symmetry while the domain is an ellipsoid. Look at the integrand again and see that it is formed of separable terms not so difficult to integrate. $\endgroup$ – Yves Daoust Jun 13 '14 at 15:33
  • $\begingroup$ It is possible to integrate this without simplifying it? Because I don't see a way to simplify it further. And this seems quite terrifying to integrate honestly $\endgroup$ – Oria Gruber Jun 13 '14 at 15:37
  • $\begingroup$ Its is not. Split the sum in three integrals and factor these. $\endgroup$ – Yves Daoust Jun 13 '14 at 15:42
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Hint: There's absolutely no reason to fear this integral. It'll start to look less terrifying if you break it down into more manageable bitesize bits. Start by bringing constant factors outside the integral, distributing terms over parentheses, and breaking the integral of the sum down into a sum of integrals:

$$\begin{align} M&=\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\ &=abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^4\sin^3\theta \cos^2\phi+b^2r^4\sin^3 \theta \sin ^2\phi+c^2r^4\sin\theta\cos^2 \theta)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\ &=abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^4\sin^3\theta \cos^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\ &~~~~+ abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(b^2r^4\sin^3 \theta \sin ^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\ &~~~~+ abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(c^2r^4\sin\theta\cos^2 \theta)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi \\ &=a^3bc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin^3\theta \cos^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\ &~~~~+ ab^3c\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin^3 \theta \sin ^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\ &~~~~+ abc^3\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin\theta\cos^2 \theta)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi \end{align}$$

Is that starting to look more manageable to you?


Hint #2: the sum of integrals arrived at above is not nearly as tedious as it looks. Notice that each of the multiple integrals you need to calculate is separable. For example, the first integral reduces to a product of one-dimensional integrals:

$$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin^3\theta \cos^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi = \left(\int_{0}^{2\pi}\cos^2\phi\mathrm{d}\phi\right) \left(\int_{0}^{\pi}\sin^3\theta\mathrm{d}\theta\right) \left(\int_{0}^{1}r^4\mathrm{d}r\right)\\ =I_1\,I_2\,I_3.$$

If you think carefully about the other two multiple integrals, you'll realize that you're with integration for this problem, since they just involve the factors $I_{1,2,3}$ again.

CORRECTION: oops, my last comment isn't exactly true. I misread the variables in the last integral. My apologies for any confusion.

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$$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta drd\theta d\phi=$$ $$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}a^3bcr^4\sin^3\theta \cos^2\phi drd\theta d\phi+\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}ab^3cr^4\sin^3 \theta \sin ^2\phi dr d\phi+\\\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}abc^3r^4\cos^2 \theta \sin\theta drd\theta d\phi=$$ $$a^3bc\int_{0}^{2\pi} sin^3\theta d\theta\int_{0}^{\pi}cos^2\phi d\phi\int_{0}^{1}r^4dr +ab^3c\int_{0}^{2\pi} sin^3 \theta d\theta \int_{0}^{\pi}\sin ^2\phi d\phi\int_{0}^{1}r^4dr +\\abc^3\int_{0}^{2\pi}cos^2 \theta \sin\theta d\theta\int_{0}^{\pi}d\phi\int_{0}^{1}r^4dr $$

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  • $\begingroup$ Yes thank you Yves I understood what you meant. This is just very very tedious. calculating 3 different triple integrals. But I suppose there is no escaping from it in this question. $\endgroup$ – Oria Gruber Jun 13 '14 at 15:59
  • $\begingroup$ Tedious starts at three weeks of work ;-) (And the integral on $r$ is a child's play.) Upvotes appreciated. $\endgroup$ – Yves Daoust Jun 13 '14 at 16:03
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You may integrate $\theta$ first, then $\phi$, and finally $r$.

$$I(r,\cos\theta,\phi):=(a^2r^2(1-\cos^2\theta) \cos^2\phi+b^2r^2(1-\cos^2 \theta) \sin ^2\phi+c^2r^2\cos^2 \theta)$$

$$J(r,\phi)=\int_{0}^{\pi}I(r,\cos\theta,\phi)\sin\theta d\theta= \int_{-1}^{1}I(r,x,\phi)dx=\frac{2}{3}r^4(a^2 + b^2 + c^2 + (a^2 - b^2)\cos(2\phi))$$

$$K(r)=\int_0^{2\pi}J(r,\phi)d\phi=\frac{4}{3}(a^2 + b^2 + c^2)\pi r^4$$

$$L=\int_0^1 K(r)r^2 dr=\frac{4}{15}(a^2 + b^2 + c^2)\pi$$

Final result is $$\frac{4}{15}abc(a^2 + b^2 + c^2)\pi$$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#66f}{\large\int_{{x^{2} \over a^{2}} + {y^{2} \over b^{2}} + {z^{2} \over c^{2}}\ \leq\ 1} \pars{x^{2} + y^{2} + z^{2}}\,\dd x\,\dd y\,\dd z} \\[3mm]&=\verts{abc}\int_{x^{2} + y^{2} + z^{2}\ \leq\ 1} \pars{a^{2}x^{2} + b^{2}y^{2} + c^{2}z^{2}}\,\dd x\,\dd y\,\dd z \\[3mm]&=\verts{abc}\int_{r\ \leq\ 1}\pars{% a^{2}\,{1 \over 3}\,r^{2} + b^{2}{1 \over 3}\,r^{2} + c^{2}{1 \over 3}\,r^{2}} \,\dd x\,\dd y\,\dd z \\[3mm]&={1 \over 3}\verts{abc}\pars{a^{2} + b^{2} + c^{2}}\int_{r\ \leq\ 1}r^{2} \,\dd x\,\dd y\,\dd z \\[3mm]&={1 \over 3}\verts{abc}\pars{a^{2} + b^{2} + c^{2}} \int_{0}^{1}r^{2}\pars{4\pi r^{2}}\,\dd r =\color{#66f}{\large{4\pi \over 15}\verts{abc}\pars{a^{2} + b^{2} + c^{2}}} \end{align}

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